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2. inverse trigonometric function — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths2. inverse trigonometric function1 Mark
MCQ
$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right) \quad$ is equal to
  • A
    $\frac{31 \pi}{12}$
  • B
    $\frac{17 \pi}{12}$
  • ✓
    $\frac{11 \pi}{12}$
  • D
    $-\frac{3 \pi}{4}$
✓

Answer

Correct option: C.
$\frac{11 \pi}{12}$
c
$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1} \tan \left(\frac{3 \pi}{4}\right)$

$\sin ^{-1} \sin \left(\frac{2 \pi}{3}\right)=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}$

$\cos ^{-1}\left(\cos \frac{2 \pi}{6}\right)=2 \pi-\frac{7 \pi}{6}=\frac{5 \pi}{6}$

$\tan ^{-1} \tan \left(\frac{3 \pi}{4}\right)=\frac{3 \pi}{4}-\pi=\frac{-\pi}{4}$

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1} \cos \frac{7 \pi}{6}+\tan ^{-1} \tan \frac{3 \pi}{4}$

$=\frac{11 \pi}{12}$

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