# Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

**Solution:**

We use the basic trigonometric identities and properties of the trigonometric ratios to solve the problem.

cosec^{2} A = 1 + cot^{2} A

sec^{2} A = 1 + tan^{2} A

Consider a ΔABC with ∠B = 90°

Using the Trigonometric Identity,

cosec^{2} A = 1 + cot^{2} A (By taking reciprocal both the sides)

1/cosec^{2} A = 1/(1 + cot^{2} A)

sin^{2} A = 1/(1 + cot^{2} A) ( As 1/cosec^{2} A = sin^{2} A)

Therefore,

sin A = ± 1/√(1 + cot^{2} A)

For any sine value with respect to an angle in a triangle, the sine value will never be negative. Since sine value will be negative for all angles greater than 180°.

Therefore, sin A = 1/√(1 + cot^{2} A)

We know that,

tan A = sin A / cos A

However, we have,

cot A = cos A/sin A

Therefore, we have,

tan A = 1/cot A

Also, sec^{2} A = 1 + tan^{2} A ((Trigonometric Identity)

= 1 + 1/cot^{2} A

= (cot^{2} A + 1) / cot^{2} A

sec A = √(cot^{2} A + 1) / cot A

Check out more on trigonometric ratios.

**Video Solution:**

## Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.4 Question 1:

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

The trigonometric ratios sin A, sec A, and tan A can be expressed in terms of cot A as 1/√(1+ cot^{2}A), √(1+ cot^{2}A)/cot A and 1/cot A respectively