Question

Let $x,y$ be real numbers such that $x{\cos }^2{39}^{\circ }=\tan {26}^{\circ }\cot {39}^{\circ }\tan {86}^{\circ }\cos {78}^{\circ }\tan {34}^{\circ }$ and $y\cos {81}^{\circ }=\cos {36}^{\circ }-\sin {36}^{\circ }.$ Then the value of $x+y^2$ is

• A. $4$
• B. $5$
• C. $10$
• D. $15$

Answer 1

The correct option is A $4$

$x=\left(\tan {26}^{\circ }\tan {86}^{\circ }\tan {34}^{\circ }\right)\frac{\cos {39}^{\circ }\cos {78}^{\circ }}{\sin {39}^{\circ }{\cos }^2{39}^{\circ }}$
We know $\tan x\tan ({60}^{\circ }-x)\tan ({60}^{\circ }+x)=\tan 3x$
$\therefore x=\frac{2\tan {78}^{\circ }\cos {78}^{\circ }}{2\sin {39}^{\circ }\cos {39}^{\circ }}=\frac{2\tan {78}^{\circ }}{\tan {78}^{\circ }}=2$

$y\cos {81}^{\circ }=\frac{\sqrt{2}}{\sqrt{2}}(\cos {36}^{\circ }-\sin {36}^{\circ })$
$=\sqrt{2}\sin ({45}^{\circ }-{36}^{\circ })$
$=\sqrt{2}\sin 9^{\circ }=\sqrt{2}\cos {81}^{\circ }$
$\Rightarrow y=\sqrt{2}$
$\therefore x+y^2=4$