Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #1671
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
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) $\frac{\pi}{2}$
Correct Answer: B

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Detailed Solution

Step 1: Evaluate the first term

We know that the principal value branch of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$. Since $\sec(\frac{\pi}{4}) = \sqrt{2}$, we have: $$\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$$

Step 2: Evaluate the second term

We use the property $\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}(x)$. Thus: $$\text{cosec}^{-1}(-\sqrt{2}) = -\text{cosec}^{-1}(\sqrt{2})$$ Since $\text{cosec}(\frac{\pi}{4}) = \sqrt{2}$, we have $\text{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$. Therefore: $$\text{cosec}^{-1}(-\sqrt{2}) = -\frac{\pi}{4}$$

Step 3: Combine the results

Substitute the values back into the original expression: $$\frac{\pi}{4} + 2\left(-\frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2\pi}{4} = -\frac{\pi}{4}$$

Final Answer: -\frac{\pi}{4}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the specific domain and range properties of inverse trigonometric functions to solve a composite expression.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a sequence of steps involving trigonometric identities and principal value definitions.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to handle negative arguments in inverse functions, a common area for conceptual errors in the Inverse Trigonometric Functions chapter.