Question

If ${\cos }^{-1}x+{\cos }^{-1}y=\frac{\pi }{2}$ and ${\tan }^{-1}x-{\tan }^{-1}y=0$ then $x^2+xy+y^2$ is equal to

• A. $\frac{1}{8}$
• B. $\frac{3}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. 0

Answer 1

The correct option is B

$\frac{3}{2}$

${\tan }^{-1}x-{\tan }^{-1}y=0\Rightarrow x=y$

${\cos }^{-1}x+{\cos }^{-1}y=\frac{\pi }{2}\Rightarrow 2{\cos }^{-1}x=\frac{\pi }{2}$

$\Rightarrow x=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\Rightarrow x^2=\frac{1}{2}$

then $x^2+xy+y^2=3x^2=\frac{3}{2}$