CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set 02

Question. If tan A = 4/3 , the value of sin A is
(a) 4/5
(b) 3/4
(c) 5/3
(d) 7/5
Answer : A

Question. If 3 tan A = 4, then the value of  3 sin A + cos A / 3 sin A - 2 cos A is:
(a) 4
(b) 11/15
(c) 7/15
(d) 3
Answer : D

Question. If sin θ + cos θ = cos θ, (θ􀀀􀉲 90°) then the value of tan θ is:
(a) √2 - 1
(b) √2 + 1
(c) √2
(d) – √2
Answer : A

Question. Given that sin a = √3 / 2 and cos B = 0, then the value of B - a is:
(a) 0°
(b) 90°
(c) 60°
(d) 30°
Answer : D

Question. If sin (A + B) = cos (A - H) = 1, then
(a) A = B = 0
(b) A = B = 45°
(c) A = 60°, B = 30°
(d) None of these
Answer : B

Question. If cos A = 5/13 , find the value of tan A + cot A
(a) 169/60
(b) 12/13
(c) 1
(d) 60/169
Answer : A

Question. If 5x = sec θ and 5/x = tan θ then find the value of (x² - 1/x²)
(a) 5
(b) 1/5
(c) 2/5
(d) 0
Answer : B

Question. If sin A + sin² A = 1, then the value of the expression (cos² A + cos⁴ A) is:
(a) 1
(b) 1/2
(c) 2
(d) 3 
Answer : A

Question. The value of (1 + cos θ) (1 - cos θ) cosec² θ =
(a) 0
(b) 1
(c) cos² θ
(d) sin² θ
Answer : B

Question. If sin θ – cos θ = 0, then the value of (sin⁴ θ + cos⁴ θ) is:
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4]
Answer : C

Question. If θ = 45° then sec θ cot θ – cosec θ tan θ is
(a) 0
(b) 1
(c) 2
(d) 3
Answer : A

Question. If x = a sin θ and y = a cos θ, then the value of x² + y² is
(a) a
(b) a²
(c) 1
(d) b²
Answer : B

Question. 4 tan² A – 4 sec² A is equal to:
(a) 2
(b) 3
(c) 4
(d) –4
Answer : D

Question. If 3 cos θ = 1, then cosec θ is equal to:
(a) 2√2
(b) 3/2√2
(c) 2√3/3
(d) 4/3√2
Answer : B

Question. If cosec θ – cot θ = 1/3, then the value of cosec θ + cot θ is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer : C

Question. If 2 sin 2θ = 3 , then the value of θ is:
(a) 90°
(b) 30°
(c) 60°
(d) 45°
Answer : B

Question. If the height and length of the shadow of a man are the same, then the angle of elevation of the sun is:
(a) 45°
(b) 60°
(c) 90°
(d) 120°
Answer : A

Question. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection. 
Answer : False.

Question. cos θ = a² + b² / 2ab, where a and b are two distinctbnumbers such that ab > 0. 
Answer : False.

Question. The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.
Answer : False.

Question. If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains unchanged.
Answer : True.

Fill in the Blanks

Fill in the blanks/tables with suitable information:

PRACTICE QUESTIONS
CLASS X: CHAPTER - 8
INTRODUCTION TO TRIGONOMETRY
T - RATIOS OF SOME PARTICULAR ANGLES

Evaluate each of the following:

1. sin 60° cos 30° + cos 60° sin 30°

Answer:
1. sin 60° cos 30° + cos 60° sin 30° = \( \frac{\sqrt{3}}{2} \) × \( \frac{\sqrt{3}}{2} \) + \( \frac{1}{2} \) × \( \frac{1}{2} \)
2. = \( \frac{3}{4} + \frac{1}{4} \) = 1
3. Hence, the value is 1.

Teacher's Note:
a) This is the expansion of sin(60° + 30°) = sin 90°.
b) Check: sin 90° = 1.

