Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 23 Trigonometrical Ratios Of Standard Angles

Exercise 23(A)

Question 1. Evaluate the following:
(i) \( \sin 30^\circ \cos 30^\circ \)
(ii) \( \tan 30^\circ \tan 60^\circ \)
(iii) \( \cos^2 60^\circ + \sin^2 30^\circ \)
(iv) \( \csc^2 60^\circ - \tan^2 30^\circ \)
(v) \( \sin^2 30^\circ + \cos^2 30^\circ + \cot^2 45^\circ \)
(vi) \( \cos^2 60^\circ + \sec^2 30^\circ + \tan^2 45^\circ \)
Answer:
(i) Substituting the standard values \( \sin 30^\circ = \frac{1}{2} \) and \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\( \sin 30^\circ \cos 30^\circ = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \)
(ii) Substituting the standard values \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \tan 60^\circ = \sqrt{3} \):
\( \tan 30^\circ \tan 60^\circ = \frac{1}{\sqrt{3}} \cdot \sqrt{3} = 1 \)
(iii) Substituting the standard values \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 30^\circ = \frac{1}{2} \):
\( \cos^2 60^\circ + \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
(iv) Substituting the standard values \( \csc 60^\circ = \frac{2}{\sqrt{3}} \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( \csc^2 60^\circ - \tan^2 30^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1 \)
(v) Substituting the standard values \( \sin 30^\circ = \frac{1}{2} \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), and \( \cot 45^\circ = 1 \):
\( \sin^2 30^\circ + \cos^2 30^\circ + \cot^2 45^\circ = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + 1^2 = \frac{1}{4} + \frac{3}{4} + 1 = 1 + 1 = 2 \)
(vi) Substituting the standard values \( \cos 60^\circ = \frac{1}{2} \), \( \sec 30^\circ = \frac{2}{\sqrt{3}} \), and \( \tan 45^\circ = 1 \):
\( \cos^2 60^\circ + \sec^2 30^\circ + \tan^2 45^\circ = \left(\frac{1}{2}\right)^2 + \left(\frac{2}{\sqrt{3}}\right)^2 + 1^2 = \frac{1}{4} + \frac{4}{3} + 1 = \frac{3 + 16 + 12}{12} = \frac{31}{12} = 2\frac{7}{12} \)
In simple words: Substitute the exact standard trigonometric ratios for the given angles into each expression and simplify the fractions to get the final answer.

Exam Tip: Memorize the standard values of basic trigonometric functions (30°, 45°, 60°) to quickly solve these multi-part questions without making calculation errors.

Question 2. Evaluate the following:
(i) \( \tan^2 30^\circ + \tan^2 45^\circ + \tan^2 60^\circ \)
(ii) \( \frac{\tan 45^\circ}{\csc 30^\circ} + \frac{\sec 60^\circ}{\cot 45^\circ} - \frac{5 \sin 90^\circ}{2 \cos 0^\circ} \)
(iii) \( 3 \sin^2 30^\circ + 2 \tan^2 60^\circ - 5 \cos^2 45^\circ \ encamp\)
Answer:
(i) Substituting \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), \( \tan 45^\circ = 1 \), and \( \tan 60^\circ = \sqrt{3} \):
\( \tan^2 30^\circ + \tan^2 45^\circ + \tan^2 60^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 + 1^2 + (\sqrt{3})^2 = \frac{1}{3} + 1 + 3 = 4 + \frac{1}{3} = \frac{13}{3} = 4\frac{1}{3} \)
(ii) Substituting \( \tan 45^\circ = 1 \), \( \csc 30^\circ = 2 \), \( \sec 60^\circ = 2 \), \( \cot 45^\circ = 1 \), \( \sin 90^\circ = 1 \), and \( \cos 0^\circ = 1 \):
\( \frac{\tan 45^\circ}{\csc 30^\circ} + \frac{\sec 60^\circ}{\cot 45^\circ} - \frac{5 \sin 90^\circ}{2 \cos 0^\circ} = \frac{1}{2} + \frac{2}{1} - \frac{5 \cdot 1}{2 \cdot 1} = \frac{1}{2} + 2 - \frac{5}{2} = \frac{1 + 4 - 5}{2} = 0 \)
(iii) Substituting \( \sin 30^\circ = \frac{1}{2} \), \( \tan 60^\circ = \sqrt{3} \), and \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\( 3 \sin^2 30^\circ + 2 \tan^2 60^\circ - 5 \cos^2 45^\circ = 3\left(\frac{1}{2}\right)^2 + 2(\sqrt{3})^2 - 5\left(\frac{1}{\sqrt{2}}\right)^2 = 3\left(\frac{1}{4}\right) + 2(3) - 5\left(\frac{1}{2}\right) = \frac{3}{4} + 6 - \frac{5}{2} = \frac{3 + 24 - 10}{4} = \frac{17}{4} = 4\frac{1}{4} \)
In simple words: Replace each trigonometric term with its corresponding numerical value, then carry out basic fraction additions and subtractions to find the final result.

Exam Tip: Be extra careful with terms that are squared. For instance, write \( \tan^2 30^\circ \) as \( \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \) to avoid missing the square.

Question 3. Prove that:
(i) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = 1 \)
(ii) \( \cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = 0 \)
(iii) \( \csc^2 45^\circ - \cot^2 45^\circ = 1 \)
(iv) \( \cos^2 30^\circ - \sin^2 30^\circ = \cos 60^\circ \)
(v) \( \left(\frac{\tan 60^\circ + 1}{\tan 60^\circ - 1}\right)^2 = \frac{1 + \cos 30^\circ}{1 - \cos 30^\circ} \)
(vi) \( 3 \csc^2 60^\circ - 2 \cot^2 30^\circ + \sec^2 45^\circ = 0 \)
Answer:
(i) \( \text{LHS} = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1 = \text{RHS} \)
(ii) \( \text{LHS} = \cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 = \text{RHS} \)
(iii) \( \text{LHS} = \csc^2 45^\circ - \cot^2 45^\circ = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1 = \text{RHS} \)
(iv) \( \text{LHS} = \cos^2 30^\circ - \sin^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
\( \text{RHS} = \cos 60^\circ = \frac{1}{2} \)
Since \( \text{LHS} = \text{RHS} \), the statement is verified.
(v) \( \text{LHS} = \left(\frac{\tan 60^\circ + 1}{\tan 60^\circ - 1}\right)^2 = \left(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right)^2 = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)^2} = \frac{3 + 1 + 2\sqrt{3}}{3 + 1 - 2\sqrt{3}} = \frac{4 + 2\sqrt{3}}{4 - 2\sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \)
\( \text{RHS} = \frac{1 + \cos 30^\circ}{1 - \cos 30^\circ} = \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}} = \frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \)
Since \( \text{LHS} = \text{RHS} \), the statement is verified.
(vi) \( \text{LHS} = 3 \csc^2 60^\circ - 2 \cot^2 30^\circ + \sec^2 45^\circ = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 2(\sqrt{3})^2 + (\sqrt{2})^2 = 3\left(\frac{4}{3}\right) - 2(3) + 2 = 4 - 6 + 2 = 0 = \text{RHS} \)
In simple words: To prove these equations, calculate the numerical value of the left side and right side separately by plugging in the standard angles. If both sides yield the same number, the statement is proved.

Exam Tip: Rationalizing denominators is key in part (v). Always simplify the numerator and denominator completely to make verification straightforward.

Question 4. If \( A = 30^\circ \), prove that:
(i) \( \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} \)
(ii) \( \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \)
(iii) \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \)
Answer:
(i) Substituting \( A = 30^\circ \), we get \( 2A = 60^\circ \).
\( \text{LHS} = \sin 2A = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
\( \text{RHS} = \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(ii) Substituting \( A = 30^\circ \), we get \( 2A = 60^\circ \).
\( \text{LHS} = \cos 2A = \cos 60^\circ = \frac{1}{2} \)
\( \text{RHS} = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{2}{4} = \frac{1}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(iii) Substituting \( A = 30^\circ \), we get \( 2A = 60^\circ \).
\( \text{LHS} = \tan 2A = \tan 60^\circ = \sqrt{3} \)
\( \text{RHS} = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: Replace the variable A with 30 degrees in both the left-hand and right-hand expressions. Show that both calculations lead to the exact same final value.

