# write all the other trigonometric ratios of angle a in terms of sec a

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

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

Medium Open in App Solution Verified by Toppr We know, cosA= secA 1 We know, sin 2 A+cos 2 A=1 sin 2 A=1−cos 2 A sinA= 1−( secA 1 ) 2 = sec 2 A sec 2 A−1 = secA sec 2 A−1 ∴sinA= secA sec 2 A−1 We know, cosec A= sinA 1 ∴cosec A= sec 2 A−1 secA We know, cotA= sinA cosA == secA sec 2 A−1 secA 1 ∴cotA= sec 2 A−1 1 We know, tanA= cotA 1 ∴tanA= sec 2 A−1 Video Explanation

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## Write all the other trigonometric ratios of ∠A in terms of sec A. We use the basic identities of trigonometric ratios, sin, cos, tan, cot, sec, cosec to solve the problem given to us.

Write all the other trigonometric ratios of ∠A in terms of sec A. We use the basic identities of trigonometric ratios, sin, cos, tan, cot, sec, cosec to solve the problem given to us.

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

**Solution:**

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

sin2 A + cos2 A = 1

cosec2 A = 1 + cot2 A

sec2 A = 1 + tan2 A We know that,

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

Also,

sin2 A + cos2 A = 1 (trigonometric identity)

sin2 A = 1 - cos2 A (By transposing)

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

sin A = √1 - (1 / sec A)2

= √(sec2 A - 1) / sec2 A

= √(sec2 A - 1) / sec A ....Equation (2)

tan2 A + 1 = sec2 A (Trigonometric identity)

tan2 A = sec2 A - 1 (By transposing)

tan A = √(sec2 A - 1) ... Equation (3)

cot A = cosA/sinA

= (1/sec A) / [√(sec2 A - 1)/sec A] .....(By substituting the values from Equations (1) and (2))

= 1 / √(sec2 A - 1) cosec A = 1/sin A

= sec A / √(sec2 A - 1) (By substituting from Equation (2) and simplifying)

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

**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

**Summary:**

All the other trigonometric ratios of ∠A in terms of sec A are cos A = 1/sec A, sin A = √(sec2A − 1)/sec A, tan A = √(sec2A − 1), cot A = 1/√(sec2A − 1), and cosec A = sec A/√(sec2A − 1).

**☛ Related Questions:**

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

## Ex 8.4, 2

Ex 8.4, 2 Write all the other trigonometric ratios of ∠ A in terms of sec A. cos A We know that cos A = 𝟏/𝒔𝒆𝒄𝑨 tan A We know that 1 + tan2 A = sec2 A tan2 A = sec2 A − 1 tan A = ± √(𝑠𝑒𝑐2 𝐴 −1) Here, A is acute (i.e. less than 90°) & tan A is positive when A is acute tan A

**Check sibling questions**

## Ex 8.4, 2 - Chapter 8 Class 10 Introduction to Trignometry (Term 1)

Last updated at May 29, 2018 by Teachoo

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### Transcript

Ex 8.4, 2 Write all the other trigonometric ratios of A in terms of sec A. cos A We know that cos A = / tan A We know that 1 + tan2 A = sec2 A tan2 A = sec2 A 1 tan A = ( 2 1) Here, A is acute (i.e. less than 90 ) & tan A is positive when A is acute tan A = ( 2 1) cot cot A = /( ) Putting value of tan A found above cot A = 1/( (sec^2 1 ) ) cosec A We know that 1 + cot2 A = cosec2 A cosec2 A = 1 + cot2 A cosec2 A = 1 + (1/( (sec^2 1 ) ))^2 cosec2 A = 1 + (1/( 2 1)) cosec2 A = ((1 ( 2 1) + 1)/( 2 1)) cosec2 A = (( 2 1 + 1)/( 2 1)) cosec2 A = ( 2 )/( 2 1) cosec A = (( 2 )/( 2 1)) cosec A = sec / ( 2 1 ) Here, A is acute (i.e. less than 90 ) & cosec A is positive when A is acute cosec A = sec / ( 2 1 ) sin A We know that sin A = /( ) sin A = 1/((sec / (sec^2 1 )) ) sin A = ( 2 1)/sec

**Next**: Ex 8.4, 3 (i) Important →

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### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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