Q1371: If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$, then prove that $...

Class CBSE Class 12 Mathematics Derivatives Q #1371

KNOWLEDGE BASED

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5 Marks 2025 AISSCE(Board Exam) LA

If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$, then prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$.

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AI Tutor Explanation

Detailed Solution

Step 1: Substitution

Let $x = \sin A$ and $y = \sin B$. Then, $\sqrt{1-x^2} = \cos A$ and $\sqrt{1-y^2} = \cos B$. The given equation becomes: $\cos A + \cos B = a(\sin A - \sin B)$

Step 2: Applying Trigonometric Identities

Using the sum-to-product identities: $2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2}) = a \cdot 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$ $\cos(\frac{A+B}{2}) \cos(\frac{A-B}{2}) = a \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$

Step 3: Simplifying the Equation

Assuming $\cos(\frac{A+B}{2}) \neq 0$, we can divide both sides by $2\cos(\frac{A+B}{2})$: $\cos(\frac{A-B}{2}) = a \sin(\frac{A-B}{2})$ $\cot(\frac{A-B}{2}) = a$ $\frac{A-B}{2} = \cot^{-1}(a)$ $A - B = 2 \cot^{-1}(a)$

Step 4: Expressing A and B in terms of x and y

Since $x = \sin A$ and $y = \sin B$, we have $A = \sin^{-1}(x)$ and $B = \sin^{-1}(y)$. Substituting these into the equation: $\sin^{-1}(x) - \sin^{-1}(y) = 2 \cot^{-1}(a)$

Step 5: Differentiating with respect to x

Differentiating both sides with respect to $x$: $\frac{d}{dx}(\sin^{-1}(x) - \sin^{-1}(y)) = \frac{d}{dx}(2 \cot^{-1}(a))$ $\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} = 0$

Step 6: Solving for dy/dx

Rearranging the equation to solve for $\frac{dy}{dx}$: $\frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$ $\frac{dy}{dx} = \sqrt{\frac{1-y^2}{1-x^2}}$

Final Answer: $\frac{dy}{dx} = \sqrt{\frac{1-y^{2}}{1-x^{2}}}$

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall trigonometric identities and differentiation rules to solve the problem.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the relationships between trigonometric functions, inverse trigonometric functions, and differentiation rules. It involves applying these concepts to derive the required result.

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 differentiation. It is a standard problem type covered in textbooks.