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Question
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
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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