MCQ
$\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$ is equal to
  • $\cot \left(7 \frac{1}{2}\right)^{\circ}$
  • B
    $\sin \left(7 \frac{1}{2}\right)^{\circ}$
  • C
    $\sin 15^{\circ}$
  • D
    $\cos 15^{\circ}$

Answer

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

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