Question

pleaae someone answer qno. 1

Answer 1

(1)

$Given : \frac{tanA + secA - 1}{tanA - secA + 1}$

We know that 1 + tan^2a = sec^2a = >  tan^2a - sec^2a = -1.

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

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

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

$= \ \textgreater \ tanA + secA$

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

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



Hope this helps!

Answer 2

Hi,

Please see the attached file!



Thanks