Question

Answer this fast plzzz

Answer 1

LHS:

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

$= \ \textgreater \ \frac{tanA + secA + tan^2A - sec^2A}{tanA - secA + 1}$

$= \ \textgreater \ \frac{tanA + secA + (tanA + secA)(tanA - secA)}{tanA - secA + 1}$

$= \ \textgreater \ \frac{tanA + secA(1 + tanA - secA)}{tanA - secA+ 1}$

$= \ \textgreater \ tanA + secA$

$= \ \textgreater \ \frac{sinA}{cosA} + \frac{1}{cosA}$

$= \ \textgreater \ \frac{sinA + 1}{cosA}$




Hope this helps!

Answer 2

Hey !!!

Solution is in this given attachment

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Hope it helps you !!!

@Rajukumar111