Question

The principal value of \(\sec^{-1}(\sqrt{2}) + 2 \text{cosec^{-1}(-\sqrt{2})\) is :}

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

Hint:

Memorize the principal value ranges for all six inverse trigonometric functions.
Remember \(\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}(x)\) but \(\sec^{-1}(-x) = \pi - \sec^{-1}(x)\).

Answer 1

The Correct Option is B

Concept:
• Principal value range of \(\sec^{-1} x\) is \([0, \pi] \setminus \{\frac{\pi}{2}\}\).
• Principal value range of \(\text{cosec}^{-1} x\) is \([-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}\).
• For \(\text{cosec}^{-1} x\), we use the identity \(\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}(x)\).

Step 1: Evaluate \(\sec^{-1}(\sqrt{2})\)
Let \(\sec^{-1}(\sqrt{2}) = y\).
Then \(\sec y = \sqrt{2}\).
Since \(\sec \frac{\pi}{4} = \sqrt{2}\) and \(\frac{\pi}{4}\) is in the principal range \([0, \pi] \setminus \{\frac{\pi}{2}\}\),
\[ \sec^{-1}(\sqrt{2}) = \frac{\pi}{4} \]

Step 2: Evaluate \(\text{cosec}^{-1}(-\sqrt{2})\)
Using the identity \(\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}(x)\):
\[ \text{cosec}^{-1}(-\sqrt{2}) = -\text{cosec}^{-1}(\sqrt{2}) \] We know \(\text{cosec} \frac{\pi}{4} = \sqrt{2}\) because \(\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\).
So, \(\text{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}\).
Thus, \(\text{cosec}^{-1}(-\sqrt{2}) = -\frac{\pi}{4}\).

Step 3: Compute the final expression value
Substitute the evaluated values into the expression:
\[ \sec^{-1}(\sqrt{2}) + 2 \text{cosec}^{-1}(-\sqrt{2}) = \frac{\pi}{4} + 2\left(-\frac{\pi}{4}\right) \] \[ = \frac{\pi}{4} - \frac{2\pi}{4} \] \[ = \frac{\pi - 2\pi}{4} = -\frac{\pi}{4} \] This matches option (B).