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

**Solution:**

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

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 acute angle in a triangle, the sine value will never be negative.

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.

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

**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

**Summary:**

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.

**☛ Related Questions:**

- Write all the other trigonometric ratios of ∠A in terms of sec A.
- Evaluate:(i) (sin² 63° + sin² 27) / (cos² 17° + cos² 73°)(ii) sin 25° cos 65° + cos 25° sin 65°
- Choose the correct option. Justify your choice.(i) 9 sec2A - 9 tan2A = _______(A) 1 (B) 9 (C) 8 (D) 0(ii) (1 + tanθ + secθ) (1 + cotθ - cosecθ)(A) 0 (B) 1 (C) 2 (D) -1(iii) (sec A + tan A) (1 - sin A) = _______(A) sec A (B) sin A (C) cosec A (D) cos A(iv) 1 + tan² A/(1 + cot² A)(A) sec 2A (B) -1 (C) cot 2A (D) tan 2A
- Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(i) (cosecθ - cotθ)² = 1 - cosθ/1 + cosθ(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A(iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ(iv) (1 + sec A)/sec A = sin² A/(1 - cos A)(v) (cos A - sin A + 1)/cos A + sin A + 1 = cosec A + cot A(vi) 1 + sin A/(1 - sin A) = sec A+ tan A(vii) (sinθ - 2sin³ θ)/(2cosθ - cosθ) = tanθ(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A(ix) (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)(x) [(1 + tan² A)/(1 + cot² A)] = [(1 - tan A/1 - cot A)²] = tan² A

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