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

N neha

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