Exam Tip: Remember these are standard double-angle trigonometric formulas. Verifying them with \( 30^\circ \) helps you cross-check both your calculation accuracy and formula recall.

Question 5. In a right-angled triangle \( ABC \), if \( AB = BC = x \), find:
(i) \( \sin 45^\circ \)
(ii) \( \cos 45^\circ \)
(iii) \( \tan 45^\circ \)
Answer:
Since \( AB = BC = x \) and angle B is a right angle (\( 90^\circ \)), the triangle is an isosceles right-angled triangle. This implies that the acute angles \( A \) and \( C \) are each equal to \( 45^\circ \).
By Pythagoras' theorem:
\( AC = \sqrt{AB^2 + BC^2} = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2} \)
Using the definitions of trigonometric ratios:
(i) \( \sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} \)
(ii) \( \cos 45^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} \)
(iii) \( \tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{x}{x} = 1 \)

In simple words: In a right-angled triangle with two equal sides, the other two angles must be 45 degrees. We use the hypotenuse length, which is side times square root of 2, to work out the basic sine, cosine, and tangent values for 45 degrees.

Question 6. Prove that:
(i) \( \sin 60^\circ = 2 \sin 30^\circ \cos 30^\circ \)
(ii) \( 4(\sin^4 30^\circ + \cos^4 60^\circ) - 3(\cos^2 45^\circ - \sin^2 90^\circ) = 2 \)
Answer:
(i) \( \text{LHS} = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
\( \text{RHS} = 2 \sin 30^\circ \cos 30^\circ = 2 \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(ii) Substituting the standard values \( \sin 30^\circ = \frac{1}{2} \), \( \cos 60^\circ = \frac{1}{2} \), \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), and \( \sin 90^\circ = 1 \):
\( \text{LHS} = 4\left[\left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4\right] - 3\left[\left(\frac{1}{\sqrt{2}}\right)^2 - 1^2\right] \)
\( = 4\left[\frac{1}{16} + \frac{1}{16}\right] - 3\left[\frac{1}{2} - 1\right] \)
\( = 4\left[\frac{2}{16}\right] - 3\left[-\frac{1}{2}\right] \)
\( = \frac{1}{2} + \frac{3}{2} = 2 = \text{RHS} \)
Since \( \text{LHS} = \text{RHS} \), the identity is proved.
In simple words: Substitute the numeric values for each trigonometric function, then evaluate the terms. Remember that subtracting a negative term will change the sign to addition.

Exam Tip: Be mindful of signs when subtracting negative quantities. In part (ii), subtracting \( -3 \times \left(-\frac{1}{2}\right) \) becomes addition, yielding \( \frac{1}{2} + \frac{3}{2} = 2 \).

Question 7. Find the value of the acute angle in each of the following equations:
(i) \( \cos^2 x + \sin^2 x = 1 \) where \( \cos x = \sin x \)
(ii) \( \sec A = \csc A \)
(iii) \( \tan \theta = \cot \theta \)
(iv) \( \sin x = \cos y \)
Answer:
(i) We are given \( \cos x = \sin x \) where \( x \) is an acute angle (\( 0^\circ < x < 90^\circ \)).
Since \( \cos^2 x + \sin^2 x = 1 \), we can substitute \( \cos x = \sin x \) into the equation:
\( \sin^2 x + \sin^2 x = 1 \)

\( \implies 2 \sin^2 x = 1 \)

\( \implies \sin^2 x = \frac{1}{2} \)

\( \implies \sin x = \frac{1}{\sqrt{2}} \) (taking the positive square root since \( x \) is acute)

\( \implies x = 45^\circ \)
(ii) Given: \( \sec A = \csc A \)
Converting to basic trigonometric ratios:
\( \frac{1}{\cos A} = \frac{1}{\sin A} \)

\( \implies \sin A = \cos A \)

\( \implies \frac{\sin A}{\cos A} = 1 \)

\( \implies \tan A = 1 \)
Since \( A \) is acute, we have:
\( A = 45^\circ \)
(iii) Given: \( \tan \theta = \cot \theta \)
Using the identity \( \cot \theta = \frac{1}{\tan \theta} \):
\( \tan \theta = \frac{1}{\tan \theta} \)

\( \implies \tan^2 \theta = 1 \)
Taking the positive root for acute angle \( \theta \):
\( \tan \theta = 1 \)

\( \implies \theta = 45^\circ \)
(iv) Given: \( \sin x = \cos y \)
We can write \( \cos y \) in terms of sine as \( \sin(90^\circ - y) \):
\( \sin x = \sin(90^\circ - y) \)
Comparing both sides for acute angles \( x \) and \( y \):
\( x = 90^\circ - y \)

\( \implies x + y = 90^\circ \)
This shows that \( x \) and \( y \) are complementary angles.
In simple words: We can solve these equations by writing all trigonometric terms in terms of a single function (like converting everything to sin or tan) and then finding the angle from standard tables.

Exam Tip: For equations like \( \sec A = \csc A \), converting to the basic ratios of \( \sin \) and \( \cos \) makes solving much simpler and prevents algebraic mistakes.

Question 8. State whether the following statements are true or false:
(i) If \( x \) and \( y \) are acute angles, then \( \sin x = \cos y \) means \( x + y = 45^\circ \).
(ii) \( \sec \theta \cdot \cot \theta = \csc \theta \).
(iii) \( \sin^2 \theta + \cos^2 \theta = 1 \).
Answer:
(i) **False**
*Explanation:* We know that \( \sin x = \cos y \) can be written as \( \sin x = \sin(90^\circ - y) \). Since \( x \) and \( y \) are acute, this simplifies to \( x = 90^\circ - y \), or \( x + y = 90^\circ \). Therefore, \( x + y = 45^\circ \) is false because the angles must be complementary (sum up to \( 90^\circ \)).
(ii) **True**
*Explanation:* Simplifying the left side:
\( \sec \theta \cdot \cot \theta = \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta} = \csc \theta \). Since LHS equals RHS, the statement is true.
(iii) **True**
*Explanation:* This is the fundamental trigonometric identity. We can verify it using \( \cos^2 \theta = 1 - \sin^2 \theta \). Substituting this gives: \( \sin^2 \theta + (1 - \sin^2 \theta) = 1 \). Thus, it is always true.
In simple words: The first statement is false because the two angles must add up to 90 degrees, not 45 degrees. The other two statements are mathematically correct identity rules.

Exam Tip: When evaluating True/False questions, always write a brief explanation or show the simplified mathematical steps to secure full marks.

Question 9. State how the following trigonometric ratios change for acute angles:
(i) How does \( \sin \theta \) change as \( \theta \) increases?
(ii) How does \( \cos \theta \) change as \( \theta \) increases?
(iii) How does \( \tan \theta \) change as \( \theta \) decreases?
Answer:
(i) **Increases**
*Explanation:* In a right-angled triangle, the sine of an acute angle is defined as the ratio of the opposite side to the hypotenuse. When the angle \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), the length of the opposite side grows larger while the hypotenuse remains constant. Consequently, the ratio \( \frac{\text{Opposite}}{\text{Hypotenuse}} \) increases, meaning \( \sin \theta \) increases.
(ii) **Decreases**
*Explanation:* The cosine of an acute angle represents the ratio of the base (adjacent side) to the hypotenuse. As the angle \( \theta \) increases, the base decreases relative to the hypotenuse. Therefore, the ratio \( \frac{\text{Base}}{\text{Hypotenuse}} \) becomes smaller, which means \( \cos \theta \) decreases.
(iii) **Decreases**
*Explanation:* The tangent of an acute angle is the ratio of the opposite side to the base. If the angle \( \theta \) decreases, the opposite side shrinks relative to the base. This causes the ratio \( \frac{\text{Opposite}}{\text{Base}} \) to get smaller, so \( \tan \theta \) decreases.
In simple words: For angles between 0 and 90 degrees, sine goes up as the angle grows, while cosine goes down. Tangent follows the same direction as sine, so it goes down as the angle shrinks.

Exam Tip: Recall the boundary values: \( \sin 0^\circ = 0 \), \( \sin 90^\circ = 1 \) (increases); \( \cos 0^\circ = 1 \), \( \cos 90^\circ = 0 \) (decreases); and \( \tan 0^\circ = 0 \), \( \tan 90^\circ = \infty \) (increases as angle increases, decreases as angle decreases).

