Question

Evaluate: \[ \cos\left[\sin^{-1}(-1)-\tan^{-1}(-\sqrt{3})\right]. \]

Hint:

• Note the signs carefully when subtracting negative angles.
• \( \sin^{-1} \) and \( \tan^{-1} \) of negative values result in negative angles in the fourth quadrant \( (-\frac{\pi}{2}, 0) \).

Answer 1

Concept:
• Inverse function values:
• \( \sin^{-1}(-1) = -\frac{\pi}{2} \).
• \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).

Step 1: Find the values of inverse functions
Let \( \alpha = \sin^{-1}(-1) \). Since \( \sin(-\frac{\pi}{2}) = -1 \): \[ \alpha = -\frac{\pi}{2} \] Let \( \beta = \tan^{-1}(-\sqrt{3}) \). Since \( \tan(-\frac{\pi}{3}) = -\sqrt{3} \): \[ \beta = -\frac{\pi}{3} \]

Step 2: Substitute values into the expression
\[ \cos [ \alpha - \beta ] = \cos \left[ -\frac{\pi}{2} - \left( -\frac{\pi}{3} \right) \right] \] \[ = \cos \left( -\frac{\pi}{2} + \frac{\pi}{3} \right) \]

Step 3: Simplify and calculate the result
\[ \cos \left( \frac{-3\pi + 2\pi}{6} \right) = \cos \left( -\frac{\pi}{6} \right) \] Since \( \cos(-\theta) = \cos \theta \): \[ = \cos \frac{\pi}{6} \] \[ = \frac{\sqrt{3}}{2} \]