Question

tan+sin/tan-sin = sec+1/sec-1

Answer 1

the ans is here............

Answer 2

Solution:

Given ,

LHS =$\frac{tanA+sinA}{tanA-sinA}$

= $\frac{(\frac{sinA}{cosA})+1}{(\frac{sinA}{cosA})-1}$

Take sinA common ,we get

=$\frac{sinA(\frac{1}{cosA}+1)}{sinA(\frac{1}{cosA}-1)}$

After cancellation of SinA , we get

= $\frac{secA+1}{secA-1}$

/* 1/cosA = secA */

= RHS

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