Question 10. Find the values of the following expressions to two decimal places (take \( \sqrt{3} = 1.732 \)):
(i) \( \sin 60^\circ \)
(ii) \( \frac{2}{\tan 30^\circ} \)
Answer:
(i) We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Substituting \( \sqrt{3} = 1.732 \):
\( \sin 60^\circ = \frac{1.732}{2} = 0.866 \approx 0.87 \) (rounded to two decimal places).
(ii) We have \( \frac{2}{\tan 30^\circ} \).
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( \frac{2}{\tan 30^\circ} = \frac{2}{\frac{1}{\sqrt{3}}} = 2\sqrt{3} \)
Substituting \( \sqrt{3} = 1.732 \):
\( 2\sqrt{3} = 2 \times 1.732 = 3.464 \approx 3.46 \) (rounded to two decimal places).
In simple words: Find the standard ratio values first, then substitute 1.732 for the square root of 3, do the division or multiplication, and round your answer to two decimal places.

Exam Tip: Pay attention to the rounding instructions. Always calculate one extra decimal digit (the third decimal place) before rounding to ensure the second decimal digit is correct.

Question 11. Evaluate the following expressions:
(i) \( \frac{\cos 3A - 2\cos 4A}{\sin 3A + 2\sin 4A} \) when \( A = 15^\circ \)
(ii) \( \frac{3 \sin 3B + 2 \cos(2B + 5^\circ)}{2 \cos 3B - \sin(2B - 10^\circ)} \) when \( B = 20^\circ \)
Answer:
(i) Substituting \( A = 15^\circ \) into the expression:
\( 3A = 3 \times 15^\circ = 45^\circ \)
\( 4A = 4 \times 15^\circ = 60^\circ \)
The expression becomes:
\( \frac{\cos 45^\circ - 2 \cos 60^\circ}{\sin 45^\circ + 2 \sin 60^\circ} = \frac{\frac{1}{\sqrt{2}} - 2\left(\frac{1}{2}\right)}{\frac{1}{\sqrt{2}} + 2\left(\frac{\sqrt{3}}{2}\right)} = \frac{\frac{1}{\sqrt{2}} - 1}{\frac{1}{\sqrt{2}} + \sqrt{3}} = \frac{\frac{1 - \sqrt{2}}{\sqrt{2}}}{\frac{1 + \sqrt{6}}{\sqrt{2}}} = \frac{1 - \sqrt{2}}{1 + \sqrt{6}} \)
Rationalizing the denominator by multiplying the numerator and denominator by \( (\sqrt{6} - 1) \):
\( \frac{1 - \sqrt{2}}{\sqrt{6} + 1} \cdot \frac{\sqrt{6} - 1}{\sqrt{6} - 1} = \frac{(1 - \sqrt{2})(\sqrt{6} - 1)}{6 - 1} = \frac{\sqrt{6} - 1 - \sqrt{12} + \sqrt{2}}{5} = \frac{1}{5}(\sqrt{6} - 1 - 2\sqrt{3} + \sqrt{2}) \)
(ii) Substituting \( B = 20^\circ \) into the expression:
\( 3B = 3 \times 20^\circ = 60^\circ \)
\( 2B + 5^\circ = 2(20^\circ) + 5^\circ = 45^\circ \)
\( 2B - 10^\circ = 2(20^\circ) - 10^\circ = 30^\circ \)
The expression becomes:
\( \frac{3 \sin 60^\circ + 2 \cos 45^\circ}{2 \cos 60^\circ - \sin 30^\circ} = \frac{3\left(\frac{\sqrt{3}}{2}\right) + 2\left(\frac{1}{\sqrt{2}}\right)}{2\left(\frac{1}{2}\right) - \frac{1}{2}} = \frac{\frac{3\sqrt{3}}{2} + \sqrt{2}}{1 - \frac{1}{2}} = \frac{\frac{3\sqrt{3} + 2\sqrt{2}}{2}}{\frac{1}{2}} = 3\sqrt{3} + 2\sqrt{2} \)
In simple words: First substitute the given angle value, calculate the resulting angles inside the functions, use the standard tables to change them to numbers, and then simplify the fractions.

Exam Tip: Be careful when simplifying fractions. Always find a common denominator for the numerator before dividing by the main denominator.

Exercise 23(B)

Question 1. If \( A = 60^\circ \) and \( B = 30^\circ \), verify that:
(i) \( \sin(A+B) = \sin A \cos B + \cos A \sin B \)
(ii) \( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
(iii) \( \cos(A-B) = \cos A \cos B + \sin A \sin B \)
(iv) \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
Answer:
(i) \( \text{LHS} = \sin(A+B) = \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1 \)
\( \text{RHS} = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(ii) \( \text{LHS} = \cos(A+B) = \cos(60^\circ + 30^\circ) = \cos 90^\circ = 0 \)
\( \text{RHS} = \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ = \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(iii) \( \text{LHS} = \cos(A-B) = \cos(60^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \text{RHS} = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ = \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(iv) \( \text{LHS} = \tan(A-B) = \tan(60^\circ - 30^\circ) = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
\( \text{RHS} = \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3}\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{3 - 1}{\sqrt{3}}}{1 + 1} = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: This question tests the standard sum and difference angle formulas. Plug in A = 60 degrees and B = 30 degrees to show that both sides of the formulas match.

Exam Tip: Be careful with the signs in sum and difference formulas - cosine sum uses a minus sign (\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)), whereas cosine difference uses a plus sign.

Question 2. If \( A = 30^\circ \), verify that:
(i) \( \sin 2A = 2 \sin A \cos A = \frac{2 \tan A}{1 + \tan^2 A} \)
(ii) \( \cos 2A = \cos^2 A - \sin^2 A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \)
(iii) \( 2 \cos^2 A - 1 = 1 - 2 \sin^2 A \)
(iv) \( \sin 3A = 3 \sin A - 4 \sin^3 A \)
Answer:
(i) Substituting \( A = 30^\circ \) into each part of the equation:
First part: \( \sin 2A = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
Second part: \( 2 \sin A \cos A = 2 \sin 30^\circ \cos 30^\circ = 2 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} \)
Third part: \( \frac{2 \tan A}{1 + \tan^2 A} = \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{\sqrt{3}}{2} \)
Since all three parts equal \( \frac{\sqrt{3}}{2} \), the identity is verified.
(ii) Substituting \( A = 30^\circ \) into each part:
First part: \( \cos 2A = \cos 60^\circ = \frac{1}{2} \)
Second part: \( \cos^2 A - \sin^2 A = \cos^2 30^\circ - \sin^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
Third part: \( \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{2}{4} = \frac{1}{2} \)
Since all three parts equal \( \frac{1}{2} \), the identity is verified.
(iii) For \( A = 30^\circ \):
\( \text{LHS} = 2 \cos^2 A - 1 = 2 \cos^2 30^\circ - 1 = 2 \left(\frac{\sqrt{3}}{2}\right)^2 - 1 = 2 \left(\frac{3}{4}\right) - 1 = \frac{3}{2} - 1 = \frac{1}{2} \)
\( \text{RHS} = 1 - 2 \sin^2 A = 1 - 2 \sin^2 30^\circ = 1 - 2 \left(\frac{1}{2}\right)^2 = 1 - 2 \left(\frac{1}{4}\right) = 1 - \frac{1}{2} = \frac{1}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(iv) For \( A = 30^\circ \):
\( \text{LHS} = \sin 3A = \sin 90^\circ = 1 \)
\( \text{RHS} = 3 \sin A - 4 \sin^3 A = 3 \sin 30^\circ - 4 \sin^3 30^\circ = 3 \left(\frac{1}{2}\right) - 4 \left(\frac{1}{2}\right)^3 = \frac{3}{2} - 4 \left(\frac{1}{8}\right) = \frac{3}{2} - \frac{1}{2} = 1 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: Substitute 30 degrees for A in each expression. Work through the calculations step by step to prove that all parts of the equations result in the same value.

Exam Tip: For multiple-equality questions (such as part i and ii), show the substitution and final value for each part separately to ensure a clear and complete proof.

