MCQ
$\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})$ is equal to
- A$\pi$
- ✓$-\frac{\pi}{2}$
- C$0$
- D$2\sqrt{3}$
✓
Answer
Correct option: B.
$-\frac{\pi}{2}$
$\tan^{-1}\sqrt{3}-\cot^{-1}\bigg(-\sqrt{3}\bigg)=\tan^{-1}\tan\frac{\pi}{3}-\cot^{-1}\bigg(-\cot\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\cot^{-1}\bigg[\cot\bigg(\pi-\frac{\pi}{6}\bigg)\bigg]=\frac{\pi}{3}-\bigg(\pi-\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}$
Therefore, option $(b)$ is correct.
$=\frac{\pi}{3}-\cot^{-1}\bigg[\cot\bigg(\pi-\frac{\pi}{6}\bigg)\bigg]=\frac{\pi}{3}-\bigg(\pi-\frac{\pi}{6}\bigg)$
$=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}$
Therefore, option $(b)$ is correct.
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