Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 26 Trigonometrical Ratios

Question 1. Find all other trigonometric ratios for each of the following cases:
(i) \( \sin A = \frac{12}{13} \)
(ii) \( \cos B = \frac{4}{5} \)
(iii) \( \cot A = \frac{1}{11} \)
(iv) \( \csc C = \frac{15}{11} \)
(v) \( \tan C = \frac{5}{12} \)
(vi) \( \sin B = \frac{\sqrt{3}}{2} \)
(vii) \( \cos A = \frac{7}{25} \)
(viii) \( \tan B = \frac{8}{15} \)
(ix) \( \sec B = \frac{15}{12} \)
(x) \( \csc C = \sqrt{10} \)
Answer:
(i) We are given \( \sin A = \frac{12}{13} \).
Since \( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \), let the perpendicular be 12 and the hypotenuse be 13.
Using Pythagoras' theorem, we find:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( \implies \text{Base} = \sqrt{(13)^2 - (12)^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \)
Now, we calculate the remaining ratios:
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{5}{13} \)
\( \sec A = \frac{1}{\cos A} = \frac{13}{5} \)
\( \cot A = \frac{1}{\tan A} = \frac{5}{12} \)
\( \csc A = \frac{1}{\sin A} = \frac{13}{12} \)

(ii) We are given \( \cos B = \frac{4}{5} \).
Since \( \cos B = \frac{\text{Base}}{\text{Hypotenuse}} \), let the base be 4 and the hypotenuse be 5.
By applying Pythagoras' theorem, we obtain:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} \)
\( \implies \text{Perpendicular} = \sqrt{(5)^2 - (4)^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \)
Therefore, the other trigonometric ratios are:
\( \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{3}{5} \)
\( \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \)
\( \sec B = \frac{1}{\cos B} = \frac{5}{4} \)
\( \cot B = \frac{1}{\tan B} = \frac{4}{3} \)
\( \csc B = \frac{1}{\sin B} = \frac{5}{3} \)

(iii) We are given \( \cot A = \frac{1}{11} \).
Since \( \cot A = \frac{\text{Base}}{\text{Perpendicular}} \), let the base be 1 and the perpendicular be 11.
Using the Pythagorean relation:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( \implies \text{Hypotenuse} = \sqrt{(11)^2 + (1)^2} = \sqrt{121 + 1} = \sqrt{122} \)
Thus, the other ratios are:
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{\sqrt{122}} \)
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = 11 \)
\( \sec A = \frac{1}{\cos A} = \sqrt{122} \)
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{11}{\sqrt{122}} \)
\( \csc A = \frac{1}{\sin A} = \frac{\sqrt{122}}{11} \)

(iv) We are given \( \csc C = \frac{15}{11} \).
Since \( \csc C = \frac{\text{Hypotenuse}}{\text{Perpendicular}} \), let the hypotenuse be 15 and the perpendicular be 11.
According to Pythagoras' theorem:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( \implies \text{Base} = \sqrt{(15)^2 - (11)^2} = \sqrt{225 - 121} = \sqrt{104} \)
Hence, the other ratios are:
\( \sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{11}{15} \)
\( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\sqrt{104}}{15} \)
\( \tan C = \frac{\text{Perpendicular}}{\text{Base}} = \frac{11}{\sqrt{104}} \)
\( \sec C = \frac{1}{\cos C} = \frac{15}{\sqrt{104}} \)
\( \cot C = \frac{1}{\tan C} = \frac{\sqrt{104}}{11} \)

(v) We are given \( \tan C = \frac{5}{12} \).
Since \( \tan C = \frac{\text{Perpendicular}}{\text{Base}} \), let the perpendicular be 5 and the base be 12.
Using Pythagoras' theorem, we find:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( \implies \text{Hypotenuse} = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)
The remaining trigonometric ratios are:
\( \cot C = \frac{1}{\tan C} = \frac{12}{5} \)
\( \sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13} \)
\( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{12}{13} \)
\( \sec C = \frac{1}{\cos C} = \frac{13}{12} \)
\( \csc C = \frac{1}{\sin C} = \frac{13}{5} \)

(vi) We are given \( \sin B = \frac{\sqrt{3}}{2} \).
Since \( \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \), let the perpendicular be \( \sqrt{3} \) and the hypotenuse be 2.
From Pythagoras' theorem, we have:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( \implies \text{Base} = \sqrt{(2)^2 - (\sqrt{3})^2} = \sqrt{4 - 3} = \sqrt{1} = 1 \)
Therefore, the ratios are:
\( \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{2} \)
\( \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \sqrt{3} \)
\( \sec B = \frac{1}{\cos B} = 2 \)
\( \cot B = \frac{1}{\tan B} = \frac{1}{\sqrt{3}} \)
\( \csc B = \frac{1}{\sin B} = \frac{2}{\sqrt{3}} \)

(vii) We are given \( \cos A = \frac{7}{25} \).
Since \( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} \), let the base be 7 and the hypotenuse be 25.
By Pythagoras' theorem:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} \)
\( \implies \text{Perpendicular} = \sqrt{(25)^2 - (7)^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \)
Now, we compute the other ratios:
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{24}{25} \)
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{24}{7} \)
\( \sec A = \frac{1}{\cos A} = \frac{25}{7} \)
\( \cot A = \frac{1}{\tan A} = \frac{7}{24} \)
\( \csc A = \frac{1}{\sin A} = \frac{25}{24} \)

(viii) We are given \( \tan B = \frac{8}{15} \).
Since \( \tan B = \frac{\text{Perpendicular}}{\text{Base}} \), let the perpendicular be 8 and the base be 15.
Applying Pythagoras' theorem, we get:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( \implies \text{Hypotenuse} = \sqrt{(8)^2 + (15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \)
The remaining trigonometric ratios are:
\( \cot B = \frac{1}{\tan B} = \frac{15}{8} \)
\( \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{8}{17} \)
\( \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{15}{17} \)
\( \sec B = \frac{1}{\cos B} = \frac{17}{15} \)
\( \csc B = \frac{1}{\sin B} = \frac{17}{8} \)

