Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #1311
KNOWLEDGE BASED
REMEMBER
2 Marks 2024 AISSCE(Board Exam) VSA
Find the principal value of $tan^{-1}(1)+cos^{-1}(-\frac{1}{2})+sin^{-1}(-\frac{1}{\sqrt{2}})$

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Detailed Solution

Step 1: Find the principal value of $tan^{-1}(1)$

The principal value of $tan^{-1}(1)$ is the angle $\theta$ in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $tan(\theta) = 1$. This occurs at $\theta = \frac{\pi}{4}$.

Step 2: Find the principal value of $cos^{-1}(-\frac{1}{2})$

The principal value of $cos^{-1}(-\frac{1}{2})$ is the angle $\theta$ in the range $[0, \pi]$ such that $cos(\theta) = -\frac{1}{2}$. This occurs at $\theta = \frac{2\pi}{3}$.

Step 3: Find the principal value of $sin^{-1}(-\frac{1}{\sqrt{2}})$

The principal value of $sin^{-1}(-\frac{1}{\sqrt{2}})$ is the angle $\theta$ in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $sin(\theta) = -\frac{1}{\sqrt{2}}$. This occurs at $\theta = -\frac{\pi}{4}$.

Step 4: Add the principal values

Now, we add the principal values we found: $$tan^{-1}(1) + cos^{-1}(-\frac{1}{2}) + sin^{-1}(-\frac{1}{\sqrt{2}}) = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{4} = \frac{2\pi}{3}$$

Final Answer: $\frac{2\pi}{3}$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the principal value ranges and standard values of inverse trigonometric functions.
Knowledge Dimension: PROCEDURAL
Justification: The question requires applying a specific procedure to find the principal values of inverse trigonometric functions and then summing them.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the knowledge of inverse trigonometric functions and their principal values as covered in the textbook.