2. cos 60° cos 30° - sin 60° sin 30°

Answer:
1. cos 60° cos 30° - sin 60° sin 30° = \( \frac{1}{2} \) × \( \frac{\sqrt{3}}{2} \) - \( \frac{\sqrt{3}}{2} \) × \( \frac{1}{2} \)
2. = \( \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \) = 0
3. Hence, the value is 0.

Teacher's Note:
a) This is cos(60° + 30°) = cos 90°, which is 0.
b) Keep the minus sign between the two products.

3. cos 45° cos 30° + sin 45° sin 30°

Answer:
1. cos 45° cos 30° + sin 45° sin 30° = \( \frac{\sqrt{2}}{2} \) × \( \frac{\sqrt{3}}{2} \) + \( \frac{\sqrt{2}}{2} \) × \( \frac{1}{2} \)
2. = \( \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \) = \( \frac{\sqrt{6} + \sqrt{2}}{4} \)
3. Hence, the value is \( \frac{\sqrt{6} + \sqrt{2}}{4} \).

Teacher's Note:
a) This equals cos (45° - 30°) = cos 15°.
b) Do not simplify the two terms before adding them.

4. sin 60° sin 45° - cos 60° cos 45°

Answer:
1. sin 60° sin 45° - cos 60° cos 45° = \( \frac{\sqrt{3}}{2} \) × \( \frac{\sqrt{2}}{2} \) - \( \frac{1}{2} \) × \( \frac{\sqrt{2}}{2} \)
2. = \( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \) = \( \frac{\sqrt{6} - \sqrt{2}}{4} \)
3. Hence, the value is \( \frac{\sqrt{6} - \sqrt{2}}{4} \).

Teacher's Note:
a) The pattern is cos (60° + 45°) with the sign reversed.
b) Check: the value is about (2.449 - 1.414)/4 = 0.259, which matches -cos 105°.

5. \( \frac{\sin 30°}{\cos 45°} + \frac{\cot 45°}{\sec 60°} - \frac{\sin 60°}{\tan 45°} - \frac{\cos 30°}{\sin 90°} \)

Answer:
1. \( \frac{\sin 30°}{\cos 45°} \) = \( \frac{1/2}{\sqrt{2}/2} \) = \( \frac{1}{\sqrt{2}} \) = \( \frac{\sqrt{2}}{2} \)
2. \( \frac{\cot 45°}{\sec 60°} \) = \( \frac{1}{2} \)
3. \( \frac{\sin 60°}{\tan 45°} \) = \( \frac{\sqrt{3}}{2} \) and \( \frac{\cos 30°}{\sin 90°} \) = \( \frac{\sqrt{3}}{2} \)
4. Total = \( \frac{\sqrt{2}}{2} + \frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \) = \( \frac{1 + \sqrt{2} - 2\sqrt{3}}{2} \)
5. Hence, the value is \( \frac{1 + \sqrt{2} - 2\sqrt{3}}{2} \).

Teacher's Note:
a) Convert every ratio to its value first, then add and subtract.
b) Check: the value is about -0.53.

6. \( \frac{\tan^2 60° + 4\cos^2 45° + 3\,\mathrm{cosec}^2 60° + 2\cos^2 90°}{2\,\mathrm{cosec}\, 30° + 3\sec 60° - \frac{7}{3}\cot^2 30°} \)

Answer:
1. Numerator = 3 + 4 × \( \frac{1}{2} \) + 3 × \( \frac{4}{3} \) + 2 × 0 = 3 + 2 + 4 + 0 = 9.
2. Denominator = 2 × 2 + 3 × 2 - \( \frac{7}{3} \) × 3 = 4 + 6 - 7 = 3.
3. Value = \( \frac{9}{3} \) = 3.
4. Hence, the value is 3.

Teacher's Note:
a) Write cosec 30° = 2 and sec 60° = 2 before substituting.
b) Check: cot² 30° = 3, so \( \frac{7}{3} \) × 3 = 7.

