Question

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

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

Answer 1

Answer: (B), (D)

The correct options are
B $x=\frac{\pi }{4}$
D $y=\frac{\pi }{6}$

Let, $\tan \theta =t$

We have, $t^2-4t+1=0$

$(t-2{)}^2-4+1=0$

$(t-2{)}^2=3$

$(t-2)=\pm \sqrt{3}$

$t=2\pm \sqrt{3}$

$\tan \theta =2+\sqrt{3}$ and $\tan \theta =2-\sqrt{3}$

Hence,
$\theta ={15}^{\circ }$ and $\theta ={75}^{\circ }$

Hence,
$x-y={15}^{\circ }$

$x+y={75}^{\circ }$

$x={45}^{\circ }=\frac{\pi }{4}$ and $y={30}^{\circ }=\frac{\pi }{6}$