# Write all the other trigonometric ratios of ∠A in terms of sec A

**Solution:**

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

sin^{2} A + cos^{2} A = 1

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

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

We know that,

cos A = 1/sec A .....Equation (1)

Also,

sin^{2} A + cos^{2} A = 1 (trigonometric identity)

sin^{2} A = 1 - cos^{2} A (By transposing)

Using value of cos A from Equation (1) and simplifying further

sin A = √1 - (1 / sec A)^{2}

= √(sec^{2} A - 1) / sec^{2} A

= √(sec^{2} A - 1) / sec A ....Equation (2)

tan^{2} A + 1 = sec^{2} A (Trigonometric identity)

tan^{2} A = sec^{2} A - 1 (By transposing)

Trigonometric Function,

tan A = √(sec^{2} A - 1) ... Equation (3)

cot A= cosA/sinA

= (1/sec A) / [√(sec^{2} A - 1)/sec A] .....(By substituting Equations (1) and (2))

= 1 / √(sec^{2} A - 1)

cosec A = 1/sin A

= sec A / √(sec^{2} A - 1) (By substituting Equation (2) and simplifying)

**Video Solution:**

## Write all the other trigonometric ratios of ∠A in terms of sec A.

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.4 Question 2:

Write all the other trigonometric ratios of ∠A in terms of sec A.

All the other trigonometric ratios of ∠A in terms of sec A are cos A = 1/sec A, sin A = √(sec^{2}A − 1)/sec A, tan A = √(sec^{2}A − 1), cot A = 1/√(sec^{2}A − 1), cosec A = sec A/√(sec^{2}A − 1).