Class CBSE Class 12 Mathematics Derivatives Q #1297
KNOWLEDGE BASED
REMEMBER
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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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: PROCEDURAL
Justification: The question requires a series of steps involving trigonometric substitutions, application of trigonometric identities, and differentiation to arrive at the solution. These are all procedural steps.
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.

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