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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
The value of $\tan \left(7 \frac{1}{2}\right)^{\circ}$ is equal to
  • A
    $\sqrt{6}+\sqrt{3}+\sqrt{2}-2$
  • ✓
    $\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
  • C
    $\sqrt{6}-\sqrt{3}+\sqrt{2}+2$
  • D
    $\sqrt{6}-\sqrt{3}-\sqrt{2}-2$
✓

Answer

Correct option: B.
$\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
(B)
Since, $\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}$
Putting $\frac{ A }{2}=\left(7 \frac{1}{2}\right)^{\circ}$, we get
$\frac{1-\cos \theta}{\sin \theta}=\tan \frac{\theta}{2}$, where $\theta \neq(2 n+1) \pi$
$\tan \left(7 \frac{1}{2}\right)^{\circ}=\frac{1-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{1-\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}$
$=\frac{2 \sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

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