(ix) We are given \( \sec B = \frac{15}{12} \).
Since \( \sec B = \frac{\text{Hypotenuse}}{\text{Base}} \), let the hypotenuse be 15 and the base be 12.
Using Pythagoras' theorem:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} \)
\( \implies \text{Perpendicular} = \sqrt{(15)^2 - (12)^2} = \sqrt{225 - 144} = \sqrt{81} = 9 \)
Hence, the other ratios are:
\( \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{9}{15} \)
\( \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{9}{12} \)
\( \cot B = \frac{1}{\tan B} = \frac{12}{9} \)
\( \csc B = \frac{1}{\sin B} = \frac{15}{9} \)
\( \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{12}{15} \)

(x) We are given \( \csc C = \sqrt{10} \).
Since \( \csc C = \frac{\text{Hypotenuse}}{\text{Perpendicular}} \), let the hypotenuse be \( \sqrt{10} \) and the perpendicular be 1.
Using the Pythagorean relation:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( \implies \text{Base} = \sqrt{(\sqrt{10})^2 - (1)^2} = \sqrt{10 - 1} = \sqrt{9} = 3 \)
Thus, the other ratios are:
\( \sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{10}} \)
\( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{3}{\sqrt{10}} \)
\( \tan C = \frac{\text{Perpendicular}}{\text{Base}} = \frac{1}{3} \)
\( \sec C = \frac{1}{\cos C} = \frac{\sqrt{10}}{3} \)
\( \cot C = \frac{1}{\tan C} = 3 \)
In simple words: Trigonometric ratios are based on the sides of a right-angled triangle. By using the given ratio and Pythagoras' theorem, we can find the length of the third side and then calculate all the other ratios.

Exam Tip: Remember the basic definitions of all six trigonometric ratios and always write down the Pythagorean formula carefully to avoid simple calculation errors while finding the third side.

Question 2. For a right-angled triangle \( \Delta ABC \), with the right angle at \( A \), if the sides are \( AB = 5 \text{ units} \) and \( AC = 12 \text{ units} \), find:
(i) \( \sin B \) and \( \cos B \)
(ii) \( \cos C \)
(iii) \( \tan B \)
Answer:
In the right triangle \( \Delta ABC \), we apply Pythagoras' theorem:
\( BC^2 = AB^2 + AC^2 \)
\( \implies BC = \sqrt{AB^2 + AC^2} \)
\( \implies BC = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)
Here, the side measures are:
\( AC = 12 \text{ units} \)
\( BC = 13 \text{ units} \)
\( AB = 5 \text{ units} \)
(i) For angle \( B \), the opposite side (perpendicular) is \( AC \) and the adjacent side (base) is \( AB \):
\( \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{BC} = \frac{12}{13} \)
\( \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{BC} = \frac{5}{13} \)
(ii) For angle \( C \), the adjacent side (base) is \( AC \) and the hypotenuse is \( BC \):
\( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AC}{BC} = \frac{12}{13} \)
(iii) For angle \( B \):
\( \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{AB} = \frac{12}{5} \)
In simple words: First, find the longest side (hypotenuse) using the other two sides and Pythagoras' theorem. After that, look at which angle you are using to decide which side is the perpendicular (opposite) and which is the base (adjacent), and write down the ratios.

Exam Tip: Be careful with the reference angle! For angle B, the opposite side AC is the perpendicular, but for angle C, the opposite side AB becomes the perpendicular.

Question 3. In a right-angled triangle \( \Delta ABC \), with the right angle at \( B \), if \( AB = 12 \text{ units} \) and \( BC = 5 \text{ units} \), find the values of:
(i) \( \sin A \)
(ii) \( \tan A \)
(iii) \( \cos C \)
(iv) \( \cot C \)
Answer:
In the triangle \( \Delta ABC \), we apply Pythagoras' theorem:
\( AC^2 = AB^2 + BC^2 \)
\( \implies AC = \sqrt{AB^2 + BC^2} \)
\( \implies AC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13 \)
So, we have:
\( AB = 12 \text{ units} \)
\( BC = 5 \text{ units} \)
\( AC = 13 \text{ units} \)
(i) For angle \( A \), the opposite side (perpendicular) is \( BC \) and the hypotenuse is \( AC \):
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5}{13} \)
(ii) For angle \( A \), the adjacent side (base) is \( AB \):
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{5}{12} \)
(iii) For angle \( C \), the adjacent side (base) is \( BC \):
\( \cos C = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5}{13} \)
(iv) For angle \( C \), the opposite side (perpendicular) is \( AB \):
\( \cot C = \frac{\text{Base}}{\text{Perpendicular}} = \frac{BC}{AB} = \frac{5}{12} \)
In simple words: First find the diagonal side (hypotenuse) using the other two sides. Then, use the definition of each ratio, making sure to look from angle A for the first two parts, and from angle C for the last two parts.

Exam Tip: Always double check that you identify the base and perpendicular relative to the specific angle asked in the question (A or C).

Question 4. If \( \sin A = \frac{3}{5} \), find the values of \( \cos A \) and \( \tan A \).
Answer:
Given that \( \sin A = \frac{3}{5} \).
Since \( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \), we can let:
Perpendicular = 3
Hypotenuse = 5
Using the Pythagorean theorem:
\( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Base})^2 \)
\( \implies (\text{Base})^2 = (\text{Hypotenuse})^2 - (\text{Perpendicular})^2 \)
\( \implies \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( \implies \text{Base} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \)
Therefore:
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5} \)
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \)
In simple words: Since sine is perpendicular over hypotenuse, we use these values to find that the base is 4. Then we write down cosine as base over hypotenuse, and tangent as perpendicular over base.

Exam Tip: (3, 4, 5) is a standard Pythagorean triplet. Memorizing common triplets like (3, 4, 5), (5, 12, 13), and (8, 15, 17) helps verify your calculations instantly.

