Question

If $x$ and $y$ are acute angles such that $x+y$ and $x-y$ satisfy the equation ${\tan }^2\theta -4\tan \theta +1=0$, then

• A. $x=\frac{\pi }{4}$
• B. $x=\frac{\pi }{6}$
• C. $y=\frac{\pi }{6}$
• D. $y=\frac{\pi }{4}$

Answer 1

Answer: (A), (C)

The correct options are
A $x=\frac{\pi }{4}$
C $y=\frac{\pi }{6}$

${\tan }^2\theta -4\tan \theta +1=0$
Now, $\tan (x+y)$ & $\tan (x-y)$
$\therefore$ Sum of roots $=4$
$\therefore \tan (x+y)+\tan (x-y)=4$

$\frac{\sin (x+y)}{\cos (x+y)}+\frac{\sin (x-y)}{\cos (x-y)}=4$
$\frac{\sin (x+y)\cos (x-y)+\sin (x-y)\cos (x+y)}{\cos (x+y)\cos (x-y)}=4$
$\frac{2\sin (x+y+x-y)}{2\cos (x+y)\cos (x-y)}=4$
$\frac{2\sin 2x}{\cos (x+y+x-y)+\cos (x+y-x+y)}=4[\because 2\cos C\cos D=\cos (C+D)+\cos (C-D)]$
$\frac{\sin 2x}{\cos 2x+\cos 2y}=2$ ........ $\left(i\right)$
Also, Product of Roots $=\tan (x+y)\tan (x-y)=1$
$\frac{2\sin (x+y)\sin (x-y)}{2\cos (x+y)\cos (x-y)}=1$
$\frac{\cos (x+y-x+y)-\cos (x+y+x-y)}{\cos (x+y+x-y)+\cos (x+y-x+y)}=1[\because 2\sin C\sin D=\cos (C-D)-\cos (C+D)]$
$\cos 2y-\cos 2x=\cos 2x+\cos 2y$
$2\cos 2x=0$
$\cos 2x=0$
$2x=\frac{\pi }{2}$
[$\because$ x is an acute angle$⟹$ it lies in 1st quadrant]
$x=\frac{\pi }{4}$
Substitute value of $x$ in equation $\left(i\right)$

$\Rightarrow \frac{\sin \frac{\pi }{2}}{\cos \frac{\pi }{2}+\cos 2y}=2$
$2\cos 2y=1⟹\cos 2y=\frac{1}{2}⟹2y=\frac{\pi }{3}$
$\therefore y=\frac{\pi }{6}$