Question

If $\tan x=\frac{1}{7},\tan y=\frac{1}{3}$ and cos2x = sin ky, then k = ______________.

Answer 1

$$\begin{gathered} \mathrm{If}\tan x=\frac{1}{7},\tan y=\frac{1}{3} \\ \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x} \\ \cos 2x=\frac{1-{\displaystyle \frac{1}{49}}}{1+{\displaystyle \frac{1}{49}}}=\frac{{\displaystyle \frac{48}{49}}}{{\displaystyle \frac{50}{49}}}=\frac{24}{25} \\ \mathrm{since},\sin 2y=\frac{2\tan y}{1+{\tan }^{2}y}=\frac{2\left({\displaystyle \frac{1}{3}}\right)}{1+{\displaystyle \frac{1}{9}}}=\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{10}{9}}}=\frac{2}{10}\times \frac{9}{3}=\frac{3}{5}.......\left(1\right) \\ \sin 4y=2\sin 2y\cos 2y \\ =2\left(\frac{3}{5}\right)\left(\frac{1-{\tan }^{2}y}{1+{\tan }^{2}y}\right) \\ \mathrm{from}\left(1\right) \\ =\frac{2\times 3}{5}\left[\frac{1-{\displaystyle \frac{1}{9}}}{1+{\displaystyle \frac{1}{9}}}\right] \\ =\frac{6}{5}\times \frac{8}{9}\times \frac{9}{10} \\ \sin 4y=\frac{24}{25} \\ \mathrm{Hence},\cos 2x=\sin 4y=\sin ky\left(\mathrm{given}\right) \\ \Rightarrow k=4 \end{gathered}$$