Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #944
KNOWLEDGE BASED
APPLY
2 Marks 2024 VSA
Find the value of \(\tan^{-1}(-\frac{1}{\sqrt{3}})+\cot^{-1}(\frac{1}{\sqrt{3}})+\tan^{-1}[\sin(-\frac{\pi}{2})].\)
Explanation
Let the given expression be $E$.$$E = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right]$$Evaluate each term separately using principal values:$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$(Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\tan^{-1}(-x) = -\tan^{-1}x$)$\cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$(Since $\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}$)For the third term, first evaluate the sine function:$\sin\left(-\frac{\pi}{2}\right) = -1$So, $\tan^{-1}(-1) = -\frac{\pi}{4}$Substitute these values back into the expression:$$E = -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4}$$Find a common denominator (which is 12):$$E = \frac{-2\pi + 4\pi - 3\pi}{12}$$$$E = \frac{-\pi}{12}$$

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Step-by-Step Solution

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

    Since \(\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}\), we have \(\tan^{-1}(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}\).

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

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

  3. Evaluate \(\tan^{-1}[\sin(-\frac{\pi}{2})]\):

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

  4. Combine the results:

    \(\tan^{-1}(-\frac{1}{\sqrt{3}})+\cot^{-1}(\frac{1}{\sqrt{3}})+\tan^{-1}[\sin(-\frac{\pi}{2})] = -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4}\)

  5. Simplify the expression:

    To simplify, find a common denominator, which is 12.

    \(-\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} = -\frac{2\pi}{12} + \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{-2\pi + 4\pi - 3\pi}{12} = \frac{-\pi}{12}\)

Correct Answer: -\(\frac{\pi}{12}\)

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires students to apply their knowledge of trigonometric identities and inverse trigonometric functions to simplify and solve the given expression.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to execute a sequence of steps involving trigonometric identities and inverse trigonometric functions to arrive at the solution.
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 standard formulas and procedures related to inverse trigonometric functions, as covered in the textbook.