Question

\(\cos^{-1}\left(-\frac{1}{2}\right)-2\sin^{-1}\left(\frac{1}{2}\right)+3\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)-4\tan^{-1}(-1)\) equals

• A. \(\frac{19\pi}{12}\)
• B. \(\frac{35\pi}{12}\)
• C. \(\frac{47\pi}{12}\)
• D. \(\frac{43\pi}{12}\)

Hint:

Always use principal values: \(\cos^{-1}(x)\in[0,\pi]\), \(\sin^{-1}(x)\in[-\pi/2,\pi/2]\), \(\tan^{-1}(x)\in[-\pi/2,\pi/2]\).

Answer 1

The Correct Option is D

Step 1: Evaluate each inverse trig value.
\[ \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3} \]
\[ \sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \]
\[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4} \]
\[ \tan^{-1}(-1)=-\frac{\pi}{4} \]
Step 2: Substitute in expression.
\[ \frac{2\pi}{3}-2\left(\frac{\pi}{6}\right)+3\left(\frac{3\pi}{4}\right)-4\left(-\frac{\pi}{4}\right) \]
Step 3: Simplify each part.
\[ \frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3} \]
\[ 3\cdot\frac{3\pi}{4}=\frac{9\pi}{4} \]
\[ -4\left(-\frac{\pi}{4}\right)=+\pi \]
So total:
\[ \frac{\pi}{3}+\frac{9\pi}{4}+\pi \]
Step 4: Take LCM 12.
\[ \frac{\pi}{3}=\frac{4\pi}{12},\quad \frac{9\pi}{4}=\frac{27\pi}{12},\quad \pi=\frac{12\pi}{12} \]
\[ \Rightarrow \frac{4\pi+27\pi+12\pi}{12}=\frac{43\pi}{12} \]
Final Answer:
\[ \boxed{\frac{43\pi}{12}} \]