Q1244: Evaluate : $sec^{2}(tan^{-1}\frac{1}{2})+cosec^{2}(cot^{-1}\...

Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #1244

KNOWLEDGE BASED

APPLY

2 Marks 2024 AISSCE(Board Exam) VSA

Evaluate : $sec^{2}(tan^{-1}\frac{1}{2})+cosec^{2}(cot^{-1}\frac{1}{3})$

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AI Tutor Explanation

Detailed Solution

Step 1: Simplify the first term

Let $tan^{-1}\frac{1}{2} = \theta$. Then, $tan\theta = \frac{1}{2}$. We need to find $sec^2\theta$. We know that $sec^2\theta = 1 + tan^2\theta$. Therefore, $sec^2(tan^{-1}\frac{1}{2}) = sec^2\theta = 1 + tan^2\theta = 1 + (\frac{1}{2})^2 = 1 + \frac{1}{4} = \frac{5}{4}$.

Step 2: Simplify the second term

Let $cot^{-1}\frac{1}{3} = \phi$. Then, $cot\phi = \frac{1}{3}$. We need to find $cosec^2\phi$. We know that $cosec^2\phi = 1 + cot^2\phi$. Therefore, $cosec^2(cot^{-1}\frac{1}{3}) = cosec^2\phi = 1 + cot^2\phi = 1 + (\frac{1}{3})^2 = 1 + \frac{1}{9} = \frac{10}{9}$.

Step 3: Add the two terms

Now, we add the two simplified terms: $sec^{2}(tan^{-1}\frac{1}{2})+cosec^{2}(cot^{-1}\frac{1}{3}) = \frac{5}{4} + \frac{10}{9} = \frac{5 \times 9 + 10 \times 4}{36} = \frac{45 + 40}{36} = \frac{85}{36}$.

Final Answer: 85/36

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Pedagogical Audit

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

Knowledge Dimension: PROCEDURAL

Justification: The student needs to follow a specific procedure involving trigonometric identities and inverse trigonometric functions to solve the problem.

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 trigonometric identities, which are core concepts in the syllabus.