Question 3. If \( A = B = 45^\circ \), verify that:
(i) \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
(ii) \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
Answer:
(i) Substituting \( A = B = 45^\circ \):
\( \text{LHS} = \sin(A - B) = \sin(45^\circ - 45^\circ) = \sin 0^\circ = 0 \)
\( \text{RHS} = \sin A \cos B - \cos A \sin B = \sin 45^\circ \cos 45^\circ - \cos 45^\circ \sin 45^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} - \frac{1}{2} = 0 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(ii) Substituting \( A = B = 45^\circ \):
\( \text{LHS} = \cos(A + B) = \cos(45^\circ + 45^\circ) = \cos 90^\circ = 0 \)
\( \text{RHS} = \cos A \cos B - \sin A \sin B = \cos 45^\circ \cos 45^\circ - \sin 45^\circ \sin 45^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} - \frac{1}{2} = 0 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: Since both A and B are 45 degrees, the difference A - B is 0 degrees and the sum A + B is 90 degrees. Calculating the left side and right side shows both expressions equal 0.

Exam Tip: Remember standard boundary ratios such as \( \sin 0^\circ = 0 \) and \( \cos 90^\circ = 0 \) to confidently prove identities of this type.

Question 4. If \( A = 30^\circ \), verify that:
(i) \( \sin 3A = 4 \sin A \sin(60^\circ - A) \sin(60^\circ + A) \)
(ii) \( (\sin A - \cos A)^2 = 1 - \sin 2A \)
(iii) \( \cos 2A = \cos^4 A - \sin^4 A \)
(iv) \( \frac{1 - \cos 2A}{\sin 2A} = \tan A \)
(v) \( \frac{1 + \sin 2A + \cos 2A}{\sin A + \cos A} = 2 \cos A \)
(vi) \( 4 \cos A \cos(60^\circ - A) \cos(60^\circ + A) = \cos 3A \)
(vii) \( \frac{\cos^3 A - \cos 3A}{\cos A} + \frac{\sin^3 A + \sin 3A}{\sin A} = 3 \)
Answer:
(i) For \( A = 30^\circ \), \( 3A = 90^\circ \).
\( \text{LHS} = \sin 3A = \sin 90^\circ = 1 \)
\( \text{RHS} = 4 \sin 30^\circ \sin(60^\circ - 30^\circ) \sin(60^\circ + 30^\circ) = 4 \sin 30^\circ \sin 30^\circ \sin 90^\circ = 4 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) (1) = 1 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(ii) For \( A = 30^\circ \):
\( \text{LHS} = (\sin 30^\circ - \cos 30^\circ)^2 = \left(\frac{1}{2} - \frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} - 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \)
\( \text{RHS} = 1 - \sin 2A = 1 - \sin 60^\circ = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(iii) For \( A = 30^\circ \):
\( \text{LHS} = \cos 2A = \cos 60^\circ = \frac{1}{2} \)
\( \text{RHS} = \cos^4 30^\circ - \sin^4 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^4 - \left(\frac{1}{2}\right)^4 = \frac{9}{16} - \frac{1}{16} = \frac{8}{16} = \frac{1}{2} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(iv) For \( A = 30^\circ \):
\( \text{LHS} = \frac{1 - \cos 60^\circ}{\sin 60^\circ} = \frac{1 - \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \)
\( \text{RHS} = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(v) For \( A = 30^\circ \):
\( \text{LHS} = \frac{1 + \sin 60^\circ + \cos 60^\circ}{\sin 30^\circ + \cos 30^\circ} = \frac{1 + \frac{\sqrt{3}}{2} + \frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} = \frac{\frac{3 + \sqrt{3}}{2}}{\frac{1 + \sqrt{3}}{2}} = \frac{3 + \sqrt{3}}{1 + \sqrt{3}} \)
Factoring \( \sqrt{3} \) in the numerator of the left-hand side:
\( \frac{\sqrt{3}(\sqrt{3} + 1)}{1 + \sqrt{3}} = \sqrt{3} \)
\( \text{RHS} = 2 \cos 30^\circ = 2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(vi) For \( A = 30^\circ \):
\( \text{LHS} = 4 \cos 30^\circ \cos 30^\circ \cos 90^\circ = 4 \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) (0) = 0 \)
\( \text{RHS} = \cos 3A = \cos 90^\circ = 0 \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
(vii) For \( A = 30^\circ \):
\( \text{LHS} = \frac{\cos^3 30^\circ - \cos 90^\circ}{\cos 30^\circ} + \frac{\sin^3 30^\circ + \sin 90^\circ}{\sin 30^\circ} = \frac{\left(\frac{\sqrt{3}}{2}\right)^3 - 0}{\frac{\sqrt{3}}{2}} + \frac{\left(\frac{1}{2}\right)^3 + 1}{\frac{1}{2}} = \left(\frac{\sqrt{3}}{2}\right)^2 + \frac{\frac{1}{8} + 1}{\frac{1}{2}} = \frac{3}{4} + \frac{\frac{9}{8}}{\frac{1}{2}} = \frac{3}{4} + \frac{9}{4} = \frac{12}{4} = 3 = \text{RHS} \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: For all these sub-parts, put 30 degrees in place of the letter A. Calculate the values of the expressions on both sides to show that they are equal.

Exam Tip: When evaluating trigonometric expressions to the fourth power (like \( \cos^4 30^\circ \)), raise the base value to the fourth power correctly: \( \left(\frac{\sqrt{3}}{2}\right)^4 = \frac{9}{16} \).

Exercise 23(C)

Question 1. Solve the following equations for A, if :
(i) 2 sin A = 1
(ii) 2 cos 2 A = 1
(iii) sin 3 A = \( \frac{\sqrt{3}}{2} \)
(iv) sec 2 A = 2
(v) \( \sqrt{3} \) tan A = 1
(vi) tan 3 A = 1
(vii) 2 sin 3 A = 1
(viii) \( \sqrt{3} \) cot 2 A = 1
Answer:
(i) Given, \( 2 \sin A = 1 \)
\( \implies \sin A = \frac{1}{2} \)
\( \implies \sin A = \sin 30^\circ \)
\( \implies A = 30^\circ \)

(ii) Given, \( 2 \cos 2A = 1 \)
\( \implies \cos 2A = \frac{1}{2} \)
\( \implies \cos 2A = \cos 60^\circ \)
\( \implies 2A = 60^\circ \)
\( \implies A = 30^\circ \)

(iii) Given, \( \sin 3A = \frac{\sqrt{3}}{2} \)
\( \implies \sin 3A = \sin 60^\circ \)
\( \implies 3A = 60^\circ \)
\( \implies A = 20^\circ \)

(iv) Given, \( \sec 2A = 2 \)
\( \implies \sec 2A = \sec 60^\circ \)
\( \implies 2A = 60^\circ \)
\( \implies A = 30^\circ \)

(v) Given, \( \sqrt{3} \tan A = 1 \)
\( \implies \tan A = \frac{1}{\sqrt{3}} \)
\( \implies \tan A = \tan 30^\circ \)
\( \implies A = 30^\circ \)

(vi) Given, \( \tan 3A = 1 \)
\( \implies \tan 3A = \tan 45^\circ \)
\( \implies 3A = 45^\circ \)
\( \implies A = 15^\circ \)

(vii) Given, \( 2 \sin 3A = 1 \)
\( \implies \sin 3A = \frac{1}{2} \)
\( \implies \sin 3A = \sin 30^\circ \)
\( \implies 3A = 30^\circ \)
\( \implies A = 10^\circ \)

(viii) Given, \( \sqrt{3} \cot 2A = 1 \)
\( \implies \cot 2A = \frac{1}{\sqrt{3}} \)
\( \implies \cot 2A = \cot 60^\circ \)
\( \implies 2A = 60^\circ \)
\( \implies A = 30^\circ \)
In simple words: To find the value of angle A, simplify each equation so that the trigonometric function is on one side, match it with the standard table angle, and then divide to get the final value.

Exam Tip: Memorizing standard trigonometric table values for \(30^\circ\), \(45^\circ\), and \(60^\circ\) is essential to solving these equations quickly.

