Question

If $\sec x+\tan x=k$, then show that $\sin x=\frac{k^2-1}{k^2+1}$.

Answer 1

Given $\sec x+\tan x=k$......(1).

Now,

$\frac{k^2-1}{k^2+1}$

$=\frac{({\sec }^{2}x+{\tan }^{2}x+2\tan x.\sec x)-1}{({\sec }^{2}x+{\tan }^{2}x+2\tan x.\sec x)+1}$ [ Using (1)]

$=\frac{(2{\tan }^{2}x+2\tan x.\sec x)}{(2{\sec }^{2}x+2\tan x.\sec x)}$ [ Since ${\sec }^{2}x=1+{\tan }^{2}x$ and ${\sec }^{2}x-1={\tan }^{2}x$]

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

$=\frac{\tan x}{\sec x}$

$=\sin x$.