7. 4(sin⁴ 30° + cos⁴ 60°) - 3(cos² 45° - sin² 90°) + 5cos² 90°

Answer:
1. sin⁴ 30° = \( \frac{1}{16} \) and cos⁴ 60° = \( \frac{1}{16} \), so 4(\( \frac{1}{16} + \frac{1}{16} \)) = \( \frac{1}{2} \).
2. cos² 45° - sin² 90° = \( \frac{1}{2} \) - 1 = \( -\frac{1}{2} \), so -3 × (\( -\frac{1}{2} \)) = \( \frac{3}{2} \).
3. 5 cos² 90° = 0.
4. Total = \( \frac{1}{2} + \frac{3}{2} + 0 \) = 2.
5. Hence, the value is 2.

Teacher's Note:
a) Square the sine and cosine values before multiplying.
b) Check: the first bracket gives \( \frac{1}{8} \), and 4 × \( \frac{1}{8} \) = \( \frac{1}{2} \).

8. \( \frac{4}{\cot^2 30°} + \frac{1}{\sin^2 30°} - 2\cos^2 45° - \sin^2 0° \)

Answer:
1. cot² 30° = 3, so \( \frac{4}{3} \).
2. sin² 30° = \( \frac{1}{4} \), so \( \frac{1}{\sin^2 30°} \) = 4.
3. 2cos² 45° = 2 × \( \frac{1}{2} \) = 1 and sin² 0° = 0.
4. Total = \( \frac{4}{3} + 4 - 1 - 0 \) = \( \frac{4}{3} + 3 \) = \( \frac{13}{3} \).
5. Hence, the value is \( \frac{13}{3} \).

Teacher's Note:
a) Use the reciprocal identity 1/sin² = cosec² to speed up the working.
b) Check: 4/3 + 3 = 13/3.

9. \( \frac{1}{\cos^2 30°} + \frac{1}{\sin^3 30°} - \frac{1}{2}\tan^2 45° - 8\sin^2 90° \)

Answer:
1. \( \frac{1}{\cos^2 30°} \) = \( \frac{1}{3/4} \) = \( \frac{4}{3} \).
2. \( \frac{1}{\sin^3 30°} \) = \( \frac{1}{1/8} \) = 8.
3. \( \frac{1}{2}\tan^2 45° \) = \( \frac{1}{2} \) and 8 sin² 90° = 8.
4. Total = \( \frac{4}{3} + 8 - \frac{1}{2} - 8 \) = \( \frac{4}{3} - \frac{1}{2} \) = \( \frac{5}{6} \).
5. Hence, the value is \( \frac{5}{6} \).

Teacher's Note:
a) Square the sine of 30° first, then take the cube of that value.
b) Check: \( \frac{4}{3} - \frac{1}{2} = \frac{8 - 3}{6} = \frac{5}{6} \).

10. cot² 30° - 2cos² 30° - \( \frac{3}{4} \)sec² 45° + \( \frac{1}{4} \)cosec² 30°

Answer:
1. cot² 30° = 3, so the first term is 3.
2. 2cos² 30° = 2 × \( \frac{3}{4} \) = \( \frac{3}{2} \).
3. \( \frac{3}{4} \)sec² 45° = \( \frac{3}{4} \) × 2 = \( \frac{3}{2} \).
4. \( \frac{1}{4} \)cosec² 30° = \( \frac{1}{4} \) × 4 = 1.
5. Total = 3 - \( \frac{3}{2} \) - \( \frac{3}{2} \) + 1 = 1.
6. Hence, the value is 1.

Teacher's Note:
a) Replace each squared ratio by its value before combining.
b) Check: 4 - 3 = 1.

11. (sin² 30° + 4cot² 45° - sec² 60°)(cosec² 45° sec² 30°)

Answer:
1. sin² 30° + 4cot² 45° - sec² 60° = \( \frac{1}{4} \) + 4 - 4 = \( \frac{1}{4} \).
2. cosec² 45° sec² 30° = 2 × \( \frac{4}{3} \) = \( \frac{8}{3} \).
3. Product = \( \frac{1}{4} \) × \( \frac{8}{3} \) = \( \frac{2}{3} \).
4. Hence, the value is \( \frac{2}{3} \).