Question 5. In a right-angled triangle \( \Delta ABC \) with the right angle at \( C \), if \( \cos B = \frac{1}{3} \), find the values of \( \sin A \), \( \tan B \), and \( \cot A \).
Answer:
We are given \( \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{1}{3} \).
Let us define \( BC = 1 \) and \( AB = 3 \).
Applying the Pythagorean theorem in the right triangle \( \Delta ABC \):
\( (AB)^2 = (AC)^2 + (BC)^2 \)
\( \implies AC = \sqrt{(AB)^2 - (BC)^2} \)
\( \implies AC = \sqrt{3^2 - 1^2} = \sqrt{9 - 1} = \sqrt{8} = 2\sqrt{2} \)
Now, we evaluate the required ratios:
For angle \( A \), the perpendicular is \( BC \) and the base is \( AC \):
\( \sin A = \frac{BC}{AB} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{3} \)
\( \cot A = \frac{1}{\tan A} = \frac{\text{Base}}{\text{Perpendicular}} = \frac{AC}{BC} = \frac{2\sqrt{2}}{1} = 2\sqrt{2} \)
For angle \( B \), the perpendicular is \( AC \) and the base is \( BC \):
\( \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC} = \frac{2\sqrt{2}}{1} = 2\sqrt{2} \)
In simple words: Use the given ratio of cosine to label the sides BC and AB. Find the third side AC with Pythagoras' theorem. Then, use these sides to write the ratios for the specified angles.

Exam Tip: Remember that in a right triangle ABC with angle C = 90°, the sine of angle A is always equal to the cosine of angle B because they are complementary angles.

Question 6. If \( \sin \theta = \frac{8}{17} \), find the other trigonometric ratios of \( \theta \).
Answer:
Given that \( \sin \theta = \frac{8}{17} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \).
Let the perpendicular be 8 and the hypotenuse be 17.
According to Pythagoras' theorem:
\( \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( \implies \text{Base} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \)
Thus, the other trigonometric ratios are:
\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{15}{17} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{8}{15} \)
\( \csc \theta = \frac{1}{\sin \theta} = \frac{17}{8} \)
\( \sec \theta = \frac{1}{\cos \theta} = \frac{17}{15} \)
\( \cot \theta = \frac{1}{\tan \theta} = \frac{15}{8} \)
In simple words: Since sine is 8 over 17, we find that the adjacent side (base) is 15 using the Pythagorean formula. Using these three sides, we write down all other five ratios.

Exam Tip: Clearly write the formula for each trigonometric ratio before substituting the values to ensure you gain step-wise marks in the examination.

Question 7. If \( \tan A = 0.75 \), find all other trigonometric ratios of the angle \( A \).
Answer: We are given that:
\( \tan A = 0.75 = \frac{75}{100} = \frac{3}{4} \)
Since the tangent ratio represents the perpendicular divided by the base:
\( \text{Perpendicular} = 3 \) and \( \text{Base} = 4 \)
Using the Pythagorean theorem, the hypotenuse is:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
Using these values, we can determine the remaining trigonometric ratios:
\( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{3}{5} = 0.6 \)
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5} = 0.8 \)
\( \csc A = \frac{1}{\sin A} = \frac{5}{3} \approx 1.67 \)
\( \sec A = \frac{1}{\cos A} = \frac{5}{4} = 1.25 \)
\( \cot A = \frac{1}{\tan A} = \frac{4}{3} \approx 1.33 \)
In simple words: Turn the decimal value of tangent into a fraction to find the base and perpendicular. Calculate the hypotenuse with the Pythagorean theorem, and then write down the rest of the ratios.

Exam Tip: Converting decimal values to simplified fractions is a helpful first step to identify the sides of the right-angled triangle directly.

Question 8. If \( \sin A = 0.8 \), find all other trigonometric ratios of the angle \( A \).
Answer: We are given:
\( \sin A = 0.8 = \frac{8}{10} = \frac{4}{5} \)
Since the sine function represents the ratio of the perpendicular to the hypotenuse:
\( \text{Perpendicular} = 4 \) and \( \text{Hypotenuse} = 5 \)
We calculate the base using the Pythagorean theorem:
\( \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \)
Now, we evaluate the rest of the trigonometric ratios:
\( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{3}{5} = 0.6 \)
\( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{4}{3} \approx 1.33 \)
\( \csc A = \frac{1}{\sin A} = \frac{5}{4} = 1.25 \)
\( \sec A = \frac{1}{\cos A} = \frac{5}{3} \approx 1.67 \)
\( \cot A = \frac{1}{\tan A} = \frac{3}{4} = 0.75 \)
In simple words: Write the sine value as a simplified fraction to find the perpendicular and hypotenuse. Solve for the base using the Pythagorean theorem, and then compute the other ratios.

Exam Tip: Always state your final ratios clearly in both fractional and decimal form if requested by the examiner.

Question 9. If \( 8\tan\theta = 15 \), find the values of:
(i) \( \sin\theta \)
(ii) \( \cot\theta \)
(iii) \( \sin^2\theta - \cot^2\theta \)
Answer: We start with the given equation:
\( 8\tan\theta = 15 \)
\( \implies \tan\theta = \frac{15}{8} \)
Since tangent is the ratio of the perpendicular to the base:
\( \text{Perpendicular} = 15 \) and \( \text{Base} = 8 \)
We find the hypotenuse using the Pythagorean theorem:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \)
Using these side lengths, we calculate the required expressions:
(i) \( \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{15}{17} \)
(ii) \( \cot\theta = \frac{1}{\tan\theta} = \frac{8}{15} \)
(iii) We can evaluate \( \sin^2\theta - \cot^2\theta \) using the difference of squares factorization:
\( \sin^2\theta - \cot^2\theta = (\sin\theta + \cot\theta)(\sin\theta - \cot\theta) \)
\( = \left(\frac{15}{17} + \frac{8}{15}\right)\left(\frac{15}{17} - \frac{8}{15}\right) \)
\( = \left(\frac{225 + 136}{255}\right)\left(\frac{225 - 136}{255}\right) \)
\( = \left(\frac{361}{255}\right)\left(\frac{89}{255}\right) = \frac{32129}{65025} \)
In simple words: Isolate tangent to find the perpendicular and the base, then compute the hypotenuse. Calculate sine and cotangent values to solve the given algebraic expression.

