Question

The value of \(\tan^{-1}(-1) + \sec^{-1}(-2) + \sin^{-1}\left( \frac{1}{\sqrt{2}} \right)\)is __________.

Hint:

When evaluating inverse trigonometric functions, recall the standard values for angles like \( \tan^{-1}(-1) \), \( \sec^{-1}(-2) \), and \( \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) \).

Answer 1

Step 1: Begin by evaluating each term separately. 

Step 2: For \( \tan^{-1}(-1) \), we know that: \[ \tan^{-1}(-1) = -\frac{\pi}{4}, \] since the tangent of \( -\frac{\pi}{4} \) is \( -1 \). 

Step 3: Next, evaluate \( \sec^{-1}(-2) \). The value of \( \sec^{-1}(-2) \) corresponds to the angle \( \theta \) such that \( \sec \theta = -2 \). 
Since \( \sec \theta = \frac{1}{\cos \theta} \), we have \( \cos \theta = -\frac{1}{2} \), and the angle that satisfies this condition is \( \theta = \frac{2\pi}{3} \) (since \( \cos \frac{2\pi}{3} = -\frac{1}{2} \)). 

Step 4: Now, evaluate \( \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) \). The angle whose sine is \( \frac{1}{\sqrt{2}} \) is \( \frac{\pi}{4} \), since \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \). 

Step 5: Now, add all the results together: \[ \tan^{-1}(-1) + \sec^{-1}(-2) + \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) = -\frac{\pi}{4} + \frac{2\pi}{3} + \frac{\pi}{4}. \] 

Step 6: Simplify the expression: \[ -\frac{\pi}{4} + \frac{\pi}{4} = 0, \] so the expression becomes: \[ \frac{2\pi}{3}. \] Thus, the final value is: \[ \tan^{-1}(-1) + \sec^{-1}(-2) + \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) = \frac{2\pi}{3}. \]