Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #1266
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2024 AISSCE(Board Exam) VSA
Find the value of $tan^{-1}(-\frac{1}{\sqrt{3}})+cot^{-1}(\frac{1}{\sqrt{3}})+tan^{-1}[sin(-\frac{\pi}{2})]$
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Detailed Solution

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

We know that $tan^{-1}(-x) = -tan^{-1}(x)$. Therefore, $tan^{-1}(-\frac{1}{\sqrt{3}}) = -tan^{-1}(\frac{1}{\sqrt{3}})$. Since $tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, we have $tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. Thus, $tan^{-1}(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}$.

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

Since $cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$, we have $cot^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{3}$.

Step 3: Evaluate $sin(-\frac{\pi}{2})$

We know that $sin(-\frac{\pi}{2}) = -sin(\frac{\pi}{2}) = -1$.

Step 4: Evaluate $tan^{-1}[sin(-\frac{\pi}{2})]$

We have $tan^{-1}[sin(-\frac{\pi}{2})] = tan^{-1}(-1)$. Since $tan(-\frac{\pi}{4}) = -1$, we have $tan^{-1}(-1) = -\frac{\pi}{4}$.

Step 5: Combine the results

We need to find the value of $tan^{-1}(-\frac{1}{\sqrt{3}})+cot^{-1}(\frac{1}{\sqrt{3}})+tan^{-1}[sin(-\frac{\pi}{2})]$. Substituting the values we found: $$-\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} = \frac{-2\pi + 4\pi - 3\pi}{12} = \frac{-\pi}{12}$$

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

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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 their values for specific angles to solve the problem.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of the concepts of inverse trigonometric functions, their properties, and the relationship between trigonometric functions and their inverses.
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, which is a core part of the syllabus.

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