Question

Prove that:
1 + cos A - sin A/1 +cos A + sin a = sec a - tan a​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{1+cos(A)-sin(A)}{1+cos(A)+sin(A)}}$

$\sf{=\dfrac{\dfrac{1+cos(A)-sin(A)}{cos(A)}}{\dfrac{1+cos(A)+sin(A)}{cos(A)}}}$

$\sf{=\dfrac{sec(A)+1-tan(A)}{sec(A)+1+tan(A)}}$

$\sf{=\dfrac{sec(A)-tan(A)+1}{1+sec(A)+tan(A)}}$

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

$\sf{=\dfrac{sec(A)-tan(A)+\{sec(A)-tan(A)\}\{sec(A)+tan(A)\}}{1+sec(A)+tan(A)}}$

$\sf{=\dfrac{sec(A)-tan(A)\{1+sec(A)+tan(A)\}}{1+sec(A)+tan(A)}}$

$\sf{=sec(A)-tan(A)}$