Question

If ${\sin }^6\theta +{\cos }^6\theta +k{\cos }^{2}2\theta =1$ then $k$ is equal to

• A. $\frac{1}{2}{\tan }^{2}2\theta$
• B. $\frac{1}{4}{\tan }^{2}2\theta$
• C. $4{\cot }^{2}2\theta$
• D. $\frac{3}{4}{\tan }^{2}2\theta$

Answer 1

The correct option is D $\frac{3}{4}{\tan }^{2}2\theta$

${\sin }^6\theta +{\cos }^6\theta +k{\cos }^{2}2\theta =1$
Substituting ${\sin }^{6}A+{\cos }^{6}A=1-3{\sin }^{2}A.{\cos }^{2}A$, we get
$1-3{\sin }^2\theta .{\cos }^2\theta +k{\cos }^{2}2\theta =1$

$\Rightarrow -\frac{3}{4}{\sin }^{2}2\theta =-k{\cos }^{2}2\theta$ ..... $[\because \sin 2\theta =2\sin \theta \cos \theta ]$

$\Rightarrow k=\frac{3}{4}{\tan }^{2}2\theta$