Question 2. Calculate the value of A, if :
(i) \( (\sin A - 1)(2\cos A - 1) = 0 \)
(ii) \( (\tan A - 1)(\text{cosec } 3A - 1) = 0 \)
(iii) \( (\sec 2A - 1)(\text{cosec } 3A - 1) = 0 \)
(iv) \( \cos 3A \cdot (2\sin 2A - 1) = 0 \)
(v) \( (\text{cosec } 2A - 2)(\cot 3A - 1) = 0 \)
Answer:
(i) Given, \( (\sin A - 1)(2\cos A - 1) = 0 \)
This means either \( \sin A - 1 = 0 \) or \( 2\cos A - 1 = 0 \).
Case 1: \( \sin A - 1 = 0 \)
\( \implies \sin A = 1 \)
\( \implies \sin A = \sin 90^\circ \)
\( \implies A = 90^\circ \)
Case 2: \( 2\cos A - 1 = 0 \)
\( \implies 2\cos A = 1 \)
\( \implies \cos A = \frac{1}{2} \)
\( \implies \cos A = \cos 60^\circ \)
\( \implies A = 60^\circ \)
Hence, \( A = 90^\circ \) or \( 60^\circ \).

(ii) Given, \( (\tan A - 1)(\text{cosec } 3A - 1) = 0 \)
This gives \( \tan A - 1 = 0 \) or \( \text{cosec } 3A - 1 = 0 \).
Case 1: \( \tan A - 1 = 0 \)
\( \implies \tan A = 1 \)
\( \implies \tan A = \tan 45^\circ \)
\( \implies A = 45^\circ \)
Case 2: \( \text{cosec } 3A - 1 = 0 \)
\( \implies \text{cosec } 3A = 1 \)
\( \implies \text{cosec } 3A = \text{cosec } 90^\circ \)
\( \implies 3A = 90^\circ \)
\( \implies A = 30^\circ \)
Hence, \( A = 45^\circ \) or \( 30^\circ \).

(iii) Given, \( (\sec 2A - 1)(\text{cosec } 3A - 1) = 0 \)
This gives \( \sec 2A - 1 = 0 \) or \( \text{cosec } 3A - 1 = 0 \).
Case 1: \( \sec 2A - 1 = 0 \)
\( \implies \sec 2A = 1 \)
\( \implies \sec 2A = \sec 0^\circ \)
\( \implies 2A = 0^\circ \)
\( \implies A = 0^\circ \)
Case 2: \( \text{cosec } 3A - 1 = 0 \)
\( \implies \text{cosec } 3A = 1 \)
\( \implies \text{cosec } 3A = \text{cosec } 90^\circ \)
\( \implies 3A = 90^\circ \)
\( \implies A = 30^\circ \)
Hence, \( A = 0^\circ \) or \( 30^\circ \).

(iv) Given, \( \cos 3A \cdot (2\sin 2A - 1) = 0 \)
This gives \( \cos 3A = 0 \) or \( 2\sin 2A - 1 = 0 \).
Case 1: \( \cos 3A = 0 \)
\( \implies \cos 3A = \cos 90^\circ \)
\( \implies 3A = 90^\circ \)
\( \implies A = 30^\circ \)
Case 2: \( 2\sin 2A - 1 = 0 \)
\( \implies 2\sin 2A = 1 \)
\( \implies \sin 2A = \frac{1}{2} \)
\( \implies \sin 2A = \sin 30^\circ \)
\( \implies 2A = 30^\circ \)
\( \implies A = 15^\circ \)
Hence, \( A = 30^\circ \) or \( 15^\circ \).

(v) Given, \( (\text{cosec } 2A - 2)(\cot 3A - 1) = 0 \)
This gives \( \text{cosec } 2A - 2 = 0 \) or \( \cot 3A - 1 = 0 \).
Case 1: \( \text{cosec } 2A - 2 = 0 \)
\( \implies \text{cosec } 2A = 2 \)
\( \implies \text{cosec } 2A = \text{cosec } 30^\circ \)
\( \implies 2A = 30^\circ \)
\( \implies A = 15^\circ \)
Case 2: \( \cot 3A - 1 = 0 \)
\( \implies \cot 3A = 1 \)
\( \implies \cot 3A = \cot 45^\circ \)
\( \implies 3A = 45^\circ \)
\( \implies A = 15^\circ \)
Hence, \( A = 15^\circ \).
In simple words: When two brackets multiply to give zero, make each bracket term zero separately, simplify for the trigonometric ratio, and find the corresponding standard angle solutions.

Exam Tip: If both factors yield the exact same solution for the variable (as seen in sub-part v), present it as a single unique answer.

Question 3. If \( 2\sin x^\circ - 1 = 0 \) and \( x^\circ \) is an acute angle; find :
(i) \( \sin x^\circ \)
(ii) \( x^\circ \)
(iii) \( \cos x^\circ \) and \( \tan x^\circ \)
Answer:
(i) Given, \( 2\sin x^\circ - 1 = 0 \)
\( \implies 2\sin x^\circ = 1 \)
\( \implies \sin x^\circ = \frac{1}{2} \)

(ii) From part (i), we have:
\( \sin x^\circ = \frac{1}{2} \)
We know that \( \sin 30^\circ = \frac{1}{2} \).
\( \implies \sin x^\circ = \sin 30^\circ \)
\( \implies x^\circ = 30^\circ \)

(iii) Substituting \( x^\circ = 30^\circ \):
\( \cos x^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2} \)
and
\( \tan x^\circ = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
In simple words: Solve the initial equation to find the sine value first. Use this value to determine the angle \( x \), and then compute the remaining trigonometric functions.

Exam Tip: Write down step-by-step substitutions for \( \cos 30^\circ \) and \( \tan 30^\circ \) to secure full marks for the last sub-part.

Question 4. If \( 4\cos^2 x^\circ - 1 = 0 \) and \( 0 \le x^\circ \le 90^\circ \), find:
(i) \( x^\circ \)
(ii) \( \sin^2 x^\circ + \cos^2 x^\circ \)
(iii) \( \frac{1}{\cos^2 x^\circ} - \tan^2 x^\circ \)
Answer:
(i) Given, \( 4\cos^2 x^\circ - 1 = 0 \)
\( \implies 4\cos^2 x^\circ = 1 \)
\( \implies \cos^2 x^\circ = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \)
Taking square root (since \( 0 \le x^\circ \le 90^\circ \), \( \cos x^\circ \) is positive):
\( \implies \cos x^\circ = \frac{1}{2} \)
We know that \( \cos 60^\circ = \frac{1}{2} \).
\( \implies \cos x^\circ = \cos 60^\circ \)
\( \implies x^\circ = 60^\circ \)

(ii) Substituting \( x^\circ = 60^\circ \):
\( \sin^2 x^\circ + \cos^2 x^\circ = \sin^2 60^\circ + \cos^2 60^\circ \)
\( \implies \sin^2 60^\circ + \cos^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \)
\( \implies \sin^2 60^\circ + \cos^2 60^\circ = \frac{3}{4} + \frac{1}{4} = 1 \)

(iii) Substituting \( x^\circ = 60^\circ \):
\( \frac{1}{\cos^2 x^\circ} - \tan^2 x^\circ = \frac{1}{\cos^2 60^\circ} - \tan^2 60^\circ \)
\( \implies \frac{1}{\cos^2 60^\circ} - \tan^2 60^\circ = \frac{1}{\left(\frac{1}{2}\right)^2} - (\sqrt{3})^2 \)
\( \implies \frac{1}{\cos^2 60^\circ} - \tan^2 60^\circ = 4 - 3 = 1 \)
In simple words: Solve the equation to get the value of the angle first. Once you have the angle, substitute it into the expressions to calculate the final values.

Exam Tip: Since \( \sin^2 \theta + \cos^2 \theta = 1 \) is a standard trigonometric identity, verifying that your result in sub-part (ii) equals 1 is a great way to check your work.

Question 5. If \( 4\sin^2 \theta - 1 = 0 \) and angle \( \theta \) is less than \( 90^\circ \), find the value of \( \theta \) and hence the value of \( \cos^2 \theta + \tan^2 \theta \).
Answer:
Given, \( 4\sin^2 \theta - 1 = 0 \)
\( \implies 4\sin^2 \theta = 1 \)
\( \implies \sin^2 \theta = \frac{1}{4} \)
Since \( \theta < 90^\circ \) is acute, we take the positive square root:
\( \implies \sin \theta = \frac{1}{2} \)
We know that \( \sin 30^\circ = \frac{1}{2} \).
\( \implies \sin \theta = \sin 30^\circ \)
\( \implies \theta = 30^\circ \)

Now, substituting \( \theta = 30^\circ \) into the given expression:
\( \cos^2 \theta + \tan^2 \theta = \cos^2 30^\circ + \tan^2 30^\circ \)
\( \implies \cos^2 30^\circ + \tan^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 \)
\( \implies \cos^2 30^\circ + \tan^2 30^\circ = \frac{3}{4} + \frac{1}{3} \)
Taking the LCM of 4 and 3, which is 12:
\( \implies \cos^2 30^\circ + \tan^2 30^\circ = \frac{9 + 4}{12} = \frac{13}{12} \)
In simple words: First find the angle \( \theta \) from the sine equation. Then place that angle value into the second expression and find the sum by taking a common denominator.

