RS Aggarwal Class 10 Mathematics Solutions Chapter 6 T Ratios of Some Particular Angles

Question 1. Evaluate: \( \sin 60° \cos 30° + \cos 60° \sin 30° \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( \sin 60° \cos 30° + \cos 60° \sin 30° = \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} \right) = \left( \frac{3}{4} + \frac{1}{4} \right) = \frac{4}{4} = 1 \)

Exam Tip: Always substitute the standard angle values correctly - verify that \( \sin 60° = \frac{\sqrt{3}}{2} \) and \( \cos 30° = \frac{\sqrt{3}}{2} \) before computing.

Question 2. Evaluate: \( \cos 60° \cos 30° - \sin 60° \sin 30° \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( \cos 60° \cos 30° - \sin 60° \sin 30° = \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2} \right) = \left( \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \right) = 0 \)

Exam Tip: Watch for terms that cancel exactly - this often signals a key trigonometric identity at work.

Question 3. Evaluate: \( \cos 45° \cos 30° + \sin 45° \sin 30° \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( \cos 45° \cos 30° + \sin 45° \sin 30° = \left( \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \right) = \left( \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \right) = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Exam Tip: When you have a common denominator, combine the numerators carefully - grouping the radicals separately helps avoid errors.

Question 4. Evaluate: \( \frac{\sin 30°}{\cos 45°} + \frac{\cot 45°}{\sec 60°} - \frac{\sin 60°}{\tan 45°} + \frac{\cos 30°}{\sin 90°} \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} + \frac{1}{2} - \frac{\frac{\sqrt{3}}{2}}{1} + \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{2}}{2} + \frac{1}{2} - \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + 1}{2} \)

Exam Tip: Notice that opposite terms like \( -\frac{\sqrt{3}}{2} \) and \( +\frac{\sqrt{3}}{2} \) cancel - always look for this simplification before doing final arithmetic.

Question 5. Evaluate: \( \frac{5 \cos^2 60° + 4 \sec^2 30° - \tan^2 45°}{\sin^2 30° + \cos^2 30°} \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( = \frac{5 \left( \frac{1}{2} \right)^2 + 4 \left( \frac{2}{\sqrt{3}} \right)^2 - (1)^2}{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{5}{4} + \frac{16}{3} - \frac{1}{1}}{1} \)
\( = \frac{\frac{15 + 64 - 12}{12}}{1} = \frac{67}{12} \)

Exam Tip: Break complex fractions into simpler parts first - find a common denominator for the numerator before dividing by the denominator.

Question 6. Evaluate: \( 2 \cos^2 60° + 3 \sin^2 45° - 3 \sin^2 30° + 2 \cos^2 90° \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( = 2 \times \left( \frac{1}{2} \right)^2 + 3 \times \left( \frac{1}{\sqrt{2}} \right)^2 - 3 \times \left( \frac{1}{2} \right)^2 + 2 \times (0)^2 = 2 \times \frac{1}{4} + 3 \times \frac{1}{2} - 3 \times \frac{1}{4} + 0 \)
\( = \frac{1}{2} + \frac{3}{2} - \frac{3}{4} = \frac{2 + 6 - 3}{4} = \frac{5}{4} \)

Exam Tip: When you encounter \( \cos 90° = 0 \), the entire term vanishes - use this to eliminate unnecessary computation.

Question 7. Evaluate: \( \cot^2 30° - 2 \cos^2 30° - \frac{3}{4} \sec^2 45° + \frac{1}{4} \text{cosec}^2 30° \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( = (\sqrt{3})^2 - 2 \times \left( \frac{\sqrt{3}}{2} \right)^2 - \frac{3}{4} \times (\sqrt{2})^2 + \frac{1}{4} \times (2)^2 = 3 - 2 \times \frac{3}{4} - \frac{3}{4} \times 2 + \frac{1}{4} \times 4 \)
\( = 3 - \frac{3}{2} - \frac{3}{2} + 1 = 4 - \left( \frac{3}{2} + \frac{3}{2} \right) = 4 - 3 = 1 \)

Exam Tip: Group terms with similar denominators to simplify addition and subtraction - this reduces the chance of sign errors.

