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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
$\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]=$
  • A
    $\pi-x$
  • B
    $2 \pi-x$
  • C
    $\frac{x}{2}$
  • ✓
    $\pi-\frac{x}{2}$
✓

Answer

Correct option: D.
$\pi-\frac{x}{2}$
(D) $\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$
$=\cot ^{-1}\left[\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}-\sqrt{1+\sin x})}\right.$ $\left.\times \frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}\right]$
$=\cot ^{-1}\left[\frac{(1-\sin x)+(1+\sin x)+2 \sqrt{1-\sin ^2 x}}{(1-\sin x)-(1+\sin x)}\right]$
$=\cot ^{-1}\left[\frac{2(1+\cos x)}{-2 \sin x}\right]$
$=\cot ^{-1}\left[-\frac{2 \cos ^2\left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}\right]$
$=\cot ^{-1}\left(-\cot \frac{x}{2}\right)=\cot ^{-1}\left[\cot \left(\pi-\frac{x}{2}\right)\right]$
$=\pi-\frac{x}{2}$

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