Question

If $\sec x+\tan x=a$,then $\frac{a^2-1}{a^2+1}$= ?

• A. $\sin x$
• B. $\cos x$
• C. $\cot x$
• D. $cosecx$

Answer 1

The correct option is A $\sin x$

Given $\sec x+\tan x=a$

then $a^2=(\sec x+\tan x{)}^2$

$\to a^2={\sec }^{2}x+{\tan }^{2}x+2\sec x\tan x$ ...... (1)

$\to a^2=2{\sec }^{2}x-1+2\sec x\tan x$

$\to a^2+1=2\sec x(\sec x+\tan x)$ .... (2)

Consider equation (1)

$\to a^2=2{\tan }^{2}x+1+2\sec x\tan x$

$\to a^2-1=2\tan x(\sec x+\tan x)$ ...(3)

Divide (2) & (3)

$\to \frac{a^2-1}{a^2+1}=\frac{2\tan x(\sec x+\tan x)}{2\sec x(\sec x+\tan x)}=\sin x$