For a>0, t ∈ (0, π 2 ), let x=√a^sin ^-1 t and y=√a^cos ^-1 t, Then, 1+ ( d y d x )^2 equals :

$$\begin{aligned} & \text { Let } x=\sqrt{a^{\sin ^{-1} t}} \\ & \Rightarrow \quad x^2=a^{\sin ^{-1} t} \Rightarrow 2 \log x=\sin ^{-1} t \cdot \log a \\ & \Rightarrow \quad \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x} \\ & \Rightarrow \quad \frac{2 \sqrt{1-t^2}}{x \log a}=\frac{d t}{d x} \end{aligned}$$​ Let x=asin−1t​⇒x2=asin−1t⇒2logx=sin−1t⋅loga⇒x2​=1−t2​loga​⋅dxdt​⇒xloga21−t2​​=dxdt​​ Now, let $y=\sqrt{a^{\cos ^{-1} t}}$y=acos−1t​ $$\begin{aligned} & \Rightarrow 2 \log y=\cos ^{-1} t \cdot \log a \\ & \Rightarrow \quad \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x} \end{aligned}$$​⇒2logy=cos−1t⋅loga⇒y2​⋅dxdy​=1−t2​−loga​⋅dxdt​​ $\Rightarrow \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \times \frac{2 \sqrt{1-t^2}}{x \log a} \quad$⇒y2​⋅dxdy​=1−t2​−loga​×xloga21−t2​​ (from(1) $$\Rightarrow \frac{d y}{d x}=-\frac{y}{x}$$⇒dxdy​=−xy​ Hence, $1+\left(\frac{d y}{d x}\right)^2=1+\left(\frac{-y}{x}\right)^2=\frac{x^2+y^2}{x^2}$1+(dxdy​)2=1+(x−y​)2=x2x2+y2​