Exam Tip: Using algebraic identities like \( a^2 - b^2 = (a+b)(a-b) \) can significantly simplify calculations when working with large fractions.

Question 14. In a right-angled triangle \( ABC \), right-angled at \( B \), a point \( D \) lies on \( BC \) such that \( BD : DC = 1 : 2 \). Find the values of:
(i) \( \frac{\tan \angle BAC}{\tan \angle BAD} \)
(ii) \( \frac{\cot \angle BAC}{\cot \angle BAD} \)
Answer: Let us set \( BD = x \). Since \( AD \) divides \( BC \) in the ratio \( 1 : 2 \), we have:
\( BD : DC = 1 : 2 \)
\( \implies DC = 2x \)
Thus, the total length of \( BC \) is:
\( BC = BD + DC = x + 2x = 3x \)
In the right-angled triangles \( ABC \) and \( ABD \) with right angle at \( B \):
(i) Evaluating the ratio of tangents:
\( \frac{\tan \angle BAC}{\tan \angle BAD} = \frac{\frac{BC}{AB}}{\frac{BD}{AB}} = \frac{BC}{BD} = \frac{3x}{x} = 3 \)
(ii) Evaluating the ratio of cotangents:
\( \frac{\cot \angle BAC}{\cot \angle BAD} = \frac{\frac{AB}{BC}}{\frac{AB}{BD}} = \frac{BD}{BC} = \frac{x}{3x} = \frac{1}{3} \)
In simple words: Represent the segments of the base BC using a variable x. Since both triangles share the vertical side AB, it cancels out when evaluating the ratios.

Exam Tip: Clearly define the segment lengths in terms of a variable like \( x \) to show logical structure in your geometric proofs.

Question 19. In a right-angled triangle \( PQR \), right-angled at \( Q \), \( PS \) is the median drawn to the side \( QR \). A point \( T \) divides the side \( PQ \) in the ratio \( 1 : 2 \) such that \( QT : PT = 1 : 2 \). Find the values of:
(i) \( \frac{\tan \angle PSQ}{\tan \angle PRQ} \)
(ii) \( \frac{\tan \angle TSQ}{\tan \angle PRQ} \)
Answer: Since \( PS \) is the median from vertex \( P \) to side \( QR \), the point \( S \) is the midpoint of \( QR \):
\( QS = SR \)
\( \implies QR = 2QS \)
Also, the point \( T \) divides the side \( PQ \) in the ratio \( 1 : 2 \):
\( \therefore QT = x \) and \( PT = 2x \)
\( \implies PQ = 3x \)
Using these geometric relations, we calculate the ratios:
(i) \( \frac{\tan \angle PSQ}{\tan \angle PRQ} = \frac{\frac{PQ}{QS}}{\frac{PQ}{QR}} = \frac{PQ}{QS} \times \frac{QR}{PQ} = \frac{QR}{QS} = \frac{2QS}{QS} = 2 \)
(ii) \( \frac{\tan \angle TSQ}{\tan \angle PRQ} = \frac{\frac{QT}{QS}}{\frac{PQ}{QR}} = \frac{QT}{QS} \times \frac{QR}{PQ} = \frac{x}{QS} \times \frac{2QS}{3x} = \frac{2}{3} \)
In simple words: Express the segments using the properties of a median and the given ratios. Substitute these into the trigonometric definitions to evaluate the final numerical answers.

Exam Tip: Remember that a median bisects a side. Write out this relationship explicitly at the start of your solution.

Question 22. If \( 24\cos\theta = 7\sin\theta \), find the value of \( \sin\theta + \cos\theta \).
Answer: We are given:
\( 24\cos\theta = 7\sin\theta \)
We rearrange the terms to find the tangent ratio:
\( \frac{\sin\theta}{\cos\theta} = \frac{24}{7} \)
\( \implies \tan\theta = \frac{24}{7} \)
Since tangent is the ratio of perpendicular to base:
\( \text{Perpendicular} = 24 \) and \( \text{Base} = 7 \)
We can find the hypotenuse using Pythagoras' theorem:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \)
Now we find the sum of sine and cosine:
\( \sin\theta + \cos\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} + \frac{\text{Base}}{\text{Hypotenuse}} \)
\( = \frac{24}{25} + \frac{7}{25} = \frac{31}{25} \)
In simple words: Rearrange the equation to determine tangent. Calculate the hypotenuse using Pythagoras' theorem, find sine and cosine, and add them together.

Exam Tip: Memorizing common Pythagorean triplets such as (7, 24, 25) can save precious time during exams.

Question 24. If \( 8\tan A = 15 \), find the value of \( \sin A - \cos A \).
Answer: We are given:
\( 8\tan A = 15 \)
\( \implies \tan A = \frac{15}{8} \)
Since tangent is the ratio of the perpendicular to the base:
\( \text{Perpendicular} = 15 \) and \( \text{Base} = 8 \)
We compute the hypotenuse using the Pythagorean theorem:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \)
Now we calculate the difference of sine and cosine:
\( \sin A - \cos A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} - \frac{\text{Base}}{\text{Hypotenuse}} \)
\( = \frac{15}{17} - \frac{8}{17} = \frac{7}{17} \)
In simple words: Find the tangent as a fraction to identify the perpendicular and base. Use Pythagoras' theorem to calculate the hypotenuse, then evaluate sine minus cosine.

Exam Tip: Always state which side is perpendicular, base, or hypotenuse with respect to the reference angle A to avoid configuration errors.