Teacher's Note:
a) The two brackets are multiplied, as the brackets show.
b) Check: sec² 60° = 4 and cot² 45° = 1.

12. In right triangle ABC, ∠B = 90°, AB = 3 cm and AC = 6 cm. Find ∠C and ∠A.

Answer:
1. AC is the hypotenuse, so sin C = \( \frac{AB}{AC} \) = \( \frac{3}{6} \) = \( \frac{1}{2} \).
2. Hence, ∠C = 30°.
3. ∠A = 180° - 90° - 30° = 60°.
4. Hence, ∠C = 30° and ∠A = 60°.

Teacher's Note:
a) The side opposite to C is AB, so use sine.
b) Check: the side opposite to 30° is half the hypotenuse, which is 3 cm.

13. If A = 30°, verify that:

(i) sin 2A = \( \frac{2\tan A}{1 + \tan^2 A} \)

Answer:
1. LHS: sin 60° = \( \frac{\sqrt{3}}{2} \).
2. RHS: tan 30° = \( \frac{1}{\sqrt{3}} \), so \( \frac{2/\sqrt{3}}{1 + 1/3} \) = \( \frac{2}{\sqrt{3}} \) × \( \frac{3}{4} \) = \( \frac{\sqrt{3}}{2} \).
3. LHS = RHS, so the identity is verified.
4. Hence, sin 2A = \( \frac{2\tan A}{1 + \tan^2 A} \) is verified.

Teacher's Note:
a) Put A = 30° on both sides and evaluate separately.
b) Check: both sides equal \( \frac{\sqrt{3}}{2} \).

(ii) cos 2A = \( \frac{1 - \tan^2 A}{1 + \tan^2 A} \)

Answer:
1. LHS: cos 60° = \( \frac{1}{2} \).
2. RHS: \( \frac{1 - 1/3}{1 + 1/3} \) = \( \frac{2/3}{4/3} \) = \( \frac{1}{2} \).
3. LHS = RHS, so the identity is verified.
4. Hence, cos 2A = \( \frac{1 - \tan^2 A}{1 + \tan^2 A} \) is verified.

Teacher's Note:
a) Compute tan² 30° = \( \frac{1}{3} \) first.
b) Check: both sides equal \( \frac{1}{2} \).

(iii) tan 2A = \( \frac{2\tan A}{1 - \tan^2 A} \)

Answer:
1. LHS: tan 60° = \( \sqrt{3} \).
2. RHS: \( \frac{2/\sqrt{3}}{1 - 1/3} \) = \( \frac{2}{\sqrt{3}} \) × \( \frac{3}{2} \) = \( \frac{3}{\sqrt{3}} \) = \( \sqrt{3} \).
3. LHS = RHS, so the identity is verified.
4. Hence, tan 2A = \( \frac{2\tan A}{1 - \tan^2 A} \) is verified.

Teacher's Note:
a) Rationalise \( \frac{3}{\sqrt{3}} \) to see the answer clearly.
b) Check: both sides equal √3.

14. If A = 45°, verify that

(i) sin 2A = 2 sin A cos A

Answer:
1. LHS: sin 90° = 1.
2. RHS: 2 × sin 45° × cos 45° = 2 × \( \frac{1}{\sqrt{2}} \) × \( \frac{1}{\sqrt{2}} \) = 1.
3. LHS = RHS, so the identity is verified.
4. Hence, sin 2A = 2 sin A cos A is verified for A = 45°.

Teacher's Note:
a) Use sin 45° = cos 45° = \( \frac{1}{\sqrt{2}} \).
b) Check: 2 × \( \frac{1}{2} \) = 1.

(ii) cos 2A = 2cos² A - 1 = 1 - 2sin² A

Answer:
1. LHS: cos 90° = 0.
2. 2cos² 45° - 1 = 2 × \( \frac{1}{2} \) - 1 = 0.
3. 1 - 2sin² 45° = 1 - 2 × \( \frac{1}{2} \) = 0.
4. All three expressions equal 0, so the identity is verified.
5. Hence, cos 2A = 2cos² A - 1 = 1 - 2sin² A is verified for A = 45°.