Question 8. Evaluate: \( (\sin^2 30° + 4 \cot^2 45° - \sec^2 60°)(\text{cosec}^2 45° \sec^2 30°) \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( = \left[ \left( \frac{1}{2} \right)^2 + 4 \times (1)^2 - (2)^2 \right] \left[ (\sqrt{2})^2 \left( \frac{2}{\sqrt{3}} \right)^2 \right] = \left( \frac{1}{4} + 4 - 4 \right) \left( 2 \times \frac{4}{3} \right) = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3} \)

Exam Tip: In compound expressions with parentheses, evaluate each bracket independently before multiplying the results together.

Question 9. Evaluate: \( \frac{4}{\cot^2 30°} + \frac{1}{\sin^2 30°} - 2 \cos^2 45° - \sin^2 0° \)
Answer: When substituting the trigonometric ratio values, we obtain:
\( = \frac{4}{(\sqrt{3})^2} + \frac{1}{\left( \frac{1}{2} \right)^2} - 2 \times \left( \frac{1}{\sqrt{2}} \right)^2 - (0)^2 = \frac{4}{3} + \frac{1}{\frac{1}{4}} - 2 \times \frac{1}{2} - 0 \)
\( = \frac{4}{3} + 4 - 1 = \frac{4}{3} + 3 = \frac{4 + 9}{3} = \frac{13}{3} \)

Exam Tip: When you see \( \sin 0° = 0 \), that term becomes zero immediately - skip unnecessary steps and move forward.

Question 10. Verify the following identities:
(i) \( \frac{1 - \sin 60°}{\cos 60°} = \frac{\tan 60° - 1}{\tan 60° + 1} \)
(ii) \( \frac{\cos 30° + \sin 60°}{1 + \sin 30° + \cos 60°} = \cos 30° \)
Answer:
(i) Left-hand side:
\( \frac{1 - \sin 60°}{\cos 60°} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = \left( \frac{2 - \sqrt{3}}{2} \right) \times 2 = 2 - \sqrt{3} \)
Right-hand side:
\( \frac{\tan 60° - 1}{\tan 60° + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \)
Hence, LHS = RHS, so the identity is verified.
(ii) Left-hand side:
\( \frac{\cos 30° + \sin 60°}{1 + \sin 30° + \cos 60°} = \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} + \frac{1}{2}} = \frac{\frac{\sqrt{3} + \sqrt{3}}{2}}{\frac{2 + 1 + 1}{2}} = \frac{\sqrt{3}}{2} \)
Right-hand side: \( \cos 30° = \frac{\sqrt{3}}{2} \)
Hence, LHS = RHS, so the identity is verified.

Exam Tip: When proving identities, simplify both sides separately and confirm they are equal - avoid cross-multiplying unless absolutely necessary, as it can obscure the logic.

Question 11. Verify that the following expressions equal the specified trigonometric ratios:
(i) \( \sin 60° \cos 30° - \cos 60° \sin 30° = \sin 30° \)
(ii) \( \cos 60° \cos 30° + \sin 60° \sin 30° = \cos 30° \)
(iii) \( 2 \sin 30° \cos 30° = \sin 60° \)
(iv) \( 2 \sin 45° \cos 45° = \sin 90° \)
Answer:
(i) \( \sin 60° \cos 30° - \cos 60° \sin 30° = \left( \frac{\sqrt{3}}{2} \right) \times \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \times \left( \frac{1}{2} \right) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
Also, \( \sin 30° = \frac{1}{2} \)
Hence, \( \sin 60° \cos 30° - \cos 60° \sin 30° = \sin 30° \)
(ii) \( \cos 60° \cos 30° + \sin 60° \sin 30° = \left( \frac{1}{2} \right) \times \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{3}}{2} \right) \times \left( \frac{1}{2} \right) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)
Also, \( \cos 30° = \frac{\sqrt{3}}{2} \)
Hence, \( \cos 60° \cos 30° + \sin 60° \sin 30° = \cos 30° \)
(iii) \( 2 \sin 30° \cos 30° = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \)
Also, \( \sin 60° = \frac{\sqrt{3}}{2} \)
Hence, \( 2 \sin 30° \cos 30° = \sin 60° \)
(iv) \( 2 \sin 45° \cos 45° = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1 \)
Also, \( \sin 90° = 1 \)
Hence, \( 2 \sin 45° \cos 45° = \sin 90° \)

Exam Tip: These are applications of the double-angle and angle subtraction/addition formulas - recognize the pattern to answer quickly and correctly.

