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Prove the following: tanA1+secA-tanA1-secA = 2cosec A - Mathematics

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Question

Prove the following:

`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A

Sum
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Solution

L.H.S:

`tanA/(1 + sec A) - tanA/(1 - sec A)`

Taking LCM of the denominators,

= `(tanA(1 - sec A) - tanA(1 + sec A))/((1 + sec A)(1 - sec A))`

Since, (1 + sec A)(1 – sec A) = 1 – sec2A

= `(tan A(1 - secA - 1 - sec A))/(1 - sec^2A)`

= `(tan A(-2 sec A))/(1 - sec^2 A)`

= `(2 tan A  *sec A)/(sec^2 A - 1)`

Since,

sec2A – tan2A = 1

sec2A – 1 = tan2A

= `(2 tan A * sec A)/(tan^2 A)` 

Since, sec A = `(1/cosA)` and tan A = `(sinA/cosA)`

= `(2secA)/tanA = (2cosA)/(cosA sinA)`

= `2/sinA`

= 2 cosec A  ...`(∵ 1/sinA = "cosec" A)`

= R.H.S

Hence proved.

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Chapter 8: Introduction To Trigonometry and Its Applications - Exercise 8.3 [Page 95]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 10
Chapter 8 Introduction To Trigonometry and Its Applications
Exercise 8.3 | Q 2 | Page 95

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