Question

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

• A. True
• B. False

Answer 1

The correct option is A True

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

$=\frac{1+2{\cos }^2\frac{x}{2}-1-2\sin \frac{x}{2}\cos \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}+1-2{\sin }^2\frac{x}{2}-1}$

$=\frac{2\cos \frac{x}{2}(\cos \frac{x}{2}-\sin \frac{x}{2})}{2\sin \frac{x}{2}(\cos \frac{x}{2}-\sin \frac{x}{2})}$

LHS = $\cot \frac{x}{2}$

$RHS=\frac{1+2{\cos }^2\frac{x}{2}-1}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\cot \frac{x}{2};$ LHS = RHS