Q1341: If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$, prove that $\frac...

Class CBSE Class 12 Mathematics Derivatives Q #1341

KNOWLEDGE BASED

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3 Marks 2024 AISSCE(Board Exam) SA

If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$, 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 $\sqrt{1-y^2} = \cos B$.

Step 2: Rewrite the given equation

Substituting these into the given equation, we have: $\cos A + \cos B = a(\sin A - \sin B)$

Step 3: Apply trigonometric identities

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

Step 4: Simplify the equation

If $\cos(\frac{A+B}{2}) \neq 0$, 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 5: Express A and B in terms of x and y

Since $x = \sin A$ and $y = \sin B$, we have $A = \sin^{-1}(x)$ and $B = \sin^{-1}(y)$. Therefore, $\sin^{-1}(x) - \sin^{-1}(y) = 2 \cot^{-1}(a)$

Step 6: Differentiate both sides 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 7: Solve for dy/dx

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

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 it requires recalling 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. It directly tests the student's knowledge of inverse trigonometric functions and differentiation. The question is based on standard textbook problems and requires direct application of formulas and techniques taught in the syllabus.