If tan x=
Question:

If $\tan x=\frac{1}{7}$, tan $y=\frac{1}{3}$ and $\cos 2 x=\sin k y$, then $k=$ _______________________

Solution:

If $\tan x=\frac{1}{7}, \tan y=\frac{1}{3}$

$\cos 2 x=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$

$\cos 2 x=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{\frac{48}{49}}{\frac{50}{49}}=\frac{24}{25}$

since, $\sin 2 y=\frac{2 \tan y}{1+\tan ^{2} y}=\frac{2\left(\frac{1}{3}\right)}{1+\frac{1}{4}}=\frac{\frac{2}{3}}{\frac{10}{4}}=\frac{2}{10} \times \frac{9}{3}=\frac{3}{5} \quad \ldots \ldots(1)$

$\sin 4 y=2 \sin 2 y \cos 2 y$

$=2\left(\frac{3}{5}\right)\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)$

from (1)

$=\frac{2 \times 3}{5}\left[\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\right]$

$=\frac{6}{5} \times \frac{8}{9} \times \frac{9}{10}$

$\sin 4 y=\frac{24}{25}$

Hence, $\cos 2 x=\sin 4 y=\sin k y$        (given)

$\Rightarrow k=4$

Administrator

Leave a comment

Please enter comment.
Please enter your name.