Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #1357
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2025 AISSCE(Board Exam) VSA
Evaluate: $\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]$.
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Detailed Solution

Step 1: Evaluate the innermost inverse trigonometric function

We start by evaluating $\cos^{-1}\frac{\sqrt{3}}{2}$. We know that $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$. Therefore, $\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$.

Step 2: Substitute the value into the expression

Substituting this value into the given expression, we get: $\tan^{-1}\left[2\sin\left(2\cdot\frac{\pi}{6}\right)\right] = \tan^{-1}\left[2\sin\left(\frac{\pi}{3}\right)\right]$.

Step 3: Evaluate the sine function

We know that $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$. Therefore, the expression becomes: $\tan^{-1}\left[2\cdot\frac{\sqrt{3}}{2}\right] = \tan^{-1}\left[\sqrt{3}\right]$.

Step 4: Evaluate the inverse tangent function

We know that $\tan\frac{\pi}{3} = \sqrt{3}$. Therefore, $\tan^{-1}\left[\sqrt{3}\right] = \frac{\pi}{3}$.

Final Answer: $\frac{\pi}{3}$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the properties of inverse trigonometric functions and apply them in a step-by-step manner to simplify and evaluate the given expression.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to evaluate the expression, involving the application of inverse trigonometric identities and simplification steps.<\/span>
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of inverse trigonometric functions and their properties, as covered in the textbook.

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