Question

If $\frac{{\cos }^4\theta }{x}+\frac{{\sin }^4\theta }{y}=\frac{1}{x+y}$ then $\frac{dy}{dx}=$

• A. $xy$
• B. ${\tan }^2\theta$
• C. $0$
• D. $(x^2+y^2){\sec }^2\theta$

Answer 1

Answer: (B)

${\tan }^2\theta$

Given $\frac{{\cos }^4\theta }{x}+\frac{{\sin }^4\theta }{y}=\frac{1}{x+y}$

$\Rightarrow (x+y)(\frac{{\cos }^4\theta }{x}+\frac{{\sin }^4\theta }{y})={({\cos }^2\theta +{\sin }^2\theta )}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$

$\therefore \frac{y}{x}{\cos }^4\theta +\frac{x}{y}{\sin }^4\theta -2{\sin }^2\theta {\cos }^2\theta =0$

$\Rightarrow {(\sqrt{\frac{y}{x}}{\cos }^2\theta -\sqrt{\frac{x}{y}}{\sin }^2\theta )}^2=0$

$\therefore {\tan }^2\theta =\frac{y}{x}\Rightarrow y=x{\tan }^2\theta$

$\Rightarrow \frac{dy}{dx}={\tan }^2\theta$