Prove that

To Prove: $\tan \left(\frac{\pi}{4}+\frac{\pi}{2}\right)=\tan x+\sec x$

Proof: Consider L.H.S,

$\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{x}{2}}\left(\because\right.$ this is of the form $\left.\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right)$

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

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

Multiply and divide L.H.S by $\cos \frac{x}{2}+\sin \frac{x}{2}$

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

$=\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}$

$=\frac{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \cos \frac{x}{2} \sin \frac{x}{2}}{\cos x}$ $\left(\because \cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}=\cos x\right)$

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

$=\frac{1+\sin x}{\cos x}\left(\because 2 \cos \frac{x}{2} \sin \frac{x}{2}=\sin x\right)$

$=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$

$\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\sec x+\tan x=$ R.H.S

$\because$ L.H.S $=$ R.H.S, Hence proved