Question

Match List-I (Inverse Trigonometric function Principal values) with List-II:

• A. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
• B. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
• C. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
• D. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Hint:

Convert sec and cosec into cos and sin before solving principal values.

Answer 1

The Correct Option is B

Concept: Use principal value ranges: \[ \sec^{-1}x \in [0,\pi],\ x\neq 0 \quad,\quad \cosec^{-1}x \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], x\neq 0 \]

Step 1: {Evaluate $\sec^{-1}(-2)$.}
\[ \sec \theta = -2 \Rightarrow \cos\theta = -\frac{1}{2} \Rightarrow \theta = \frac{2\pi}{3} \]

Step 2: {Evaluate $\cosec^{-1}(-\sqrt{2})$.}
\[ \csc\theta = -\sqrt{2} \Rightarrow \sin\theta = -\frac{1}{\sqrt{2}} \Rightarrow \theta = -\frac{\pi}{4} \]

Step 3: {Evaluate $\cosec^{-1}(2)$.}
\[ \sin\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} \]

Step 4: {Evaluate $\sec^{-1}\left(-\frac{2}{\sqrt{3}}\right)$.}
\[ \cos\theta = -\frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{5\pi}{6} \]