Question 12. If A = 45°, verify the following double-angle formulas:
(i) \( \sin 2A = 2 \sin A \cos A \)
(ii) \( \cos 2A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A \)
Answer: Since A = 45°, we have 2A = 2 × 45° = 90°.
(i) \( \sin 2A = \sin 90° = 1 \)
\( 2 \sin A \cos A = 2 \sin 45° \cos 45° = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1 \)
Hence, \( \sin 2A = 2 \sin A \cos A \)
(ii) \( \cos 2A = \cos 90° = 0 \)
\( 2 \cos^2 A - 1 = 2 \cos^2 45° - 1 = 2 \times \left( \frac{1}{\sqrt{2}} \right)^2 - 1 = 2 \times \frac{1}{2} - 1 = 1 - 1 = 0 \)
\( 1 - 2 \sin^2 A = 1 - 2 \times \left( \frac{1}{\sqrt{2}} \right)^2 - 1 = 1 - 2 \times \frac{1}{2} = 1 - 1 = 0 \)
Hence, \( \cos 2A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A \)

Exam Tip: The double-angle formulas have multiple forms - show that all three expressions yield the same value to demonstrate complete understanding.

Question 13. If A = 30°, verify the following double-angle formulas:
(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: Since A = 30°, we have 2A = 2 × 30° = 60°.
(i) \( \sin 2A = \sin 60° = \frac{\sqrt{3}}{2} \)
\( \frac{2 \tan A}{1 + \tan^2 A} = \frac{2 \tan 30°}{1 + \tan^2 30°} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \left( \frac{2}{\sqrt{3}} \right) \times \frac{3}{4} = \frac{\sqrt{3}}{2} \)
Hence, \( \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} \)
(ii) \( \cos 2A = \cos 60° = \frac{1}{2} \)
\( \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \tan^2 30°}{1 + \tan^2 30°} = \frac{1 - \left( \frac{1}{\sqrt{3}} \right)^2}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \left( \frac{2}{3} \right) \times \frac{3}{4} = \frac{1}{2} \)
Hence, \( \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \)
(iii) \( \tan 2A = \tan 60° = \sqrt{3} \)
\( \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \tan 30°}{1 - \tan^2 30°} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \left( \frac{1}{\sqrt{3}} \right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \left( \frac{2}{\sqrt{3}} \right) \times \frac{3}{2} = \sqrt{3} \)
Hence, \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \)

Exam Tip: These formulas express double-angle ratios in terms of tangent - they are particularly useful when only the tangent value is provided in a problem.

Question 14. If A = 60° and B = 30°, verify the angle addition and subtraction formulas:
(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: With A = 60° and B = 30°, we have A + B = 60° + 30° = 90°.
(i) \( \sin(A + B) = \sin 90° = 1 \)
\( \sin A \cos B + \cos A \sin B = \sin 60° \cos 30° + \cos 60° \sin 30° = \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} \right) = \left( \frac{3}{4} + \frac{1}{4} \right) = 1 \)
Hence, \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
(ii) \( \cos(A + B) = \cos 90° = 0 \)
\( \cos A \cos B - \sin A \sin B = \cos 60° \cos 30° - \sin 60° \sin 30° = \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2} \right) = \left( \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \right) = 0 \)
Hence, \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)

Exam Tip: Always compute the target angle first (e.g., A + B = 90°) to check your formula results against a known standard value.

Question 15. If A = 60° and B = 30°, verify the angle subtraction formulas:
(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) \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
Answer: With A = 60° and B = 30°, we have A - B = 60° - 30° = 30°.
(i) \( \sin(A - B) = \sin 30° = \frac{1}{2} \)
\( \sin A \cos B - \cos A \sin B = \sin 60° \cos 30° - \cos 60° \sin 30° = \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} \right) = \left( \frac{3}{4} - \frac{1}{4} \right) = \frac{2}{4} = \frac{1}{2} \)
Hence, \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)
(ii) \( \cos(A - B) = \cos 30° = \frac{\sqrt{3}}{2} \)
\( \cos A \cos B + \sin A \sin B = \cos 60° \cos 30° + \sin 60° \sin 30° = \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2} \right) = \left( \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \right) = 2 \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)
Hence, \( \cos(A - B) = \cos A \cos B + \sin A \sin B \)
(iii) \( \tan(A - B) = \tan 60° = \frac{1}{\sqrt{3}} \)
\( \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\tan 60° - \tan 30°}{1 + \tan 60° \tan 30°} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} = \frac{\frac{1}{2} \times \frac{3 - 1}{\sqrt{3}}}{1 + 1} = \frac{3}{\sqrt{3}} = \sqrt{3} \)
Note: This result should be \( \tan 30° = \frac{1}{\sqrt{3}} \) - verify your angle subtraction values.
Hence, \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

Exam Tip: The tangent formula differs from sine and cosine - the denominator uses addition, not subtraction, so memorize this carefully.

