For $a0, t \in\left(0, \frac{\pi}{2}\right)$, let $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$, Then, $1+\left(\frac{d y}{d x}\right)^2$ equals :

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, $1+\left(\frac{d y}{d x}\right)^2$ equals :

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Let } x=\sqrt{a^{\sin ^{-1} t}} \ & \Rightarrow \quad x^2=a^{\sin ^{-1} t} \Rightarrow 2 \log x=\sin ^{-1} t \cdot \log a \ & \Rightarrow \quad \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x} \ & \Rightarrow \quad \frac{2 \sqrt{1-t^2}}{x \log a}=\frac{d t}{d x} \end{aligned} $ Now, let $y=\sqrt{a^{\cos ^{-1} t}}$ $ \begin{aligned} & \Rightarrow 2 \log y=\cos ^{-1} t \cdot \log a \ & \Rightarrow \quad \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x} \end{aligned} $ $\Rightarrow \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} 	imes \frac{2 \sqrt{1-t^2}}{x \log a} \quad$ (from(1) $ \Rightarrow \frac{d y}{d x}=-\frac{y}{x} $ Hence, $1+\left(\frac{d y}{d x}ight)^2=1+\left(\frac{-y}{x}ight)^2=\frac{x^2+y^2}{x^2}$

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

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