# If  $x= \cos t+\log \tan \left ( \frac{t}{2} \right ),y= \sin t,$ then find the values of $\frac{d^{2}y}{dt^{2}} \, and\, \frac{d^{2}y}{dx^{2}}\, at\: t= \frac{\pi }{4}$

$given:\, x= \cos t+\log \tan \left ( \frac{t}{2} \right )$
Differentiating wrt t,we get
$\frac{dx}{dt}= -\sin t+\frac{1}{tan\left ( \frac{t}{2} \right )}\cdot \frac{1}{2}\sec ^{2}\frac{t}{2}$
$\frac{dx}{dt}= -\sin t+\frac{\cos \left ( \frac{t}{2} \right )}{2\sin \left ( \frac{t}{2} \right )}\times \frac{1}{\cos ^{2}\left ( \frac{t}{2} \right )}$
$= -\sin t+\frac{1}{2\sin \left ( \frac{t}{2} \right )\cos \left ( \frac{t}{2} \right )}$
$\frac{dx}{dt}= -\sin t+\frac{1}{\sin t}\left [ \sin A= 2\sin \frac{A}{2} \cos \frac{A}{2}\right ]$

$\frac{dx}{dt}=\frac{-\sin ^{2}t+1}{\sin t}$
$\Rightarrow \frac{dx}{dt}= \frac{\cos ^{2}t}{\sin t} \: --\left ( i\right )\left [ \therefore 1-\sin ^{2}t= \cos ^{2} t\right ]$
$and \: y= \sin t\left ( given \right )$
Differentiating wrt,we get
$\frac{dy}{dt}= \cos t$
$now,\frac{dy}{dx}= \frac{dy}{dt}\times \frac{dt}{dx}$
$\Rightarrow \frac{dy}{dx}=\cos t\times \frac{\sin t}{\cos ^{2}t} \left [ using\, i \ \right ]$
$\Rightarrow \frac{dy}{dx}= \tan t$
Again Differentiating both side wrt x we get
$\frac{d^{2}y}{dx^{2}}= \sec ^{2}t\frac{dt}{dx}$
$\Rightarrow \frac{d^{2}y}{dx^{2}}= \sec ^{2}t\frac{\sin t}{\cos ^{2}t} \left [ using\, equation \ 1 \right ]$
$\Rightarrow \frac{d^{2}y}{dx^{2}}= \sec ^{2}\frac{\pi }{4} \cdot \frac{\sin \frac{\pi }{4}}{\cos ^{2}\frac{\pi }{4}}$     at $t= \frac{\pi }{4}$
$\Rightarrow \left ( \sqrt{2} \right )^{2} \cdot\frac{\frac{\sqrt{2}}{2}}{\left ( \frac{1}{\sqrt{2}} \right )^{2}}\rightarrow \frac{2\times \frac{\sqrt{2}}{2}}{\frac{1}{2}}= 2\sqrt{2}$

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