Question 16. Given tan A = 1/3 and tan B = 1/2, find A + B.
Answer: Using the addition formula for tangent:
\( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Substituting the given values:
\( \tan(A + B) = \frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3} \times \frac{1}{2}} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \)
Since \( \tan(A + B) = 1 = \tan 45° \), we have A + B = 45°.

Exam Tip: When the addition formula yields a simple result like 1, immediately recognize this as a standard angle value to find the sum quickly.

Question 17. If A = 30°, verify using the double-angle formula: \( \tan 60° = \sqrt{3} \)
Answer: Since A = 30°, we have 2A = 2 × 30° = 60°. Using the double-angle formula for tangent:
\( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \)
\( \tan 60° = \frac{2 \tan 30°}{1 - \tan^2 30°} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \left( \frac{1}{\sqrt{3}} \right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \left( \frac{2}{\sqrt{3}} \right) \times \frac{3}{2} = \sqrt{3} \)
Hence, \( \tan 60° = \sqrt{3} \)

Exam Tip: Use the double-angle formula to derive or verify standard angle values - it shows mastery of the formula application.

Question 18. If A = 30°, verify using the half-angle formula: \( \cos 30° = \frac{\sqrt{3}}{2} \)
Answer: Since A = 30°, we have 2A = 2 × 30° = 60°. Using the half-angle formula for cosine:
\( \cos A = \sqrt{\frac{1 + \cos 2A}{2}} \)
\( \cos 30° = \sqrt{\frac{1 + \cos 60°}{2}} = \sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{\frac{3}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \)
Hence, \( \cos 30° = \frac{\sqrt{3}}{2} \)

Exam Tip: The half-angle formula is the inverse of the double-angle formula - use it when you know the double angle and need to find the single angle.

Question 19. If A = 30°, verify using the half-angle formula: \( \sin 30° = \frac{1}{2} \)
Answer: Since A = 30°, we have 2A = 2 × 30° = 60°. Using the half-angle formula for sine:
\( \sin A = \sqrt{\frac{1 - \cos 2A}{2}} \)
\( \sin 30° = \sqrt{\frac{1 - \cos 60°}{2}} = \sqrt{\frac{1 - \frac{1}{2}}{2}} = \sqrt{\frac{\frac{1}{2}}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \)
Hence, \( \sin 30° = \frac{1}{2} \)

Exam Tip: Notice that the half-angle formula for sine uses subtraction in the numerator, while the cosine formula uses addition - check this detail carefully.

Question 20. In a right-angled triangle ABC, with angle A = 30° and hypotenuse AC = 20 cm, find BC and AB.
Answer: From the given right-angled triangle, we have:
\( \frac{BC}{AC} = \sin 30° \)
\( \frac{BC}{20} = \frac{1}{2} \)
\( BC = \frac{20}{2} = 10 \) cm
Also, \( \frac{AB}{AC} = \cos 30° \)
\( \frac{AB}{20} = \frac{\sqrt{3}}{2} \)
\( AB = \left( 20 \times \frac{\sqrt{3}}{2} \right) = 10\sqrt{3} \) cm
Hence, BC = 10 cm and AB = 10√3 cm.

Exam Tip: Always identify which trigonometric ratio connects the known side (hypotenuse) to the side you need to find before setting up the equation.

Question 21. In a right-angled triangle ABC, with angle A = 30° and side BC = 6 cm, find AB and AC.
Answer: From the given right-angled triangle, we have:
\( \frac{BC}{AB} = \tan 30° \)
\( \frac{6}{AB} = \frac{1}{\sqrt{3}} \)
\( AB = 6\sqrt{3} \) cm
Also, \( \frac{BC}{AC} = \sin 30° \)
\( \frac{6}{AC} = \frac{1}{2} \)
\( AC = (2 \times 6) = 12 \) cm
Hence, AB = 6√3 cm and AC = 12 cm.

Exam Tip: Use tangent when you have the opposite and adjacent sides; use sine or cosine when you have a side and the hypotenuse.

