Question

Question is attached.Step by step explanation needed!!​

Answer 1

Your answer is attached.

• Hence Proved.

Answer 2

LHS

$\\ \sf\longmapsto \dfrac{tanA+secA-1}{tanA-secA+1}$

As we know

$\\ \boxed{\sf sec^2A-tan^2A=1}$

$\\ \sf\longmapsto \dfrac{tanA+secA-(sec^2A-tan^2A)}{tanA-secA+1}$

$\\ \sf\longmapsto \dfrac{tanA+secA\cancel{(1-secA+tanA)}}{\cancel{(1-secA+tanA)}}$

$\\ \sf\longmapsto tanA+secA$

$\\ \sf\longmapsto \dfrac{sinA}{cosA}+\dfrac{1}{cosA}$

$\\ \sf\longmapsto \dfrac{1+sinA}{cosA}$

$\\ \sf\longmapsto RHS$

Hence proved