Q585: $[\sec^{-1}(-\sqrt{2})-\tan^{-1}(\frac{1}{\sqrt{3}})]$ is ...

Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #585

KNOWLEDGE BASED

APPLY

1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE

$[\sec^{-1}(-\sqrt{2})-\tan^{-1}(\frac{1}{\sqrt{3}})]$ is equal to:

(A) $\frac{11\pi}{12}$

(B) $\frac{5\pi}{12}$

(D) $\frac{7\pi}{12}$

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AI Tutor Explanation

Detailed Solution

Step 1: Evaluate $\sec^{-1}(-\sqrt{2})$

We need to find the angle whose secant is $-\sqrt{2}$. Since the range of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$, we are looking for an angle in the second quadrant where secant is negative. $\sec(\frac{3\pi}{4}) = -\sqrt{2}$, so $\sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}$.

Step 2: Evaluate $\tan^{-1}(\frac{1}{\sqrt{3}})$

We need to find the angle whose tangent is $\frac{1}{\sqrt{3}}$. Since the range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, we are looking for an angle in the first quadrant where tangent is positive. $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, so $\tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.

Step 3: Calculate the difference

Now, we subtract the two values: $\sec^{-1}(-\sqrt{2}) - \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{3\pi}{4} - \frac{\pi}{6}$

Step 4: Simplify the expression

To subtract the fractions, we need a common denominator, which is 12. $\frac{3\pi}{4} - \frac{\pi}{6} = \frac{9\pi}{12} - \frac{2\pi}{12} = \frac{7\pi}{12}$

Final Answer: $\frac{7\pi}{12}$

AI Suggestion: Option D

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply their knowledge of inverse trigonometric functions and their ranges to evaluate the given expression.

Knowledge Dimension: PROCEDURAL

Justification: The student needs to follow a specific procedure to evaluate the inverse trigonometric functions and then perform the subtraction. This involves knowing the ranges of the inverse trigonometric functions and applying the correct formulas.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of inverse trigonometric functions, which is a core topic in the syllabus.