Exam Tip: Be careful when adding fractions with different denominators. Always find the lowest common multiple (LCM) first to avoid arithmetic mistakes.

Question 6. If \( \sin 3A = 1 \) and \( 0 \le A \le 90^\circ \), find:
(i) \( \sin A \)
(ii) \( \cos 2A \)
(iii) \( \tan^2 A - \frac{1}{\cos^2 A} \)
Answer:
Given, \( \sin 3A = 1 \)
Since \( \sin 90^\circ = 1 \):
\( \implies \sin 3A = \sin 90^\circ \)
\( \implies 3A = 90^\circ \)
\( \implies A = 30^\circ \)

(i) Substituting \( A = 30^\circ \):
\( \sin A = \sin 30^\circ = \frac{1}{2} \)

(ii) Substituting \( A = 30^\circ \):
\( \cos 2A = \cos 2(30^\circ) = \cos 60^\circ = \frac{1}{2} \)

(iii) Substituting \( A = 30^\circ \):
\( \tan^2 A - \frac{1}{\cos^2 A} = \tan^2 30^\circ - \frac{1}{\cos^2 30^\circ} \)
\( \implies \tan^2 30^\circ - \frac{1}{\cos^2 30^\circ} = \left(\frac{1}{\sqrt{3}}\right)^2 - \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} \)
\( \implies \tan^2 30^\circ - \frac{1}{\cos^2 30^\circ} = \frac{1}{3} - \frac{4}{3} = \frac{-3}{3} = -1 \)
In simple words: Find the value of A first by matching the sine function with its standard value of 1. Then use A to find the value of each sub-part.

Exam Tip: The term \( \frac{1}{\cos^2 A} \) can also be simplified as \( \sec^2 A \) before substitution to make calculations faster.

Question 7. If \( 2\cos 2A = \sqrt{3} \) and A is acute, find:
(i) A
(ii) \( \sin 3A \)
(iii) \( \sin^2 (75^\circ - A) + \cos^2 (45^\circ + A) \)
Answer:
(i) Given, \( 2\cos 2A = \sqrt{3} \)
\( \implies \cos 2A = \frac{\sqrt{3}}{2} \)
We know that \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).
\( \implies \cos 2A = \cos 30^\circ \)
\( \implies 2A = 30^\circ \)
\( \implies A = 15^\circ \)

(ii) Substituting \( A = 15^\circ \):
\( \sin 3A = \sin 3(15^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}} \)

(iii) Substituting \( A = 15^\circ \):
\( \sin^2 (75^\circ - A) + \cos^2 (45^\circ + A) = \sin^2 (75^\circ - 15^\circ) + \cos^2 (45^\circ + 15^\circ) \)
\( \implies \sin^2 (75^\circ - 15^\circ) + \cos^2 (45^\circ + 15^\circ) = \sin^2 60^\circ + \cos^2 60^\circ \)
\( \implies \sin^2 60^\circ + \cos^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \)
\( \implies \sin^2 60^\circ + \cos^2 60^\circ = \frac{3}{4} + \frac{1}{4} = 1 \)
In simple words: Solve the cosine equation to find that A is \( 15^\circ \). Then plug this angle into the other parts to find their numerical values.

Exam Tip: Be careful with brackets inside trigonometric terms like \( \sin 3A \); you must multiply the angle \( A \) by 3 before finding its sine value.

Question 8. Solve the following:
(i) If \( \sin x + \cos y = 1 \) and \( x = 30^\circ \), find the value of y.
(ii) If \( 3\tan A - 5\cos B = \sqrt{3} \) and \( B = 90^\circ \), find the value of A.
Answer:
(i) Given, \( \sin x + \cos y = 1 \) and \( x = 30^\circ \).
Substituting \( x = 30^\circ \) in the equation:
\( \implies \sin 30^\circ + \cos y = 1 \)
\( \implies \frac{1}{2} + \cos y = 1 \)
\( \implies \cos y = 1 - \frac{1}{2} \)
\( \implies \cos y = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos y = \cos 60^\circ \)
\( \implies y = 60^\circ \)

(ii) Given, \( 3\tan A - 5\cos B = \sqrt{3} \) and \( B = 90^\circ \).
Substituting \( B = 90^\circ \) in the equation:
\( \implies 3\tan A - 5\cos 90^\circ = \sqrt{3} \)
Since \( \cos 90^\circ = 0 \):
\( \implies 3\tan A - 5(0) = \sqrt{3} \)
\( \implies 3\tan A = \sqrt{3} \)
\( \implies \tan A = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \)
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( \implies \tan A = \tan 30^\circ \)
\( \implies A = 30^\circ \)
In simple words: Substitute the given angle value into the equation first, then simplify to get a single trigonometric term and solve for the unknown angle.

Exam Tip: Remember that \( \cos 90^\circ = 0 \), which simplifies the equation by making the second term completely zero.

Question 9. In the given right-angled triangle ABC, \( \angle B = 90^\circ \), \( BC = 10 \text{ units} \), \( AC = 20 \text{ units} \), \( AB = y \text{ units} \) and \( \angle ACB = x^\circ \). Find:
(i) \( \cos x^\circ \)
(ii) \( x^\circ \)
(iii) \( \frac{1}{\tan^2 x^\circ} - \frac{1}{\sin^2 x^\circ} \)
(iv) \( y \)
Answer:
(i) From the right-angled triangle ABC:
\( \cos x^\circ = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
\( \implies \cos x^\circ = \frac{10}{20} = \frac{1}{2} \)

(ii) From part (i):
\( \cos x^\circ = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos x^\circ = \cos 60^\circ \)
\( \implies x^\circ = 60^\circ \)

(iii) Substituting \( x^\circ = 60^\circ \):
\( \frac{1}{\tan^2 x^\circ} - \frac{1}{\sin^2 x^\circ} = \frac{1}{\tan^2 60^\circ} - \frac{1}{\sin^2 60^\circ} \)
\( \implies \frac{1}{\tan^2 60^\circ} - \frac{1}{\sin^2 60^\circ} = \frac{1}{(\sqrt{3})^2} - \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} \)
\( \implies \frac{1}{\tan^2 60^\circ} - \frac{1}{\sin^2 60^\circ} = \frac{1}{3} - \frac{4}{3} = \frac{-3}{3} = -1 \)

(iv) Using the tangent ratio for angle \( x^\circ \):
\( \tan x^\circ = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{BC} \)
\( \implies \tan x^\circ = \frac{y}{10} \)
\( \implies y = 10\tan x^\circ \)
Substituting \( x^\circ = 60^\circ \):
\( \implies y = 10\tan 60^\circ \)
Since \( \tan 60^\circ = \sqrt{3} \):
\( \implies y = 10\sqrt{3} \)

In simple words: Use the sides of the triangle to find the cosine of the angle first, which gives you the angle value. Then substitute the angle to calculate the algebraic expression and the unknown height.

Exam Tip: Label the sides of the triangle clearly as 'Opposite', 'Adjacent', and 'Hypotenuse' with respect to the angle \(x^\circ\) before applying the trigonometric ratios.

Question 10. In a right-angled triangle, the perpendicular is \( 5 \text{ units} \), the base is \( 5 \text{ units} \), the hypotenuse is \( x \text{ units} \), and the acute angle between the base and hypotenuse is \( \theta \). Find:
(i) \( \tan \theta \)
(ii) \( \theta \)
(iii) \( \sin^2 \theta - \cos^2 \theta \)
(iv) \( x \)
Answer:
(i) From the given right-angled triangle, we have:
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{5} = 1 \)

(ii) From part (i):
\( \tan \theta = 1 \)
Since \( \tan 45^\circ = 1 \):
\( \implies \tan \theta = \tan 45^\circ \)
\( \implies \theta = 45^\circ \)

(iii) Substituting \( \theta = 45^\circ \):
\( \sin^2 \theta - \cos^2 \theta = \sin^2 45^\circ - \cos^2 45^\circ \)
\( \implies \sin^2 45^\circ - \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} - \frac{1}{2} = 0 \)

(iv) Using the sine ratio:
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
\( \implies \sin \theta = \frac{5}{x} \)
Substituting \( \theta = 45^\circ \):
\( \implies \sin 45^\circ = \frac{5}{x} \)
\( \implies \frac{1}{\sqrt{2}} = \frac{5}{x} \)
\( \implies x = 5\sqrt{2} \)

In simple words: Find the tangent ratio to get the value of the angle, which is \( 45^\circ \). Then plug it into the other formulas to calculate the remaining variables.

