Question

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

Answer 1

${\sec }^{2}x-{\tan }^{2}x=1$ (identity)
$⟹(\sec x-\tan x)(\sec x+\tan x)=1$
$⟹\sec x-\tan x=\frac{1}{k}$
Solving the above simultaneously with $\sec x+\tan x=k$ gives:
$\sec x=\frac{1}{2}(k+\frac{1}{k})$ (1)
$\tan x=\frac{1}{2}(k-\frac{1}{k})$ (2)
Using trigonometry definitions:
$\tan x=\frac{\sin x}{\cos x}=\sin x\sec x⟹\sin x=\frac{\tan x}{\sec x}$
Substituting (1) and (2):
$\sin x=\frac{k-1/k}{k+1/k}=\frac{k^2-1}{k^2+1}$
Hence, proved.