Question 25. If \( 3\cos\theta - 4\sin\theta = 2\cos\theta + \sin\theta \), find the value of \( \tan\theta \).
Answer: We begin with the given equation:
\( 3\cos\theta - 4\sin\theta = 2\cos\theta + \sin\theta \)
Rearranging the equation to group cosine terms on one side and sine terms on the other:
\( 3\cos\theta - 2\cos\theta = \sin\theta + 4\sin\theta \)
\( \implies \cos\theta = 5\sin\theta \)
Dividing both sides by \( \cos\theta \) and by 5:
\( \implies \frac{\sin\theta}{\cos\theta} = \frac{1}{5} \)
\( \implies \tan\theta = \frac{1}{5} \)
In simple words: Group like terms together to get cosine on one side and sine on the other. Divide sine by cosine to find tangent.

Exam Tip: Take extra care with the arithmetic signs when moving terms from one side of the equation to the other.

Question 26. If \( 5\cos\theta = 3 \), evaluate \( \frac{4\cos\theta - \sin\theta}{2\cos\theta + \sin\theta} \).
Answer: We are given:
\( 5\cos\theta = 3 \)
\( \implies \cos\theta = \frac{3}{5} \)
Since cosine is the ratio of base to hypotenuse:
\( \text{Base} = 3 \) and \( \text{Hypotenuse} = 5 \)
Using the Pythagorean relation:
\( \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} \)
\( = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \)
This gives:
\( \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{4}{5} \)
Now, we substitute these ratios into the given expression:
\( \frac{4\cos\theta - \sin\theta}{2\cos\theta + \sin\theta} = \frac{4\left(\frac{3}{5}\right) - \frac{4}{5}}{2\left(\frac{3}{5}\right) + \frac{4}{5}} \)
\( = \frac{\frac{12}{5} - \frac{4}{5}}{\frac{6}{5} + \frac{4}{5}} \)
\( = \frac{\frac{8}{5}}{\frac{10}{5}} = \frac{8}{10} = \frac{4}{5} \)
In simple words: Find the perpendicular side using Pythagoras' theorem from the cosine value. Evaluate sine, and substitute both values into the expression to calculate the answer.

Exam Tip: Alternatively, you can divide both the numerator and denominator by \( \cos\theta \) to express the equation entirely in terms of \( \tan\theta \) to save time.

Question 28. If \( 5\tan\theta = 12 \), evaluate \( \frac{2\sin\theta - 3\cos\theta}{4\sin\theta - 9\cos\theta} \).
Answer: We are given:
\( 5\tan\theta = 12 \)
\( \implies \tan\theta = \frac{12}{5} \)
Since tangent is the ratio of perpendicular to base:
\( \text{Perpendicular} = 12 \) and \( \text{Base} = 5 \)
Using Pythagoras' theorem:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \)
Using these values, we get:
\( \sin\theta = \frac{12}{13} \) and \( \cos\theta = \frac{5}{13} \)
Substituting these values into the given expression:
\( \frac{2\sin\theta - 3\cos\theta}{4\sin\theta - 9\cos\theta} = \frac{2\left(\frac{12}{13}\right) - 3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right) - 9\left(\frac{5}{13}\right)} \)
\( = \frac{\frac{24 - 15}{13}}{\frac{48 - 45}{13}} = \frac{24 - 15}{48 - 45} = \frac{9}{3} = 3 \)
In simple words: Find the hypotenuse first to write sine and cosine as fractions. Put these values in the expression and simplify to get the answer.

Exam Tip: If the expression consists of homogeneous terms in sine and cosine, dividing by \( \cos\theta \) to get tangent terms is a faster, error-free method.

Question 30. If \( \cot\theta = \frac{1}{\sqrt{3}} \), show that \( \frac{1 - \cos^2\theta}{2 - \sin^2\theta} = \frac{3}{5} \).
Answer: We are given:
\( \cot\theta = \frac{1}{\sqrt{3}} \)
\( \implies \tan\theta = \sqrt{3} \)
Since cotangent is the ratio of base to perpendicular:
\( \text{Base} = 1 \) and \( \text{Perpendicular} = \sqrt{3} \)
We compute the hypotenuse:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2 \)
This gives:
\( \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{2} \)
\( \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{\sqrt{3}}{2} \)
Now we evaluate the left-hand side (L.H.S.) of the relation:
\( \text{L.H.S.} = \frac{1 - \cos^2\theta}{2 - \sin^2\theta} \)
\( = \frac{1 - \left(\frac{1}{2}\right)^2}{2 - \left(\frac{\sqrt{3}}{2}\right)^2} \)
\( = \frac{1 - \frac{1}{4}}{2 - \frac{3}{4}} \)
\( = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5} = \text{R.H.S.} \)
Therefore, the expression is verified.
In simple words: Use cotangent to find the hypotenuse and calculate sine and cosine. Put these into the equation and simplify to prove it equals 3/5.

Exam Tip: Be meticulous when squaring irrational terms like \( \sqrt{3} \) to ensure your arithmetic remains completely correct.

Question 31. If \( \csc\theta = 1\frac{9}{20} \), show that \( \frac{1 - \sin\theta + \cos\theta}{1 + \sin\theta + \cos\theta} = \frac{3}{7} \).
Answer: We are given:
\( \csc\theta = 1\frac{9}{20} = \frac{29}{20} \)
Sine is the reciprocal of cosecant:
\( \sin\theta = \frac{20}{29} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
So, \( \text{Perpendicular} = 20 \) and \( \text{Hypotenuse} = 29 \).
We calculate the base using Pythagoras' theorem:
\( \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( = \sqrt{29^2 - 20^2} = \sqrt{841 - 400} = \sqrt{441} = 21 \)
Thus, cosine is:
\( \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{21}{29} \)
Substituting sine and cosine into the L.H.S. of the identity:
\( \text{L.H.S.} = \frac{1 - \sin\theta + \cos\theta}{1 + \sin\theta + \cos\theta} \)
\( = \frac{1 - \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} \)
\( = \frac{\frac{29 - 20 + 21}{29}}{\frac{29 + 20 + 21}{29}} \)
\( = \frac{29 - 20 + 21}{29 + 20 + 21} = \frac{30}{70} = \frac{3}{7} = \text{R.H.S.} \)
Hence verified.
In simple words: Convert the mixed fraction to an improper fraction to get the values of sine and cosecant. Find cosine using Pythagoras' theorem, then substitute these into the expression to simplify.

