Question

If ${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{y}{x}$
• B. $\frac{x}{y}$
• C. $\frac{x^2}{y^2}$
• D. $\frac{y^2}{x^2}$

Answer 1

Answer: (A)

$\frac{y}{x}$

Given ${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$

$\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\cos \log a=A$ (Say)

Substitute $u=\frac{y}{x}$ and applying componendo and dividendo, we have

${\left(\frac{y}{x}\right)}^2=u^2=\left(\frac{1-A}{1+A}\right)$

$\Rightarrow \frac{y}{x}=\sqrt{\frac{1-A}{1+A}}$

$\Rightarrow x\frac{dy}{dx}-y=0$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}.$