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KCET 2017 Math Solved Paper

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Question : 34 of 60

Marks: +1, -0

If sin x =

\frac{2t}{1+t^2}

\frac{2t}{1+t^2}

\frac{dy}{dx}

is equal to

1
0
-1
2

Solution:

Given that,

\sin x = \frac{2t}{1+t^2}

and

\tan y = \frac{2t}{1-t^2}

So,

x = \sin^{-1}\left(\frac{2t}{1+t^2}\right)

→ (1)

and

y = \tan^{-1}\left(\frac{2t}{1-t^2}\right)

→ (2)

We know that,

\sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1} t

and

\tan^{-1}\left(\frac{2t}{1-t^2}\right) = 2\tan^{-1} t

So, Eqs. (1) and (2) becomes

x = 2\tan^{-1} t

and

y = 2\tan^{-1} t

Differentiating Eqs. (1) and (2) with respect to t, we get

\frac{dy}{dx} = \frac{d\left(\frac{2\tan^{-1}\theta}{d\theta}\right)}{\frac{d(2\tan^{-1}\theta)}{d\theta}} = 1

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