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Mathematics Question on Inverse Trigonometric Functions

cos−1(−12)−2sin−1(12)+3cos−1(−12)−4tan−1(−1)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 }\left(-1\right) equals

A

19π12\frac{19\pi}{12}

B

35π12\frac{35\pi}{12}

C

47π12\frac{47\pi}{12}

D

43π12\frac{43\pi}{12}

Answer

43π12\frac{43\pi}{12}

Explanation

Solution

The correct answer is D:43π12\frac{43\pi}{12}
cos⁡−1(−12)−2sin⁡−1(12)+3cos⁡−1(−12)−4tan⁡−1(−1)\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)
=π−cos⁡−1(12)−2(π6)+3(π−cos⁡−1(12))+4tan⁡−1(1)=\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)+4 \tan ^{-1}(1)
=π−π3−π3+3(π−π4)+4.π4=\pi-\frac{\pi}{3}-\frac{\pi}{3}+3\left(\pi-\frac{\pi}{4}\right)+4 . \frac{\pi}{4}
=π3+3π−3π4+π=\frac{\pi}{3}+3 \pi-\frac{3 \pi}{4}+\pi
=43π12=\frac{43 \pi}{12}