• # question_answer The angle of intersection of the curves ${{y}^{2}}=2x/\pi$ and $y=\sin x$, is                [Roorkee 1998] A)            ${{\cot }^{-1}}(-1/\pi )$           B)            ${{\cot }^{-1}}\pi$ C)            ${{\cot }^{-1}}(-\pi )$              D)            ${{\cot }^{-1}}(1/\pi )$

The curves ${{y}^{2}}=2x/\pi$ and $y=\sin x$ intersect at $(0,0)$ and $(\pi /2,\,1)$. Let the gradients of the tangents to the curves be ${{m}_{1}}$ and ${{m}_{2}}$respectively. Then ${{m}_{1}}=\frac{dy}{dx}=\frac{1}{\pi y}$ and ${{m}_{2}}=\frac{dy}{dx}=\cos x$            At $(\pi /2\,,1)$, ${{m}_{1}}=\frac{1}{\pi }$,                   ${{m}_{2}}=\cos \frac{\pi }{2}=0$                    Thus $\tan \theta =\frac{(1/\pi )-0}{1+(1/\pi )(0)}=\frac{1}{\pi }$Þ$\theta ={{\cot }^{-1}}\pi$.