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Consider

$f(x)=\tan ^{-1}\left ( \sqrt{\frac{1+\sin x}{1-\sin x}} \right ),\: x\: \epsilon \left ( 0,\frac{\pi }{2} \right )$

A normal to $y= f(x)\: at\: x=\frac{\pi }{6}$

also passes through the point :

Option 1)
$(0,0)$

Option 2)
$\left ( 0,\frac{2\pi }{3} \right )$

Option 3)
$\left ( \frac{\pi }{6},0 \right )$

Option 4)
$\left ( \frac{\pi }{4} ,0\right )$

Answers (2)

As we learnt in

y-intercept -

The distance on the y-axis from the origin where the straight line cuts it.

- wherein

$y= \tan ^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}}$ $= \tan ^{-1} \sqrt{\left ( \frac{cos\frac{x}{2}+sin\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2} }\right )^2}$

$= \tan ^{-1} \: \frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}$ $= \tan ^{-1} \tan \left ( \frac{\pi }{4}+\frac{x}{2} \right )$

$y= \frac{\pi }{4}+\frac{x}{2}$

$\frac{dy}{dt}=\frac{1}{2}$

Slope of normal = $-2$

if $x= \frac{\pi }{6}$ ; $y= \frac{\pi }{3}$

$\left ( y-\frac{\pi }{3} \right )= -2 \left ( x-\frac{\pi }{6} \right )$

$\Rightarrow 2x+y=\frac{2\pi }{3}$

Option 1)

$(0,0)$

Incorrect

Option 2)

$\left ( 0,\frac{2\pi }{3} \right )$

Correct

Option 3)

$\left ( \frac{\pi }{6},0 \right )$

Incorrect

Option 4)

$\left ( \frac{\pi }{4} ,0\right )$

Incorrect

Posted by

Sabhrant Ambastha

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