If ex=1+t-1-t1+t+1-t and tan y/2=1-t1+t then dydx at t=1/2 is

A
$- \frac{1}{2}$
B
$\frac{1}{2}$
C
0
D
2

Solution:

$\begin{array}{r} e^{x} = \frac{1 - \tan \frac{y}{2}}{1 + \tan \frac{y}{2}} = \tan (\frac{π}{4} - \frac{y}{2}) \Rightarrow y = 2 (\frac{π}{4} - \tan^{- 1} (e^{x})) \end{array}$ $\frac{d y}{d x} = 2 \cdot \frac{- 1}{1 + {(e^{x})}^{2}} \cdot e^{x} At t = \frac{1}{2}, \tan \frac{y}{2} = \sqrt{\frac{1 - \frac{1}{2}}{1 + \frac{1}{2}}} = \frac{1}{\sqrt{3}} \Rightarrow \frac{y}{2} = \frac{π}{6} \Rightarrow y = \frac{π}{3}, e^{x} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$ $\begin{array}{r} ∴ \frac{dy}{dx} = \frac{- 2 (2 - \sqrt{3})}{1 + (2 - \sqrt{3})^{2}} = \frac{- 2 (2 - \sqrt{3})}{8 - 4 \sqrt{3}} = - \frac{2}{4} = - \frac{1}{2} \end{array}$

If $e^{x} = \frac{\sqrt{1 + t} - \sqrt{1 - t}}{\sqrt{1 + t} + \sqrt{1 - t}}$ and $\tan y / 2 = \sqrt{\frac{1 - t}{1 + t}}$ then $\frac{dy}{dx}$ at $t = 1 / 2$ is

Solution:

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