Teacher's Note:
a) Check each form separately; all three must give the same value.
b) Check: the three values are 0, 0 and 0.

15. Using the formula, cos A = \( \sqrt{\frac{1 + \cos 2A}{2}} \), find the value of cos 30°, it being given that cos 60° = \( \frac{1}{2} \).

Answer:
1. Put A = 30° so that 2A = 60°.
2. cos 30° = \( \sqrt{\frac{1 + \cos 60°}{2}} \) = \( \sqrt{\frac{1 + 1/2}{2}} \) = \( \sqrt{\frac{3}{4}} \).
3. Hence, cos 30° = \( \frac{\sqrt{3}}{2} \).

Teacher's Note:
a) Double the angle to get the value you already know.
b) Check: \( \frac{\sqrt{3}}{2} \) is the known value of cos 30°.

16. Using the formula, sin A = \( \sqrt{\frac{1 - \cos 2A}{2}} \), find the value of sin 30°, it being given that cos 60° = \( \frac{1}{2} \).

Answer:
1. Put A = 30° so that 2A = 60°.
2. sin 30° = \( \sqrt{\frac{1 - \cos 60°}{2}} \) = \( \sqrt{\frac{1 - 1/2}{2}} \) = \( \sqrt{\frac{1}{4}} \).
3. Hence, sin 30° = \( \frac{1}{2} \).

Teacher's Note:
a) The same method works for sine with the minus sign inside the root.
b) Check: \( \frac{1}{2} \) is the known value of sin 30°.

17. Using the formula, tan 2A = \( \frac{2\tan A}{1 - \tan^2 A} \), find the value of tan 60°, it being given that tan 30° = \( \frac{1}{\sqrt{3}} \).

Answer:
1. Put A = 30° so that 2A = 60°.
2. tan 60° = \( \frac{2 \tan 30°}{1 - \tan^2 30°} \) = \( \frac{2/\sqrt{3}}{1 - 1/3} \) = \( \frac{2}{\sqrt{3}} \) × \( \frac{3}{2} \) = \( \sqrt{3} \).
3. Hence, tan 60° = √3.

Teacher's Note:
a) Use tan² 30° = \( \frac{1}{3} \) in the denominator.
b) Check: √3 is the known value of tan 60°.

18. If sin (A - B) = \( \frac{1}{2} \) and cos (A + B) = \( \frac{1}{2} \), then find the value of A and B.

Answer:
1. sin (A - B) = \( \frac{1}{2} \), so A - B = 30°.
2. cos (A + B) = \( \frac{1}{2} \), so A + B = 60°.
3. Adding: 2A = 90°, so A = 45°. Subtracting: 2B = 30°, so B = 15°.
4. Hence, A = 45° and B = 15°.

Teacher's Note:
a) Add and subtract the two equations to find A and B.
b) Check: 45° - 15° = 30° and 45° + 15° = 60°.

19. If sin (A + B) = 1 and cos (A - B) = 1, then find the value of A and B.

Answer:
1. sin (A + B) = 1, so A + B = 90°.
2. cos (A - B) = 1, so A - B = 0°.
3. Adding: 2A = 90°, so A = 45°. Then B = 45°.
4. Hence, A = 45° and B = 45°.

Teacher's Note:
a) Cos 0° equals 1, so A and B must be equal.
b) Check: sin 90° = 1 and cos 0° = 1.

20. If tan (A - B) = \( \frac{1}{\sqrt{3}} \) and tan (A + B) = √3, then find the value of A and B.

Answer:
1. tan (A - B) = \( \frac{1}{\sqrt{3}} \), so A - B = 30°.
2. tan (A + B) = √3, so A + B = 60°.
3. 2A = 90°, so A = 45°; 2B = 30°, so B = 15°.
4. Hence, A = 45° and B = 15°.

Teacher's Note:
a) Use the known tangent values for 30° and 60°.
b) Check: tan 30° = \( \frac{1}{\sqrt{3}} \) and tan 60° = √3.

