Question

If $x=ta{n}^{-1}\left(\frac{1}{7}\right)$ and $y=ta{n}^{-1}\left(\frac{1}{3}\right)$, then :

• A. $cos2x=sin4y$
• B. $cos4y=cos2x$
• C. $cos2uy=sin4x$
• D. None of these

Answer 1

The correct option is A $cos2x=sin4y$

$x=ta{n}^{-1}\left(\frac{1}{7}\right)\Rightarrow tanx=\frac{1}{7}$
and $y=ta{n}^{-1}\left(\frac{1}{3}\right)=tany=\frac{1}{3}$
Now $cos2x=\frac{(1-ta{n}^{2}x)}{(1+ta{n}^{2}x)}=\frac{24}{25}$
$sin4y=2sin2ycos2y$
$=\frac{2.2tany.(1-ta{n}^{2}y)}{(1+ta{n}^{2}x{)}^2}=\frac{24}{25}$
$\therefore$ $cos2x=sin4y$