If tan

(d) 2

Given:

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

$\Rightarrow \frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \times \tan x}+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \times \tan x}=\lambda \sec 2 x$

$\Rightarrow \frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}=\lambda$ sec $2 x$

$\Rightarrow \frac{(1+\tan x)^{2}+(1-\tan x)^{2}}{(1-\tan x)(1+\tan x)}=\lambda$ sec $2 x$

$\Rightarrow \frac{2\left(1+\tan ^{2} x\right)}{1-\tan ^{2} x}=\lambda \sec 2 x$

$\Rightarrow \frac{2 \sec ^{2} x}{1-\tan ^{2} x}=\lambda \sec 2 x$

$\Rightarrow \frac{2}{\cos ^{2} x\left(1-\tan ^{2} x\right)}=\lambda \sec 2 x$

$\Rightarrow \frac{2}{\cos ^{2} x\left(1-\frac{\sin ^{2} x}{\cos ^{2} x}\right)}=\lambda \sec 2 x$

$\Rightarrow \frac{2}{\cos ^{2} x-\sin ^{2} x}=\lambda \sec 2 x$

$\Rightarrow \frac{2}{\cos 2 x}=\lambda \sec 2 x$

$\Rightarrow 2 \sec 2 x=\lambda \sec 2 x$

$\Rightarrow 2=\lambda$

$\therefore \lambda=2$