Question

Which of the following trigonometric statement does hold good?

• A. $\tan (\frac{\pi }{4}+x)=\cot (\frac{\pi }{4}-x)$
• B. $\tan (\frac{\pi }{4}+x)=\frac{1+\tan x}{1-\tan x}$
• C. $\tan (\frac{\pi }{4}+x)=\sec 2x+\tan 2x$
• D. all of these

Answer 1

The correct option is D all of these

$\tan (\frac{\pi }{4}+x)=\tan (\frac{\pi }{2}-\frac{\pi }{4}+x)=\tan (\frac{\pi }{2}-(\frac{\pi }{4}-x\left)\right)=\cot (\frac{\pi }{4}-x)$

So option $A$ is true

$\tan (\frac{\pi }{4}+x)=\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}=\frac{1+\tan x}{1-\tan x}$

So option $B$ is true

$\tan (\frac{\pi }{4}+x)=\frac{1+\tan x}{1-\tan x}=\frac{(1+\tan x{)}^2}{(1-{\tan }^{2}x)}$

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

So option $C$ is true