21. If cos (A - B) = \( \frac{\sqrt{3}}{2} \) and sin (A + B) = 1, then find the value of A and B.

Answer:
1. cos (A - B) = \( \frac{\sqrt{3}}{2} \), so A - B = 30°.
2. sin (A + B) = 1, so A + B = 90°.
3. 2A = 120°, so A = 60°; 2B = 60°, so B = 30°.
4. Hence, A = 60° and B = 30°.

Teacher's Note:
a) Read sin 90° = 1 to get the sum angle.
b) Check: 60° - 30° = 30° and 60° + 30° = 90°.

22. If A and B are acute angles such that tan A = \( \frac{1}{3} \), tan B = \( \frac{1}{2} \) and tan (A + B) = \( \frac{\tan A + \tan B}{1 - \tan A \tan B} \), show that A + B = 45°.

Answer:
1. tan (A + B) = \( \frac{1/3 + 1/2}{1 - (1/3)(1/2)} \) = \( \frac{5/6}{5/6} \) = 1.
2. tan (A + B) = 1, so A + B = 45° (A and B are acute, so A + B lies between 0° and 180°).
3. Hence, A + B = 45°, as required.

Teacher's Note:
a) Use the given formula for tan (A + B) exactly.
b) Check: 1/3 + 1/2 = 5/6 and 1 - 1/6 = 5/6.

23. If A = B = 45°, verify that:

(a) sin (A + B) = sin A cos B + cos A sin B

Answer:
1. LHS: sin 90° = 1.
2. RHS: \( \frac{1}{\sqrt{2}} \) × \( \frac{1}{\sqrt{2}} \) + \( \frac{1}{\sqrt{2}} \) × \( \frac{1}{\sqrt{2}} \) = \( \frac{1}{2} + \frac{1}{2} \) = 1.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Both sides equal 1.
b) Check: sin 45° cos 45° = \( \frac{1}{2} \).

(b) sin (A - B) = sin A cos B - cos A sin B

Answer:
1. LHS: sin 0° = 0.
2. RHS: \( \frac{1}{2} - \frac{1}{2} \) = 0.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Equal angles give a zero difference.
b) Check: both sides are 0.

(c) cos (A + B) = cos A cos B - sin A sin B

Answer:
1. LHS: cos 90° = 0.
2. RHS: \( \frac{1}{2} - \frac{1}{2} \) = 0.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Use cos 90° = 0.
b) Check: both sides are 0.

(d) cos (A - B) = cos A cos B + sin A sin B

Answer:
1. LHS: cos 0° = 1.
2. RHS: \( \frac{1}{2} + \frac{1}{2} \) = 1.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Cos 0° is 1, which matches the sum of the two halves.
b) Check: both sides are 1.

(e) tan (A + B) = \( \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

Answer:
1. LHS: tan 90° is not defined.
2. RHS: \( \frac{1 + 1}{1 - 1 × 1} \) = \( \frac{2}{0} \), which is also not defined.
3. Both sides are undefined, so the identity is verified in the only sense it can be.

Teacher's Note:
a) Division by zero makes both sides undefined.
b) Check: 1 - tan 45° × tan 45° = 0.

(f) tan (A - B) = \( \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

Answer:
1. LHS: tan 0° = 0.
2. RHS: \( \frac{1 - 1}{1 + 1} \) = 0.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) The numerator is zero because the two angles are equal.
b) Check: both sides are 0.

24. If A = 60° and B = 30°, verify that:

(a) sin (A + B) = sin A cos B + cos A sin B

Answer:
1. LHS: sin 90° = 1.
2. RHS: \( \frac{\sqrt{3}}{2} \) × \( \frac{\sqrt{3}}{2} \) + \( \frac{1}{2} \) × \( \frac{1}{2} \) = \( \frac{3}{4} + \frac{1}{4} \) = 1.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) The values of sin and cos for 60° and 30° are all needed.
b) Check: both sides equal 1.