Exam Tip: For an isosceles right-angled triangle (where both legs are equal), the acute angles are always \( 45^\circ \), and the hypotenuse is \( \text{leg} \times \sqrt{2} \).

Question 11. Solve the following equations for A, if \( 0^\circ \le A \le 90^\circ \):
(i) \( 2\sin A\cos A - \cos A - 2\sin A + 1 = 0 \)
(ii) \( \tan A - 2\cos A\tan A + 2\cos A - 1 = 0 \)
(iii) \( 2\cos^2 A - 3\cos A + 1 = 0 \)
(iv) \( 2\tan 3A\cos 3A - \tan 3A + 1 = 2\cos 3A \)
Answer:
(i) Given, \( 2\sin A\cos A - \cos A - 2\sin A + 1 = 0 \)
Grouping the terms:
\( \implies \cos A(2\sin A - 1) - 1(2\sin A - 1) = 0 \)
\( \implies (2\sin A - 1)(\cos A - 1) = 0 \)
This gives:
\( \implies 2\sin A - 1 = 0 \) or \( \cos A - 1 = 0 \)
Case 1: \( 2\sin A - 1 = 0 \)
\( \implies \sin A = \frac{1}{2} \)
\( \implies \sin A = \sin 30^\circ \)
\( \implies A = 30^\circ \)
Case 2: \( \cos A - 1 = 0 \)
\( \implies \cos A = 1 \)
\( \implies \cos A = \cos 0^\circ \)
\( \implies A = 0^\circ \)
Hence, \( A = 30^\circ \) or \( 0^\circ \).

(ii) Given, \( \tan A - 2\cos A\tan A + 2\cos A - 1 = 0 \)
Rearranging and grouping:
\( \implies \tan A(1 - 2\cos A) - 1(1 - 2\cos A) = 0 \)
\( \implies (1 - 2\cos A)(\tan A - 1) = 0 \)
This gives:
\( \implies 1 - 2\cos A = 0 \) or \( \tan A - 1 = 0 \)
Case 1: \( 1 - 2\cos A = 0 \)
\( \implies 2\cos A = 1 \)
\( \implies \cos A = \frac{1}{2} \)
\( \implies \cos A = \cos 60^\circ \)
\( \implies A = 60^\circ \)
Case 2: \( \tan A - 1 = 0 \)
\( \implies \tan A = 1 \)
\( \implies \tan A = \tan 45^\circ \)
\( \implies A = 45^\circ \)
Hence, \( A = 60^\circ \) or \( 45^\circ \).

(iii) Given, \( 2\cos^2 A - 3\cos A + 1 = 0 \)
Splitting the middle term:
\( \implies 2\cos^2 A - 2\cos A - \cos A + 1 = 0 \)
\( \implies 2\cos A(\cos A - 1) - 1(\cos A - 1) = 0 \)
\( \implies (2\cos A - 1)(\cos A - 1) = 0 \)
This gives:
\( \implies 2\cos A - 1 = 0 \) or \( \cos A - 1 = 0 \)
Case 1: \( 2\cos A - 1 = 0 \)
\( \implies \cos A = \frac{1}{2} \)
\( \implies \cos A = \cos 60^\circ \)
\( \implies A = 60^\circ \)
Case 2: \( \cos A - 1 = 0 \)
\( \implies \cos A = 1 \)
\( \implies \cos A = \cos 0^\circ \)
\( \implies A = 0^\circ \)
Hence, \( A = 60^\circ \) or \( 0^\circ \).

(iv) Given, \( 2\tan 3A\cos 3A - \tan 3A + 1 = 2\cos 3A \)
Rearranging the equation:
\( \implies 2\tan 3A\cos 3A - \tan 3A - 2\cos 3A + 1 = 0 \)
Grouping terms:
\( \implies \tan 3A(2\cos 3A - 1) - 1(2\cos 3A - 1) = 0 \)
\( \implies (2\cos 3A - 1)(\tan 3A - 1) = 0 \)
This gives:
\( \implies 2\cos 3A - 1 = 0 \) or \( \tan 3A - 1 = 0 \)
Case 1: \( 2\cos 3A - 1 = 0 \)
\( \implies \cos 3A = \frac{1}{2} \)
\( \implies \cos 3A = \cos 60^\circ \)
\( \implies 3A = 60^\circ \)
\( \implies A = 20^\circ \)
Case 2: \( \tan 3A - 1 = 0 \)
\( \implies \tan 3A = 1 \)
\( \implies \tan 3A = \tan 45^\circ \)
\( \implies 3A = 45^\circ \)
\( \implies A = 15^\circ \)
Hence, \( A = 20^\circ \) or \( 15^\circ \).
In simple words: Factorise each expression by grouping similar terms together, solve each bracket for its trigonometric ratio, and find the corresponding angle values.

Exam Tip: Factoring trigonometric equations is very similar to factoring quadratic polynomial equations. Treat the trig function as a single variable when grouping terms.

Question 12. Solve for x:
(i) \( 2\cos 3x - 1 = 0 \)
(ii) \( \cos \frac{x}{3} - 1 = 0 \)
(iii) \( \sin (x + 10^\circ) = \frac{1}{2} \)
(iv) \( \cos (2x - 30^\circ) = 0 \)
(v) \( 2\cos (3x - 15^\circ) = 1 \)
(vi) \( \tan^2 (x - 5^\circ) = 3 \)
(vii) \( 3\tan^2 (2x - 20^\circ) = 1 \)
(viii) \( \cos \left(\frac{x}{2} + 10^\circ\right) = \frac{\sqrt{3}}{2} \)
(ix) \( \sin^2 x + \sin^2 30^\circ = 1 \)
(x) \( \cos^2 30^\circ + \cos^2 x = 1 \)
(xi) \( \cos^2 30^\circ + \sin^2 2x = 1 \)
(xii) \( \sin^2 60^\circ + \cos^2 (3x - 9^\circ) = 1 \)
Answer:
(i) Given, \( 2\cos 3x - 1 = 0 \)
\( \implies \cos 3x = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos 3x = \cos 60^\circ \)
\( \implies 3x = 60^\circ \)
\( \implies x = 20^\circ \)

(ii) Given, \( \cos \frac{x}{3} - 1 = 0 \)
\( \implies \cos \frac{x}{3} = 1 \)
Since \( \cos 0^\circ = 1 \):
\( \implies \cos \frac{x}{3} = \cos 0^\circ \)
\( \implies \frac{x}{3} = 0^\circ \)
\( \implies x = 0^\circ \)

(iii) Given, \( \sin (x + 10^\circ) = \frac{1}{2} \)
Since \( \sin 30^\circ = \frac{1}{2} \):
\( \implies \sin (x + 10^\circ) = \sin 30^\circ \)
\( \implies x + 10^\circ = 30^\circ \)
\( \implies x = 20^\circ \)

(iv) Given, \( \cos (2x - 30^\circ) = 0 \)
Since \( \cos 90^\circ = 0 \):
\( \implies \cos (2x - 30^\circ) = \cos 90^\circ \)
\( \implies 2x - 30^\circ = 90^\circ \)
\( \implies 2x = 120^\circ \)
\( \implies x = 60^\circ \)

(v) Given, \( 2\cos (3x - 15^\circ) = 1 \)
\( \implies \cos (3x - 15^\circ) = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos (3x - 15^\circ) = \cos 60^\circ \)
\( \implies 3x - 15^\circ = 60^\circ \)
\( \implies 3x = 75^\circ \)
\( \implies x = 25^\circ \)

