Q66 / 160 · TS EAMCET 2009 · Mathematics
$\begin{aligned} \cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right) & +3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) -4 \tan ^{-1}(-1) \text { equals }\end{aligned}$
- A$\frac{19 \pi}{12}$
- B$\frac{35 \pi}{12}$
- C$\frac{47 \pi}{12}$
- Correct answerD$\frac{43 \pi}{12}$
Solution
$\begin{aligned} & \cos ^{-1}\left(-\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\begin{array}{r}\left.-\frac{1}{\sqrt{2}}\right) \\ -4 \tan ^{-1}(-1)\end{array}\right. \\ & =\pi-\cos ^{-1}\left(\frac{1}{2}\right)-2\left(\frac{\pi}{6}\right)+3\left(\pi-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) \\ & \quad+4 \tan ^{-1}(1) \\ & =\pi-\frac{\pi}{3}-\frac{\pi}{3}+3\left(\pi-\frac{\pi}{4}\right)+4 \cdot \frac{\pi}{4} \\ & =\frac{\pi}{3}+3 \cdot \frac{3 \pi}{4}+\pi \\ & =\frac{43 \pi}{12}\end{aligned}$
Hence, the correct answer is option: 'D'.