# Introduction to Trigonometry – Class X – Part 6 | Exercise 8.3 Q1-7 Solved

### Watch the sixth video session on “Introduction To Trigonometry” for Class Xth – concepts and solved questions

Introduction To Trigonometry – Class X – Part 6

1. How to solve trigonometric ratios for complementary angles?

a) Identify the complementary angles by adding (any) two angles – it is always a pair of angles that are complementary to each other.

b) Convert one of the trigonometric ratio to another

— If the question statement consists of a mixture of sin and cos, convert sin into cos (or the other way round) and evaluate further.
— If the question statement consists of a mixture of tan and cot, convert cot into tan (or the other way round) and evaluate further. Product of tan and cot of the same angle is always one.
— If the question statement consists of a mixture of cosec and sec, convert sec into cosec (or the other way round) and evaluate further.

c) Please remember that the trigonometric ratios for specific angles: 0°,30°,45°, 60°, 90° are generally not involved as then finding the value would be simple. The questions usually involve different involves – for example : angles 18 and 72 or 24 and 66.

NCERT Exercise 8.3 Q1-7 Solved

1. Evaluate

(i) sin 18°/ cos 72°

(i) tan 26°/ cot 64°

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

2. Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

4. If tan A = cot B, prove that A + B = 90°.

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

6. If A, B and C are interior angles of a triangle ABC, then show that

Sin (B+C)/2 = Cos A/2

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

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