Question

If $\frac{{\cos }^4\theta }{{\theta }_1}+\frac{{\sin }^4\theta }{{\theta }_2}=\frac{1}{{\theta }_1+{\theta }_2}$ ,then $\frac{{\theta }_2+{\theta }_1}{{\theta }_1}$ equals

• A. ${\sin }^2\theta$
• B. ${\sec }^2\theta$
• C. ${\cos }^2\theta$
• D. ${\tan }^2\theta$

Answer 1

Answer: (B)

${\sec }^2\theta$

$\frac{{\cos }^4\theta }{{\theta }_1}+\frac{{\sin }^4\theta }{{\theta }_2}=\frac{1}{{\theta }_1+{\theta }_2}\Rightarrow \frac{{\cos }^4\theta }{{\theta }_1}+\frac{{\sin }^4\theta }{{\theta }_2}=\frac{{({\cos }^2\theta +{\sin }^2\theta )}^2}{{\theta }_1+{\theta }_2}\Rightarrow \frac{{\cos }^4\theta }{{\theta }_1}+\frac{{\sin }^4\theta }{{\theta }_2}=\frac{{\cos }^4\theta +{\sin }^4\theta +2{\cos }^2\theta {\sin }^2\theta }{{\theta }_1+{\theta }_2}\Rightarrow {\cos }^4\theta (\frac{1}{{\theta }_1}-\frac{1}{{\theta }_1+{\theta }_2})+{\sin }^4\theta (\frac{1}{{\theta }_2}-\frac{1}{{\theta }_1+{\theta }_2})=\frac{2{\cos }^2\theta {\sin }^2\theta }{{\theta }_1+{\theta }_2}\Rightarrow \frac{{\theta }_2}{{\theta }_1}{\cos }^4\theta +\frac{{\theta }_1}{{\theta }_2}{\sin }^4\theta -2{\cos }^2\theta {\sin }^2\theta =0\Rightarrow {{\theta }_2}^2{\cos }^4\theta +{{\theta }_1}^2{\sin }^4\theta -2{\theta }_1{\theta }_2{\cos }^2\theta {\sin }^2\theta =0\Rightarrow {({\theta }_2{\cos }^2\theta -{\theta }_1{\sin }^2\theta )}^2=0\Rightarrow {\tan }^2\theta =\frac{{\theta }_2}{{\theta }_1}$
$\frac{{\theta }_1+{\theta }_2}{{\theta }_1}=\frac{{\theta }_1+{\theta }_1{\tan }^2\theta }{{\theta }_1}={\sec }^2\theta$
Ans: B