Q. The value of the expression $\sin^{6}{\theta + \cos^{6}{\theta + 3\sin^{2}{\theta.\cos^{2}\theta}}}$ equals

A) 0
B) 2
C) 3
D) 1
Explanation

(d)
$\left( \sin^{2}\theta \right)^{3} + ({\cos^{2}{\theta)}}^{3} + 3\sin^{2}{\theta\cos^{2}\theta}$
$= \left( \sin^{2}{\theta + \cos^{2}\theta} \right)\left( \sin^{4}{\theta + \cos^{4}\theta} - \sin^{2}{\theta\cos^{2}\theta} \right) + 3\sin^{2}{\theta\cos^{2}\theta}$
$= 1\left\lbrack \left( \sin^{2}{\theta + \cos^{2}\theta} \right)^{2} - 3\sin^{2}{\theta\cos^{2}\theta} \right\rbrack + 3\sin^{2}{\theta\cos^{2}{\theta = 1}}$
$= 1 - 3\sin^{2}{\theta\cos^{2}{\theta + 3\sin^{2}{\theta\cos^{2}{\theta = 1}}}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.