Question

If $\tan \left(\frac{\pi }{4}+x\right).\tan \left(\frac{\pi }{4}-x\right)=1$, then $x$ equals

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. independent of $x$

Answer 1

Answer: (D)

independent of $x$

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

$or,\left(\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}.\tan x}\right)$ $\left(\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}.\tan x}\right)=1$

$or,\left(\frac{1+\tan x}{1-\tan x}\right)$ $\left(\frac{1-\tan x}{1+\tan x}\right)=1$

$or,1=1$

Which is an identity.

So, it is true for all $x$.