Question

If $\frac{1-{\tan }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)}{1+{\tan }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)}=\sin kx,$ then k = __________.

Answer 1

$$\begin{gathered} \mathrm{Given},\frac{1-{\tan }^2\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4-x$}\right.}\right)}{1+{\tan }^2\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4-x$}\right.}\right)}=\sin kx \\ \left[\mathrm{using}\mathrm{identity}:\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta \right] \\ \mathrm{we}\mathrm{have}, \\ \frac{1-{\tan }^2\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4-x$}\right.}\right)}{1+{\tan }^2\left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4-x$}\right.}\right)}=\cos 2\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4-x$}\right.\right) \\ =\cos \left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2-2x$}\right.\right) \\ =\sin 2x\left(\because \cos \left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2-\theta $}\right.\right)=\sin \theta \right) \\ =\sin kx\left(\mathrm{given}\right) \\ \Rightarrow \sin kx=\sin 2x \\ \Rightarrow kx=2x{\left(-1\right)}^n+n\mathrm{\pi } \\ \mathrm{or} \\ =k=2 \end{gathered}$$