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

The Correct Option is 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}} $