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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

^ -1 2+ ^ -1 (-3)+ ^ -1 (1)/(3)+ ^ -1 (-(1)/(2))=

\tan ^{-1} 2+\cot ^{-1}(-3)+\cot ^{-1} \frac{1}{3}+\tan ^{-1}(−12)=\left(-\frac{1}{2}\right)=

Options

  1. A−3π2-\frac{3 \pi}{2}
  2. Bπ2\frac{\pi}{2}
  3. C−π2-\frac{\pi}{2}
  4. D3π2\frac{3 \pi}{2}

Correct answer

D. 3π2\frac{3 \pi}{2}

Step-by-step solution

tan⁡−12+cot⁡−1(−3)+cot⁡−1(13)+tan⁡−1(−12)=tan⁡−1(2)−tan⁡−1(12)+π−cot⁡−1(3)+cot⁡−1(13)=tan⁡−1(2)−tan⁡−1(12)+π−tan⁡−1(13)+tan⁡−1(3)=tan⁡−1(2−121+2⋅12)+π+tan⁡−1(3)−tan⁡−1(13)=tan⁡−1(32)+π+tan⁡−1(3−131+3⋅13)=tan⁡−1(34)+π+tan⁡−1(86)=π+tan⁡−1(34)+tan⁡−1(43)=π+tan⁡−1(34)+cot⁡−1(34)=π+π/2=3π/2\begin{aligned} & \tan ^{-1} 2+\cot ^{-1}(-3)+\cot ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(-\frac{1}{2}\right) \\ & =\tan ^{-1}(2)-\tan ^{-1}\left(\frac{1}{2}\right)+\pi-\cot ^{-1}(3)+\cot ^{-1}\left(\frac{1}{3}\right) \\ & =\tan ^{-1}(2)-\tan ^{-1}\left(\frac{1}{2}\right)+\pi-\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}(3) \\ & =\tan ^{-1}\left(\frac{2-\frac{1}{2}}{1+2 \cdot \frac{1}{2}}\right)+\pi+\tan ^{-1}(3)-\tan ^{-1}\left(\frac{1}{3}\right) \\ & =\tan ^{-1}\left(\frac{3}{2}\right)+\pi+\tan ^{-1}\left(\frac{3-\frac{1}{3}}{1+3 \cdot \frac{1}{3}}\right) \\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\pi+\tan ^{-1}\left(\frac{8}{6}\right) \\ & =\pi+\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{4}{3}\right) \\ & =\pi+\tan ^{-1}\left(\frac{3}{4}\right)+\cot ^{-1}\left(\frac{3}{4}\right)=\pi+\pi / 2 \\ & =3 \pi / 2\end{aligned}
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