Question

Prove that $\frac{\sec x+1-\tan x}{\tan x+\sec x+1}=\frac{1-\sin x}{\cos x}$

Answer 1

$L.H.S=\frac{1+\sec x-\tan x}{1+\sec x+\tan x}\frac{(1+\sec x-\tan x)}{(1+\sec x-\tan x)}$

$\frac{1+{\sec }^{2}x+{\tan }^{2}x+2\sec x-2\tan x-2\sec x\tan x}{(1+\sec x{)}^2-{\tan }^{2}x}$

$\frac{{\sec }^{2}x+{\sec }^{2}x+2\sec x-2\tan x-2\sec x\tan x}{1+{\sec }^{2}x+2\sec x-{\tan }^{2}x}$

$\frac{2\left(\sec x\right(1+\sec x)-\tan x(1+\sec x\left)\right)}{1+1+{\tan }^{2}x+2\sec x-{\tan }^{2}x}$

$\frac{2(1+\sec x)(\sec x-\tan x)}{2(1+\sec x)}$

$\sec x-\tan x$

$\frac{1-\sin x}{\cos x}=RHS$

$\therefore QED$