Question

\[ \tan^{-1}(1)+\cos^{-1}\left(-\frac12\right)+\sin^{-1}\left(-\frac12\right) = \]

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{3\pi}{4}\)
• D. \(\frac{3\pi}{2}\)

Hint:

Always remember: \[ \sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}. \] This identity frequently appears in objective questions involving inverse trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Apply the inverse trigonometric identity. Using \[ \sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}, \] with \[ x=-\frac12, \] we obtain \[ \cos^{-1}\left(-\frac12\right) + \sin^{-1}\left(-\frac12\right) = \frac{\pi}{2}. \]

Step 2: Evaluate \(\tan^{-1}(1)\). We know that \[ \tan\left(\frac{\pi}{4}\right)=1. \] Therefore, \[ \tan^{-1}(1)=\frac{\pi}{4}. \]

Step 3: Add the obtained values. \[ \frac{\pi}{4}+\frac{\pi}{2} = \frac{\pi}{4}+\frac{2\pi}{4} = \frac{3\pi}{4}. \]

Step 4: Verification. \[ \cos^{-1}\left(-\frac12\right)=\frac{2\pi}{3}, \qquad \sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}. \] Hence, \[ \frac{2\pi}{3}-\frac{\pi}{6} = \frac{\pi}{2}, \] which confirms the result. Conclusion: Therefore, \[ \tan^{-1}(1) + \cos^{-1}\left(-\frac12\right) + \sin^{-1}\left(-\frac12\right) = {\frac{3\pi}{4}}. \] Hence, the correct option is \[ {(C)}. \]