Exam Tip: Double-check your calculations when evaluating multi-term rational expressions to avoid minor addition or subtraction mistakes.

Question 32. If \( b\tan\theta = a \), evaluate \( \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \).
Answer: We are given:
\( b\tan\theta = a \)
\( \implies \tan\theta = \frac{a}{b} \)
Consider the given trigonometric expression:
\( \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \)
Dividing both the numerator and denominator by \( \cos\theta \):
\( = \frac{1 + \frac{\sin\theta}{\cos\theta}}{1 - \frac{\sin\theta}{\cos\theta}} \)
\( = \frac{1 + \tan\theta}{1 - \tan\theta} \)
Substituting \( \tan\theta = \frac{a}{b} \) into this expression:
\( = \frac{1 + \frac{a}{b}}{1 - \frac{a}{b}} = \frac{\frac{b + a}{b}}{\frac{b - a}{b}} = \frac{b + a}{b - a} \)
In simple words: Divide the numerator and denominator by cosine to write the expression using tangent. Substitute the given tangent value to find the final algebraic answer.

Exam Tip: Expressing expressions in terms of tangent avoids calculating the hypotenuse, which significantly reduces the algebra required.

Question 33. If \( a\cot\theta = b \), prove that \( \frac{a\sin\theta - b\cos\theta}{a\sin\theta + b\cos\theta} = \frac{a^2 - b^2}{a^2 + b^2} \).
Answer: We are given:
\( a\cot\theta = b \)
\( \implies \cot\theta = \frac{b}{a} \)
We can find the tangent ratio by taking the reciprocal:
\( \tan\theta = \frac{1}{\cot\theta} = \frac{a}{b} \)
Now, let us consider the left-hand side (L.H.S.) of the identity:
\( \text{L.H.S.} = \frac{a\sin\theta - b\cos\theta}{a\sin\theta + b\cos\theta} \)
Dividing both the numerator and the denominator by \( \cos\theta \):
\( = \frac{a\left(\frac{\sin\theta}{\cos\theta}\right) - b}{a\left(\frac{\sin\theta}{\cos\theta}\right) + b} \)
\( = \frac{a\tan\theta - b}{a\tan\theta + b} \)
Substituting \( \tan\theta = \frac{a}{b} \):
\( = \frac{a\left(\frac{a}{b}\right) - b}{a\left(\frac{a}{b}\right) + b} = \frac{\frac{a^2}{b} - b}{\frac{a^2}{b} + b} = \frac{\frac{a^2 - b^2}{b}}{\frac{a^2 + b^2}{b}} = \frac{a^2 - b^2}{a^2 + b^2} = \text{R.H.S.} \)
Hence, the expression is proven.
In simple words: Divide both parts of the fraction by cosine to rewrite it using tangent. Substitute the value of tangent and simplify.

Exam Tip: This method works beautifully for any fractional expression with linear terms of sine and cosine of the same degree.

Question 34. If \( \cot\theta = \sqrt{7} \), show that \( \frac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta} = \frac{3}{4} \).
Answer: We are given:
\( \cot\theta = \sqrt{7} \)
\( \implies \frac{\cos\theta}{\sin\theta} = \frac{\sqrt{7}}{1} \)
Since cotangent is the ratio of base to perpendicular:
\( \text{Base} = \sqrt{7} \) and \( \text{Perpendicular} = 1 \)
We find the hypotenuse using Pythagoras' theorem:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{1^2 + (\sqrt{7})^2} = \sqrt{1 + 7} = \sqrt{8} = 2\sqrt{2} \)
Evaluating cosecant and secant:
\( \csc\theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{2\sqrt{2}}{1} \)
\( \sec\theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{2\sqrt{2}}{\sqrt{7}} \)
Now, we substitute these values into the L.H.S. of the identity:
\( \text{L.H.S.} = \frac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta} \)
\( = \frac{(2\sqrt{2})^2 - \left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2}{(2\sqrt{2})^2 + \left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2} \)
\( = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} \)
\( = \frac{\frac{56 - 8}{7}}{\frac{56 + 8}{7}} = \frac{48}{64} = \frac{3}{4} = \text{R.H.S.} \)
Hence proved.
In simple words: Use Pythagoras' theorem to find the hypotenuse from cotangent. Evaluate cosecant and secant, square them, and substitute them into the formula to prove the relation.

Exam Tip: When dealing with square roots in rational terms, keep your steps clear to make sure no squaring operations are missed.

Question 35. If \( 12\csc\theta = 13 \), find the value of \( \frac{\sin^2\theta - \cos^2\theta}{2\sin\theta\cos\theta} \times \frac{1}{\tan^2\theta} \).
Answer: We are given:
\( 12\csc\theta = 13 \)
\( \implies \csc\theta = \frac{13}{12} \)
Since sine is the reciprocal of cosecant:
\( \sin\theta = \frac{12}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
This gives: \( \text{Perpendicular} = 12 \) and \( \text{Hypotenuse} = 13 \).
We calculate the base using the Pythagorean theorem:
\( \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \)
Now, we write down the cosine and tangent values:
\( \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{5}{13} \)
\( \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{12}{5} \)
Substituting these values into the given expression:
\( \text{Expression} = \frac{\sin^2\theta - \cos^2\theta}{2\sin\theta\cos\theta} \times \frac{1}{\tan^2\theta} \)
\( = \frac{\left(\frac{12}{13}\right)^2 - \left(\frac{5}{13}\right)^2}{2\left(\frac{12}{13}\right)\left(\frac{5}{13}\right)} \times \frac{1}{\left(\frac{12}{5}\right)^2} \)
\( = \frac{\frac{144}{169} - \frac{25}{169}}{\frac{120}{169}} \times \frac{25}{144} \)
\( = \frac{\frac{119}{169}}{\frac{120}{169}} \times \frac{25}{144} \)
\( = \frac{119}{120} \times \frac{25}{144} = \frac{595}{3456} \)
In simple words: Find the base of the triangle using the cosecant value. Calculate sine, cosine, and tangent, and substitute these into the equation to compute the final fraction.

