Q. 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) 0
B) $\frac{1}{\sqrt{2}}$
C) $\frac{3}{2}$
D) $\frac{1}{8}$
Explanation
(c)
$\therefore\tan^{- 1}{x - \tan^{- 1}{y = 0 \Rightarrow x = y}}$
Also, $\cos^{- 1}{x + \cos^{- 1}{y = \frac{\pi}{2} \Rightarrow 2\cos^{- 1}{x = \frac{\pi}{2}}}}$
$\Rightarrow \cos^{- 1}{x = \frac{\pi}{4} \Rightarrow x = \frac{1}{\sqrt{2}} \Rightarrow x^{2} = \frac{1}{2}}$
Hence, $x^{2} + xy + y^{2} = 3x^{2} = \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.