Question

$\frac{tana}{seca - 1} + \frac{tana}{seca + 1} = 2coseca$
solve this fast​

Answer 1

In this question and all trignometery questions it is easier to simplify LHS and RHS seperately

First if you simplify LHS it is possible that you find your answer directly so we start the question with the simplification of LHS

LHS

In LHS take tanA common then solve (it is not compulsory it is my opinion)

$tana( \frac{1}{seca - 1} + \frac{1}{seca + 1} ) \\ = tana( \frac{seca + 1 + seca - 1}{ {sec}^{2} a - 1 })$

now using identity of sec^2A-1=tan^2A

$tana \times \frac{2seca}{ {tan}^{2} a} = \frac{2seca}{tana} \\ = \frac{ \frac{2}{cosa} }{ \frac{sina}{cosa} } = \frac{2}{cosa} \times \frac{cosa}{sina} = \frac{2}{sina} \\ = 2coseca$

LHS=2cosecA

RHS=2cosecA

⠀⠀⠀⠀⠀⠀⠀⠀⠀LHS=RHS

⠀⠀⠀⠀⠀⠀⠀⠀⠀HENCE PROVED