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MHTCET 26 Apr 2025 Shift 1 Paper

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Question : 145 of 150

Marks: +1, -0

The number of positive integral solutions of

\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^{2}}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)

are

1
2
3
4

Solution:

Concept:

Convert inverse trigonometric terms to

\tan^{-1}

form and solve for

y

in terms of

x

.

Explanation:

Let

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

.

Then

\cos \theta = \frac{y}{\sqrt{1+y^2}}

, so

\tan \theta = \frac{1}{y}

.

Thus

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

.

Let

\sin^{-1}\left(\frac{3}{\sqrt{10}}\right) = \phi

.

Then

\sin \phi = \frac{3}{\sqrt{10}}

, so

\tan \phi = 3

.

Thus

\phi = \tan^{-1} 3

.

The given equation becomes:

\tan^{-1} x + \tan^{-1}\left(\frac{1}{y}\right) = \tan^{-1} 3

\Rightarrow \tan^{-1}\left(\frac{1}{y}\right) = \tan^{-1} 3 - \tan^{-1} x

Using the formula

\tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)

:

\tan^{-1}\left(\frac{1}{y}\right) = \tan^{-1}\left(\frac{3-x}{1+3x}\right)

\Rightarrow y = \frac{1+3x}{3-x}

For

y

to be positive,

x = 1

or

x = 2

.

When

x = 1

,

y = 2

.

When

x = 2

,

y = 7

.

Hence there are two positive integral solutions.

Answer:

B. 2

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