• # question_answer For $>0,\operatorname{t}\in \left( 0,\frac{\pi }{2} \right),$ let $x=\sqrt{{{a}^{\sin -1}}t}$  and $y=\sqrt{a{{\cos }^{-1}}\operatorname{t}}.$ Then , $1+{{\left( \frac{\operatorname{dy}}{dx} \right)}^{2}}$equals:     JEE Main  Online Paper (Held On 22 April 2013) A)  $\frac{{{x}^{2}}}{{{y}^{2}}}$                                              B)  $\frac{{{y}^{2}}}{{{x}^{2}}}$ C)  $\frac{{{x}^{2}}+{{y}^{2}}}{{{y}^{2}}}$                             D)  $\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}}$

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$$\frac{2}{y}.\frac{dy}{dx}=\frac{-\log }{\sqrt{1-{{t}^{2}}}}.\frac{dt}{dx}$ $\Rightarrow$$\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 , $1+{{\left( \frac{dy}{dx} \right)}^{2}}=1+{{\left( \frac{-y}{x} \right)}^{2}}=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}}$