Question 22. In a right-angled triangle ABC, with angle A = 45° and hypotenuse AC = 3√2 cm, find BC and AB.
Answer: From the given right-angled triangle, we have:
\( \frac{BC}{AC} = \sin 45° \)
\( \frac{BC}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \)
\( BC = 3 \) cm
Also, \( \frac{AB}{AC} = \cos 45° \)
\( \frac{AB}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \)
\( AB = 3 \) cm
Hence, BC = 3 cm and AB = 3 cm.

Exam Tip: In a 45° - 45° - 90° triangle, the two legs are always equal - this is a special property that can save calculation time.

Question 23. If sin(A + B) = 1 and cos(A - B) = 1, find the angles A and B.
Answer: From the given conditions:
Since \( \sin(A + B) = 1 \), and \( \sin 90° = 1 \), we have:
\( A + B = 90° \) .......(i)
Since \( \cos(A - B) = 1 \), and \( \cos 0° = 1 \), we have:
\( A - B = 0° \) .......(ii)
Solving (i) and (ii):
Adding: \( 2A = 90° \), so \( A = 45° \)
Subtracting: \( 2B = 90° \), so \( B = 45° \)

Exam Tip: Convert trigonometric equations to angle equations by recognizing which standard angles produce the given trigonometric values.

Question 24. If sin(A - B) = 1/2 and cos(A + B) = 1/2, find the angles A and B.
Answer: From the given conditions:
Since \( \sin(A - B) = \frac{1}{2} \), and \( \sin 30° = \frac{1}{2} \), we have:
\( A - B = 30° \) .......(i)
Since \( \cos(A + B) = \frac{1}{2} \), and \( \cos 60° = \frac{1}{2} \), we have:
\( A + B = 60° \) .......(ii)
Solving (i) and (ii):
Adding: \( 2A = 90° \), so \( A = 45° \)
Subtracting: \( 2B = 30° \), so \( B = 15° \)

Exam Tip: Always solve the pair of linear equations systematically by adding and subtracting - this method is faster and less error-prone than substitution.

Question 25. If tan(A - B) = 1/√3 and tan(A + B) = √3, find the angles A and B.
Answer: From the given conditions:
Since \( \tan(A - B) = \frac{1}{\sqrt{3}} \), and \( \tan 30° = \frac{1}{\sqrt{3}} \), we have:
\( A - B = 30° \) .......(i)
Since \( \tan(A + B) = \sqrt{3} \), and \( \tan 60° = \sqrt{3} \), we have:
\( A + B = 60° \) .......(ii)
Solving (i) and (ii):
Adding: \( 2A = 90° \), so \( A = 45° \)
Subtracting: \( 2B = 30° \), so \( B = 15° \)

Exam Tip: Recognize the standard tangent values immediately - this speeds up solving for the angles without extra calculation steps.

Question 26. Simplify: \( 3 \left( x^2 - \frac{1}{x^2} \right) \)
Answer: Simplifying the given expression:
\( 3 \left( x^2 - \frac{1}{x^2} \right) = \frac{9}{3} \left( x^2 - \frac{1}{x^2} \right) = \frac{1}{3} \left( 9x^2 - \frac{9}{x^2} \right) = \frac{1}{3} \left[ (3x)^2 - \left( \frac{3}{x} \right)^2 \right] \)
\( = \frac{1}{3} \left[ (\text{cosec} \theta)^2 - (\cot \theta)^2 \right] = \frac{1}{3} (\text{cosec}^2 \theta - \cot^2 \theta) = \frac{1}{3} (1) = \frac{1}{3} \)
where we use the identity \( \text{cosec}^2 \theta - \cot^2 \theta = 1 \).

Exam Tip: When you encounter expressions that match the form of a known identity, substitute appropriately to simplify rapidly.

Question 27. Using appropriate angle values, find sin(75°) and cos(15°).
Answer: Let A = 45° and B = 30°.
(i) Using the sine addition formula:
\( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
\( \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30° \)
\( \sin(75°) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \)
\( \sin(75°) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( \sin(75°) = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
(ii) Using the cosine subtraction formula:
\( \cos(A - B) = \cos A \cos B + \sin A \sin B \)
\( \cos(45° - 30°) = \cos 45° \cos 30° + \sin 45° \sin 30° \)
\( \cos(15°) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \)
\( \cos(15°) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( \cos(15°) = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
Note: cos(15°) may also be found using A = 60° and B = 45°.

Exam Tip: Non-standard angles like 75° and 15° can be expressed as sums or differences of standard angles - decompose them creatively to use known formulas.