Question

For $a>0,t\in \left(0,\frac{\pi }{2}\right),$ let $x=\sqrt{a^{{\sin }^{-1}t}}$ and $y=\sqrt{a^{{\cos }^{-1}t}},$ Then $1+{\left(\frac{dy}{dx}\right)}^2$ equals :

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

Answer 1

Answer: (D)

Let $x=\sqrt{a^{{\sin }^{-1}t}}$
$\Rightarrow x^2=a^{{\sin }^{-1}t}\Rightarrow 2\log x={\sin }^{-1}t.\log a$
$\Rightarrow \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}}.\frac{dt}{dx}$
$\Rightarrow \frac{2\sqrt{1-t^2}}{x\log a}=\frac{dt}{dx}$ ....(1)
Now, let $y=\sqrt{a^{{\cos }^{-1}t}}$
$\Rightarrow 2\log y={\cos }^{-1}t.\log a$
$\Rightarrow \frac{2}{y}.\frac{dy}{dx}=\frac{-\log a}{\sqrt{1-t^2}}.\frac{dt}{dx}$
$\Rightarrow \frac{2}{y}.\frac{dy}{dx}=\frac{-\log a}{\sqrt{1-t^2}}\times \frac{2\sqrt{1-t^2}}{x\log a}$ (from (1)
$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$
Hence , $1+{\left(\frac{dy}{dx}\right)}^2=1+{\left(\frac{-y}{x}\right)}^2=\frac{x^2+y^2}{x^2}$