(vi) Given, \( \tan^2 (x - 5^\circ) = 3 \)
Taking the positive square root (for acute angles):
\( \implies \tan (x - 5^\circ) = \sqrt{3} \)
Since \( \tan 60^\circ = \sqrt{3} \):
\( \implies \tan (x - 5^\circ) = \tan 60^\circ \)
\( \implies x - 5^\circ = 60^\circ \)
\( \implies x = 65^\circ \)

(vii) Given, \( 3\tan^2 (2x - 20^\circ) = 1 \)
\( \implies \tan^2 (2x - 20^\circ) = \frac{1}{3} \)
Taking square root:
\( \implies \tan (2x - 20^\circ) = \frac{1}{\sqrt{3}} \)
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( \implies \tan (2x - 20^\circ) = \tan 30^\circ \)
\( \implies 2x - 20^\circ = 30^\circ \)
\( \implies 2x = 50^\circ \)
\( \implies x = 25^\circ \)

(viii) Given, \( \cos \left(\frac{x}{2} + 10^\circ\right) = \frac{\sqrt{3}}{2} \)
Since \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \cos \left(\frac{x}{2} + 10^\circ\right) = \cos 30^\circ \)
\( \implies \frac{x}{2} + 10^\circ = 30^\circ \)
\( \implies \frac{x}{2} = 20^\circ \)
\( \implies x = 40^\circ \)

(ix) Given, \( \sin^2 x + \sin^2 30^\circ = 1 \)
\( \implies \sin^2 x = 1 - \sin^2 30^\circ \)
Since \( \sin 30^\circ = \frac{1}{2} \):
\( \implies \sin^2 x = 1 - \left(\frac{1}{2}\right)^2 \)
\( \implies \sin^2 x = 1 - \frac{1}{4} = \frac{3}{4} \)
\( \implies \sin x = \frac{\sqrt{3}}{2} \)
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \sin x = \sin 60^\circ \)
\( \implies x = 60^\circ \)

(x) Given, \( \cos^2 30^\circ + \cos^2 x = 1 \)
\( \implies \cos^2 x = 1 - \cos^2 30^\circ \)
Since \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \cos^2 x = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \implies \cos^2 x = 1 - \frac{3}{4} = \frac{1}{4} \)
\( \implies \cos x = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos x = \cos 60^\circ \)
\( \implies x = 60^\circ \)

(xi) Given, \( \cos^2 30^\circ + \sin^2 2x = 1 \)
\( \implies \sin^2 2x = 1 - \cos^2 30^\circ \)
Since \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \sin^2 2x = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \implies \sin^2 2x = 1 - \frac{3}{4} = \frac{1}{4} \)
\( \implies \sin 2x = \frac{1}{2} \)
Since \( \sin 30^\circ = \frac{1}{2} \):
\( \implies \sin 2x = \sin 30^\circ \)
\( \implies 2x = 30^\circ \)
\( \implies x = 15^\circ \)

(xii) Given, \( \sin^2 60^\circ + \cos^2 (3x - 9^\circ) = 1 \)
\( \implies \cos^2 (3x - 9^\circ) = 1 - \sin^2 60^\circ \)
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \cos^2 (3x - 9^\circ) = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \implies \cos^2 (3x - 9^\circ) = 1 - \frac{3}{4} = \frac{1}{4} \)
Taking square root:
\( \implies \cos (3x - 9^\circ) = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos (3x - 9^\circ) = \cos 60^\circ \)
\( \implies 3x - 9^\circ = 60^\circ \)
\( \implies 3x = 69^\circ \)
\( \implies x = 23^\circ \)
In simple words: Simplify the given equations to find the value of the trigonometric ratio first. Then, compare it with the standard table values to determine the angle and solve for x.

Exam Tip: When taking square roots in trigonometric equations, remember that the ratios are positive for acute angles (from \( 0^\circ \) to \( 90^\circ \)).

Question 13. If \( 4\cos^2 x = 3 \) and x is an acute angle; find the value of :
(i) x
(ii) \( \cos^2 x + \cot^2 x \)
(iii) \( \cos 3x \)
(iv) \( \sin 2x \)
Answer:
(i) Given, \( 4\cos^2 x = 3 \)
\( \implies \cos^2 x = \frac{3}{4} \)
Since x is acute, taking positive square root:
\( \implies \cos x = \frac{\sqrt{3}}{2} \)
Since \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \cos x = \cos 30^\circ \)
\( \implies x = 30^\circ \)

(ii) Substituting \( x = 30^\circ \):
\( \cos^2 x + \cot^2 x = \cos^2 30^\circ + \cot^2 30^\circ \)
\( \implies \cos^2 30^\circ + \cot^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 + (\sqrt{3})^2 \)
\( \implies \cos^2 30^\circ + \cot^2 30^\circ = \frac{3}{4} + 3 = \frac{15}{4} = 3\frac{3}{4} \)

(iii) Substituting \( x = 30^\circ \):
\( \cos 3x = \cos 3(30^\circ) = \cos 90^\circ = 0 \)

(iv) Substituting \( x = 30^\circ \):
\( \sin 2x = \sin 2(30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
In simple words: Find the angle x from the first equation, and then substitute it into the expressions in the next parts to compute their numerical values.

Exam Tip: Be sure to write the final answers in mixed fraction format (like \( 3\frac{3}{4} \)) if the sub-part (ii) solution yields an improper fraction.

Question 14. In \( \Delta ABC \), \( \angle B = 90^\circ \), \( AB = y \text{ units} \), \( BC = \sqrt{3} \text{ units} \), \( AC = 2 \text{ units} \) and angle \( \angle A = x^\circ \). Find:
(i) \( \sin x^\circ \)
(ii) \( x^\circ \)
(iii) \( \tan x^\circ \)
(iv) Use \( \cos x^\circ \) to find the value of y.
Answer:
(i) From right-angled \( \Delta ABC \):
\( \sin x^\circ = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
\( \implies \sin x^\circ = \frac{\sqrt{3}}{2} \)

(ii) From part (i):
\( \sin x^\circ = \frac{\sqrt{3}}{2} \)
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\( \implies \sin x^\circ = \sin 60^\circ \)
\( \implies x^\circ = 60^\circ \)

(iii) Substituting \( x^\circ = 60^\circ \):
\( \tan x^\circ = \tan 60^\circ \)
Since \( \tan 60^\circ = \sqrt{3} \):
\( \implies \tan x^\circ = \sqrt{3} \)

(iv) Using the cosine ratio for angle \( x^\circ \):
\( \cos x^\circ = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
\( \implies \cos x^\circ = \frac{y}{2} \)
Substituting \( x^\circ = 60^\circ \):
\( \implies \cos 60^\circ = \frac{y}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \frac{1}{2} = \frac{y}{2} \)
\( \implies y = 1 \)
In simple words: Find the sine ratio from the given triangle dimensions to identify the angle. Once you know the angle, use the other trig ratios to calculate the remaining side and tangent values.

Exam Tip: Pay close attention to which angle is being used. Opposite and adjacent sides swap places depending on whether you are analyzing angle A or angle C.

Question 15. If \( 2\cos (A + B) = 2\sin (A - B) = 1 \); find the values of A and B.
Answer:
Given, \( 2\cos (A + B) = 1 \)
\( \implies \cos (A + B) = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies \cos (A + B) = \cos 60^\circ \)
\( \implies A + B = 60^\circ \) - (1)

Also given, \( 2\sin (A - B) = 1 \)
\( \implies \sin (A - B) = \frac{1}{2} \)
Since \( \sin 30^\circ = \frac{1}{2} \):
\( \implies \sin (A - B) = \sin 30^\circ \)
\( \implies A - B = 30^\circ \) - (2)

Now, we solve equations (1) and (2) simultaneously.
Adding equation (1) and (2):
\( \implies (A + B) + (A - B) = 60^\circ + 30^\circ \)
\( \implies 2A = 90^\circ \)
\( \implies A = 45^\circ \)

Substitute \( A = 45^\circ \) in equation (1):
\( \implies 45^\circ + B = 60^\circ \)
\( \implies B = 60^\circ - 45^\circ \)
\( \implies B = 15^\circ \)
Hence, \( A = 45^\circ \) and \( B = 15^\circ \).
In simple words: Turn the two given equations into a pair of linear equations in terms of A and B by using standard angle values, then add them together to solve for both angles.

Exam Tip: When solving simultaneous linear equations of the form \( A+B=x \) and \( A-B=y \), always add them to eliminate \( B \) first and find \( A \).