Exam Tip: Be cautious when simplifying fractions with large numbers. Always double-check your arithmetic steps.

Question 36. If \( \cot\theta = \frac{13}{12} \), evaluate \( \frac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta} \).
Answer: We are given:
\( \cot\theta = \frac{13}{12} = \frac{\text{Base}}{\text{Perpendicular}} \)
Thus, \( \text{Base} = 13 \) and \( \text{Perpendicular} = 12 \).
Using the Pythagorean relation:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{12^2 + 13^2} = \sqrt{144 + 169} = \sqrt{313} \)
This gives:
\( \sin\theta = \frac{12}{\sqrt{313}} \) and \( \cos\theta = \frac{13}{\sqrt{313}} \)
Substituting these values into the expression:
\( \frac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta} = \frac{2\left(\frac{12}{\sqrt{313}}\right)\left(\frac{13}{\sqrt{313}}\right)}{\left(\frac{13}{\sqrt{313}}\right)^2 - \left(\frac{12}{\sqrt{313}}\right)^2} \)
\( = \frac{\frac{312}{313}}{\frac{169}{313} - \frac{144}{313}} = \frac{\frac{312}{313}}{\frac{25}{313}} = \frac{312}{25} \)
In simple words: Compute the hypotenuse first using Pythagoras' theorem. Write sine and cosine as fractions, then substitute them into the formula to evaluate the answer.

Exam Tip: Keep the radical in the hypotenuse \( \sqrt{313} \); it will naturally cancel out when squaring the terms.

Question 37. If \( \sec A = \frac{5}{4} \), show that \( \frac{3\sin A - 4\sin^3 A}{4\cos^3 A - 3\cos A} = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \).
Answer: We are given:
\( \sec A = \frac{5}{4} \)
\( \implies \cos A = \frac{4}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \)
Thus, \( \text{Base} = 4 \) and \( \text{Hypotenuse} = 5 \).
Using the Pythagorean relation, we find the perpendicular:
\( \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} \)
\( = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \)
This gives:
\( \sin A = \frac{3}{5} \) and \( \tan A = \frac{3}{4} \)
Now, let us evaluate both sides of the identity:
Left-Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac{3\sin A - 4\sin^3 A}{4\cos^3 A - 3\cos A} \)
\( = \frac{3\left(\frac{3}{5}\right) - 4\left(\frac{3}{5}\right)^3}{4\left(\frac{4}{5}\right)^3 - 3\left(\frac{4}{5}\right)} \)
\( = \frac{\frac{9}{5} - 4\left(\frac{27}{125}\right)}{4\left(\frac{64}{125}\right) - \frac{12}{5}} \)
\( = \frac{\frac{9}{5} - \frac{108}{125}}{\frac{256}{125} - \frac{12}{5}} \)
\( = \frac{\frac{225 - 108}{125}}{\frac{256 - 300}{125}} = \frac{117}{-44} = -\frac{117}{44} \)
Right-Hand Side (R.H.S.):
\( \text{R.H.S.} = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \)
\( = \frac{3\left(\frac{3}{4}\right) - \left(\frac{3}{4}\right)^3}{1 - 3\left(\frac{3}{4}\right)^2} \)
\( = \frac{\frac{9}{4} - \frac{27}{64}}{1 - 3\left(\frac{9}{16}\right)} \)
\( = \frac{\frac{9}{4} - \frac{27}{64}}{1 - \frac{27}{16}} \)
\( = \frac{\frac{144 - 27}{64}}{\frac{16 - 27}{16}} = \frac{\frac{117}{64}}{\frac{-11}{16}} = \frac{117}{64} \times \frac{16}{-11} = -\frac{117}{44} \)
Since L.H.S. = R.H.S., the identity is verified.
In simple words: Find sine, cosine, and tangent using Pythagoras' theorem from the given secant. Solve both sides of the equation separately to show they produce the same fraction.

Exam Tip: Work out both LHS and RHS separately and label them clearly during your exams to make the evaluation easy for the examiner to follow.

Question 38. If \( \sin\theta = \frac{3}{4} \), prove that \( \sqrt{\frac{\csc^2\theta - \cot^2\theta}{\sec^2\theta - 1}} = \frac{\sqrt{7}}{3} \).
Answer: We are given:
\( \sin\theta = \frac{3}{4} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
So, \( \text{Perpendicular} = 3 \) and \( \text{Hypotenuse} = 4 \).
We find the base of the triangle using Pythagoras' theorem:
\( \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} \)
\( = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7} \)
Now we find cosecant, cotangent, and secant:
\( \csc\theta = \frac{4}{3} \)
\( \cot\theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{\sqrt{7}}{3} \)
\( \sec\theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{4}{\sqrt{7}} \)
Substituting these values into the left-hand side (L.H.S.):
\( \text{L.H.S.} = \sqrt{\frac{\csc^2\theta - \cot^2\theta}{\sec^2\theta - 1}} \)
\( = \sqrt{\frac{\left(\frac{4}{3}\right)^2 - \left(\frac{\sqrt{7}}{3}\right)^2}{\left(\frac{4}{\sqrt{7}}\right)^2 - 1}} \)
\( = \sqrt{\frac{\frac{16}{9} - \frac{7}{9}}{\frac{16}{7} - 1}} \)
\( = \sqrt{\frac{\frac{9}{9}}{\frac{9}{7}}} = \sqrt{\frac{1}{\frac{9}{7}}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3} = \text{R.H.S.} \)
Hence, the identity is proved.
In simple words: Find the base of the triangle using the sine ratio. Get cosecant, cotangent, and secant values, plug them into the root expression, and simplify to get the final answer.

