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MHTCET 5 May 2025 Shift 2 Paper

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Question : 107 of 150

Marks: +1, -0

If

x=\tan ^{-1}\left\{\frac{\sqrt{1+t^{2}}-1}{t}\right\}, y=\cos ^{-1}\left\{\frac{1-t^{2}}{1+t^{2}}\right\}, \quad

\frac{d y}{d x}

is equal to

2
$\frac{1}{2}$
4
$\frac{1}{4}$

Solution:

Concept:

Use the substitution

t = \tan\theta

to simplify both

x

and

y

, then apply the chain rule to find

\frac{dy}{dx}

.

Explanation:

Let

t = \tan\theta

, so

\theta = \tan^{-1} t

.

For

x

:

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

= \tan^{-1}\left(\frac{\sec\theta - 1}{\tan\theta}\right)

= \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)

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

= \frac{\theta}{2}

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

Therefore,

\frac{dx}{dt} = \frac{1}{2(1+t^2)}

For

y

:

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

= \cos^{-1}(\cos 2\theta)

= 2\theta

= 2\tan^{-1}t

Therefore,

\frac{dy}{dt} = \frac{2}{1+t^2}

Using the chain rule:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

= \frac{\frac{2}{1+t^2}}{\frac{1}{2(1+t^2)}}

= 4

Answer:

\frac{dy}{dx} = 4

Hence, the correct option is C.

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