MHTCET 13 Apr 2026 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

\tan 105^{\circ}=\cdots \cdots

Solution:

Concept:

Use the tangent addition formula for

A = 60^\circ

and

B = 45^\circ

Explanation:

Write

105^\circ

60^\circ + 45^\circ

Apply the formula:

\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

So,

\tan 105^\circ = \tan(60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ}

Substitute

\tan 60^\circ = \sqrt{3}

and

\tan 45^\circ = 1

\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}

Since

\sqrt{3} + 1 = 1 + \sqrt{3}

, this equals

\frac{1 + \sqrt{3}}{1 - \sqrt{3}}

Answer:

Option C:

\frac{1+\sqrt{3}}{1-\sqrt{3}}

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