# Evaluate:

(i) sin 18°/cos 72° (ii) tan 26°/cot 64°

(iii) cos 48° - sin 42°(iv) cosec 31° - sec 59°

**Solution:**

We will be using the trigonometric ratios of complementary angles to solve the given question.

sin (90° - θ) = cos θ

tan (90° - θ) = cot θ

sec (90° - θ) = cosec θ

(i) sin 18°/cos 72°

Since, sin (90° - θ) = cos θ

Here, θ = 72°

sin 18°/cos 72° = sin (90° - 72°)/cos 72°

= cos 72° / cos 72°

= 1

(ii) tan 26°/cot 64°

tan (90° - θ) = cot θ

Here, θ = 64°

tan 26°/cot 64° = tan (90° - 64°) / cot 64°

= cot 64°/cot 64°

= 1

(iii) cos 48° - sin 42°

Since, sin (90° - θ) = cos θ

Here, θ = 48°

cos 48° - sin 42° = cos 48° - sin (90° - 48°)

= cos 48° - cos 48°

= 0

(iv) cosec 31° - sec 59°

sec (90° - θ) = cosec θ

Here, θ = 31°

cosec 31° - sec 59° = cosec 31° - sec (90° - 31°)

= cosec 31° - cosec 31°

= 0

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 8

**Video Solution:**

## Evaluate: (i) sin 18°/cos 72° (ii) tan 26°/cot 64° (iii) cos 48° - sin 42° (iv) cosec 31° - sec 59°

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.3 Question 1

**Summary:**

The values are as follows: (i) sin 18°/cos 72° = 1, (ii) tan 26°/cot 64° = 1, (iii) cos 48° − sin 42° = 0, (iv) cosec 31° − sec 59° = 0

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