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MHTCET 10 May 2024 Shift 2 Paper

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Question : 115 of 150

Marks: +1, -0

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

1
2
3
4

Solution:

Concept:

Use the tangent addition formula after simplifying the inverse trigonometric angles.

Explanation:

Let

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

and

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

.

Since

\cos \theta = \frac{1}{\sqrt{2}}

, we get

\theta = \frac{\pi}{4}

, so

\tan \theta = 1

.

Also,

\tan \phi = \frac{1}{2}

.

Now, the given expression is

\tan(\theta + \phi)

.

Using the formula

\tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}

:

\tan(\theta + \phi) = \frac{1 + \frac{1}{2}}{1 - 1 \cdot \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3

.

Answer:

3

, i.e. Option C.

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