Question:
If $\frac{1-\tan ^{2}\left(\frac{\pi}{4}-x\right)}{1+\tan ^{2}\left(\frac{\pi}{4}-x\right)}=\sin k x$, then $k=$ ____________________
Solution:
Given, $\frac{1-\tan ^{2}(\pi / 4-x)}{1+\tan ^{2}(\pi / 4-x)}=\sin k x$
$\left[\right.$ using identity $\left.: \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta\right]$
We have,
$\frac{1-\tan ^{2}(\pi / 4-x)}{1+\tan ^{2}(\pi / 4-x)}=\cos 2(\pi / 4-x)$
$=\cos (\pi / 2-2 x)$
$=\sin 2 x(\because \cos (\pi / 2-\theta)=\sin \theta)$
$=\sin k x$ (given)
$\Rightarrow \sin k x=\sin 2 x$
$\Rightarrow k x=2 x(-1)^{n}+n \pi$
Or
$=k=2$