Exam Tip: Familiarize yourself with standard identities like \( \csc^2\theta - \cot^2\theta = 1 \) to verify your calculated steps quickly.

Question 39. If \( \sec A = \frac{17}{8} \), prove that \( \frac{3 - 4\sin^2 A}{4\cos^2 A - 3} = \frac{3 - \tan^2 A}{1 - 3\tan^2 A} \).
Answer: We are given:
\( \sec A = \frac{17}{8} \)
\( \implies \cos A = \frac{8}{17} = \frac{\text{Base}}{\text{Hypotenuse}} \)
Thus, \( \text{Base} = 8 \) and \( \text{Hypotenuse} = 17 \).
Using the Pythagorean theorem, the perpendicular is:
\( \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} \)
\( = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \)
Now, we find sine and tangent:
\( \sin A = \frac{15}{17} \) and \( \tan A = \frac{15}{8} \)
Let us substitute these values into the equation:
Left-Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac{3 - 4\sin^2 A}{4\cos^2 A - 3} \)
\( = \frac{3 - 4\left(\frac{15}{17}\right)^2}{4\left(\frac{8}{17}\right)^2 - 3} \)
\( = \frac{3 - 4\left(\frac{225}{289}\right)}{4\left(\frac{64}{289}\right) - 3} \)
\( = \frac{3 - \frac{900}{289}}{\frac{256}{289} - 3} \)
\( = \frac{\frac{867 - 900}{289}}{\frac{256 - 867}{289}} = \frac{-33}{-611} = \frac{33}{611} \)
Right-Hand Side (R.H.S.):
\( \text{R.H.S.} = \frac{3 - \tan^2 A}{1 - 3\tan^2 A} \)
\( = \frac{3 - \left(\frac{15}{8}\right)^2}{1 - 3\left(\frac{15}{8}\right)^2} \)
\( = \frac{3 - \frac{225}{64}}{1 - 3\left(\frac{225}{64}\right)} \)
\( = \frac{3 - \frac{225}{64}}{1 - \frac{675}{64}} \)
\( = \frac{\frac{192 - 225}{64}}{\frac{64 - 675}{64}} = \frac{-33}{-611} = \frac{33}{611} \)
Since L.H.S. = R.H.S., the equation is proven.
In simple words: Find the perpendicular first. Put the values of sine, cosine, and tangent into both sides of the equation to show they both equal 33/611.

Exam Tip: Keeping the common denominators instead of evaluating them as decimals avoids mistakes with rounded figures.

Question 40. If \( 3\tan\theta = 4 \), prove that \( \sqrt{\frac{\sec\theta - \csc\theta}{\sec\theta + \csc\theta}} = \frac{1}{\sqrt{7}} \).
Answer: We are given:
\( 3\tan\theta = 4 \)
\( \implies \tan\theta = \frac{4}{3} = \frac{\text{Perpendicular}}{\text{Base}} \)
Thus, \( \text{Perpendicular} = 4 \) and \( \text{Base} = 3 \).
We calculate the hypotenuse using Pythagoras' theorem:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} \)
\( = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
Now we find secant and cosecant:
\( \sec\theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{5}{3} \)
\( \csc\theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{5}{4} \)
Substituting these values into the left-hand side (L.H.S.):
\( \text{L.H.S.} = \sqrt{\frac{\sec\theta - \csc\theta}{\sec\theta + \csc\theta}} \)
\( = \sqrt{\frac{\frac{5}{3} - \frac{5}{4}}{\frac{5}{3} + \frac{5}{4}}} \)
\( = \sqrt{\frac{\frac{20 - 15}{12}}{\frac{20 + 15}{12}}} = \sqrt{\frac{5}{35}} = \sqrt{\frac{1}{7}} = \frac{1}{\sqrt{7}} = \text{R.H.S.} \)
Hence, the relation is verified.
In simple words: Find the hypotenuse to calculate secant and cosecant. Put these into the expression under the square root, and simplify to get 1 over the square root of 7.

Exam Tip: Since the question asks you to prove that the expression is equal to \( \frac{1}{\sqrt{7}} \), you do not need to rationalize the final denominator.

Question 41. If \( \tan\theta = \frac{m}{n} \), show that \( \frac{m\sin\theta - n\cos\theta}{m\sin\theta + n\cos\theta} = \frac{m^2 - n^2}{m^2 + n^2} \).
Answer: We are given:
\( \tan\theta = \frac{m}{n} = \frac{\text{Perpendicular}}{\text{Base}} \)
Thus, \( \text{Perpendicular} = m \) and \( \text{Base} = n \).
Using the Pythagorean relation:
\( \text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2} = \sqrt{m^2 + n^2} \)
Now we find sine and cosine:
\( \sin\theta = \frac{m}{\sqrt{m^2 + n^2}} \) and \( \cos\theta = \frac{n}{\sqrt{m^2 + n^2}} \)
Substituting these values into the left-hand side (L.H.S.):
\( \text{L.H.S.} = \frac{m\sin\theta - n\cos\theta}{m\sin\theta + n\cos\theta} \)
\( = \frac{m\left(\frac{m}{\sqrt{m^2+n^2}}\right) - n\left(\frac{n}{\sqrt{m^2+n^2}}\right)}{m\left(\frac{m}{\sqrt{m^2+n^2}}\right) + n\left(\frac{n}{\sqrt{m^2+n^2}}\right)} \)
\( = \frac{\frac{m^2 - n^2}{\sqrt{m^2+n^2}}}{\frac{m^2 + n^2}{\sqrt{m^2+n^2}}} = \frac{m^2 - n^2}{m^2 + n^2} = \text{R.H.S.} \)
Hence verified.
In simple words: Find the hypotenuse, then calculate sine and cosine. Plug these into the expression and cancel out the common square-root denominators to get the final algebraic identity.

Exam Tip: An elegant alternative is to divide the numerator and denominator of the LHS by \( \cos\theta \) to express the identity entirely in terms of \( \tan\theta \), avoiding the use of radicals.