Question

If $co{s}^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=loga$ 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

The correct option is A $\frac{y}{x}$

$\frac{x^2-y^2}{x^2+y^2}=cos\left(loga\right)=k\left(say\right)$
Putting $u=\frac{y}{x}$ and applying componendo and dividendo, we have
${\left(\frac{y}{x}\right)}^2=u^2=\frac{1-k}{1+k}$
$\Rightarrow \frac{y}{x}=\sqrt{\frac{1-k}{1+k}}\Rightarrow x\frac{dy}{dx}-y=0$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$