(b) sin (A - B) = sin A cos B - cos A sin B

Answer:
1. LHS: sin 30° = \( \frac{1}{2} \).
2. RHS: \( \frac{3}{4} - \frac{1}{4} \) = \( \frac{1}{2} \).
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Subtraction gives the difference angle.
b) Check: both sides are \( \frac{1}{2} \).

(c) cos (A + B) = cos A cos B - sin A sin B

Answer:
1. LHS: cos 90° = 0.
2. RHS: \( \frac{1}{2} \) × \( \frac{\sqrt{3}}{2} \) - \( \frac{\sqrt{3}}{2} \) × \( \frac{1}{2} \) = 0.
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) The two products are equal, so the difference is zero.
b) Check: both sides are 0.

(d) cos (A - B) = cos A cos B + sin A sin B

Answer:
1. LHS: cos 30° = \( \frac{\sqrt{3}}{2} \).
2. RHS: \( \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \) = \( \frac{\sqrt{3}}{2} \).
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Add the two equal products.
b) Check: both sides are \( \frac{\sqrt{3}}{2} \).

(e) tan (A + B) = \( \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

Answer:
1. LHS: tan 90° is not defined.
2. RHS: \( \frac{\sqrt{3} + 1/\sqrt{3}}{1 - \sqrt{3} × 1/\sqrt{3}} \) = \( \frac{4/\sqrt{3}}{0} \), which is not defined.
3. Both sides are undefined, so the identity is verified in the only sense it can be.

Teacher's Note:
a) The denominator becomes zero, as with A = B = 45°.
b) Check: tan 60° × tan 30° = 1.

(f) tan (A - B) = \( \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

Answer:
1. LHS: tan 30° = \( \frac{1}{\sqrt{3}} \).
2. RHS: \( \frac{\sqrt{3} - 1/\sqrt{3}}{1 + 1} \) = \( \frac{2/\sqrt{3}}{2} \) = \( \frac{1}{\sqrt{3}} \).
3. LHS = RHS, so the identity is verified.

Teacher's Note:
a) Simplify the numerator first.
b) Check: both sides are \( \frac{1}{\sqrt{3}} \).

25. Evaluate: \( \frac{\sin^2 45° + \frac{3}{4}\,\mathrm{cosec}^2 30° - \cos 60° + \tan^2 60°}{\sin^2 30° + \cos^2 60° + \frac{1}{2}\sec^2 45°} \)

Answer:
1. Numerator = \( \frac{1}{2} \) + \( \frac{3}{4} \) × 4 - \( \frac{1}{2} \) + 3 = \( \frac{1}{2} + 3 - \frac{1}{2} + 3 \) = 6.
2. Denominator = \( \frac{1}{4} + \frac{1}{4} + \frac{1}{2} \) × 2 = \( \frac{1}{2} + 1 \) = \( \frac{3}{2} \).
3. Value = 6 ÷ \( \frac{3}{2} \) = 4.
4. Hence, the value is 4.

Teacher's Note:
a) Evaluate each term before adding.
b) Check: 6 × \( \frac{2}{3} \) = 4.

Question. Simplest form of (1 - cos² A) (1 + cot²A) is .................... . 
Answer : 1

Question. If 3 sec θ - 5 = 0 then cot θ = .................... .
Answer : 3/4

Question. If sin θ – cos θ = 0, 0 ≤ θ ≤ 90° then the value of q is .................... .
Answer : 45º

Question. cos 1° cos 2° cos 3° ......... cos 180° = .................... .
Answer : 0

Question. If tan θ = √3 , then sec θ = ................... .
Answer : tan θ = √3 gives θ = 60°
So, sec θ = sec 60° = 2

Question. Maximum value of 1/cosecθ is .................... .
Answer : 1/cosec θ = 1/1 = sin θ
As, sin θ ≤ 1, 1/cosec θ ≤ 1
Thus, maximum value of 1/cosec θ is 1.

Question. If tan θ + cot θ = 2, then the value of tan² θ + cot² θ is .................... .
Answer : tan² θ + cot² θ = (tan θ + cot θ)² – 2 tan θ cot θ
= (2)² – 2 × 1
= 2