Q99 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics

$\tan 105^\circ = $ ......

  • A
    $\dfrac{\sqrt{3} - 1}{2\sqrt{2}}$
  • B
    $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}$
  • Correct answer
    C
    $\dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}$
  • D
    $\dfrac{2\sqrt{2}}{\sqrt{3} + 1}$

Solution

$\tan 105^\circ = \tan(60^\circ + 45^\circ)$

Using the identity $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$

Substituting $A = 60^\circ$ and $B = 45^\circ$:

$\tan 105^\circ = \dfrac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ}$

Since $\tan 60^\circ = \sqrt{3}$ and $\tan 45^\circ = 1$:

$\tan 105^\circ = \dfrac{\sqrt{3} + 1}{1 - \sqrt{3}(1)} = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}$

Answer: $\dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}$

Hence, the correct answer is option: 'C'.

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