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,

cosec² A = 1 + cot² A (By taking reciprocal both the sides)

1/cosec² A = 1/(1 + cot² A)

sin² A = 1/(1 + cot² A) ( As 1/cosec² A = sin² A)

Therefore,

sin A = ± 1/√(1 + cot² 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 + cot² 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, sec² A = 1 + tan² A (Trigonometric Identity)

= 1 + 1/cot² A

= (cot² A + 1) / cot² A

sec A = √(cot² A + 1) / cot A

Check out more on trigonometric ratios.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 8

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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+ cot²A), √(1+ cot²A)/cot A and 1/cot A respectively.

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☛ Related Questions:

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• 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