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CBSE Sample Paper 2022-23 · Section B · 2 marks

Q24: Prove dydx=−1−y21−x2\frac{dy}{dx} = -\sqrt{\frac{1 - y^2}{1 - x^2}} if y1−x2+x1−y2=1y\sqrt{1 - x^2} + x\sqrt{1 - y^2} = 1

Sample 2022-23Ch 5 · Continuity and DifferentiabilityCh 2 · Inverse Trigonometric Functions
The question

If y1−x2+x1−y2=1y\sqrt{1 - x^2} + x\sqrt{1 - y^2} = 1⁠, then prove that dydx=−1−y21−x2\frac{dy}{dx} = -\sqrt{\frac{1 - y^2}{1 - x^2}}

Exam answer

Let sin⁡−1x=A\sin^{-1}x = A and sin⁡−1y=B\sin^{-1}y = B⁠. Then x=sin⁡Ax = \sin A and y=sin⁡By = \sin B⁠, and 1−x2=cos⁡A\sqrt{1 - x^2} = \cos A⁠, 1−y2=cos⁡B\sqrt{1 - y^2} = \cos B⁠.

Why?

The roots 1−x2\sqrt{1 - x^2} and 1−y2\sqrt{1 - y^2} suggest putting x=sin⁡Ax = \sin A⁠, because

1−sin⁡2A=cos⁡2A1 - \sin^2A = \cos^2A

A=sin⁡−1xA = \sin^{-1}x lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, where cos⁡A≥0\cos A \ge 0⁠. So the root is cos⁡A\cos A⁠, not −cos⁡A-\cos A⁠. The same holds for BB⁠.

y1−x2+x1−y2=1⇒ sin⁡Bcos⁡A+sin⁡Acos⁡B=1⇒ sin⁡(A+B)=1⇒ A+B=sin⁡−11=π2⇒ sin⁡−1x+sin⁡−1y=π2\begin{aligned} &y\sqrt{1 - x^2} + x\sqrt{1 - y^2} = 1 \\ \Rightarrow\ &\sin B\cos A + \sin A\cos B = 1 \\ \Rightarrow\ &\sin(A + B) = 1 \\ \Rightarrow\ &A + B = \sin^{-1}1 = \frac{\pi}{2} \\ \Rightarrow\ &\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2} \end{aligned}
Why?

The left side is the expansion of a compound angle:

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\begin{aligned} &\sin(A + B) \\ &= \sin A\cos B + \cos A\sin B \end{aligned}

Put in y=sin⁡By = \sin B⁠, 1−x2=cos⁡A\sqrt{1 - x^2} = \cos A⁠, x=sin⁡Ax = \sin A⁠, 1−y2=cos⁡B\sqrt{1 - y^2} = \cos B⁠:

y1−x2=sin⁡Bcos⁡Ax1−y2=sin⁡Acos⁡B\begin{aligned} y\sqrt{1 - x^2} &= \sin B\cos A \\ x\sqrt{1 - y^2} &= \sin A\cos B \end{aligned}

Their sum is sin⁡(A+B)\sin(A + B)⁠. Then A+B=sin⁡−1x+sin⁡−1yA + B = \sin^{-1}x + \sin^{-1}y by the first line.

Both AA and BB lie in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so A+BA + B lies in [−π,π][-\pi, \pi]⁠. There the only angle with sine 1 is π2\frac{\pi}{2}⁠.

Differentiating w.r.t. xx⁠,

11−x2+11−y2dydx=0\frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} = 0
Why?

The standard derivative is

ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}

yy is a function of xx⁠, so by the chain rule the derivative of sin⁡−1y\sin^{-1}y carries an extra factor dydx\frac{dy}{dx}⁠. The derivative of the constant π2\frac{\pi}{2} is 0.

⇒dydx=−1−y21−x2\Rightarrow \frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}
Why?

Move the first term to the right side, then multiply both sides by 1−y2\sqrt{1 - y^2}⁠.

11−y2dydx=−11−x2dydx=−11−x2×1−y2=−1−y21−x2\begin{aligned} \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} &= -\frac{1}{\sqrt{1 - x^2}} \\ \frac{dy}{dx} &= -\frac{1}{\sqrt{1 - x^2}} \times \sqrt{1 - y^2} \\ &= -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \end{aligned}

A quotient of two square roots is the square root of the quotient (for p≥0p \ge 0⁠, q>0q > 0⁠):

pq=pq\frac{\sqrt p}{\sqrt q} = \sqrt{\frac{p}{q}}

∴ dydx=−1−y21−x2\therefore\ \dfrac{dy}{dx} = -\sqrt{\dfrac{1 - y^2}{1 - x^2}}⁠. Hence proved.

Where this comes from

Chapter 5, Continuity and Differentiability: Derivative using an inverse trig substitution.

Question type: Implicit

How to spot it: click to see

Type 5. Implicit: xx and yy mixed in one equation

Spot it: an equation such as x2+xy+y2=100x^2 + xy + y^2 = 100 that you cannot easily solve for yy⁠.

Here: xx and yy are mixed under roots. Put x=sin⁡Ax = \sin A⁠, y=sin⁡By = \sin B⁠: the left side is sin⁡(A+B)=1\sin(A + B) = 1⁠, so sin⁡−1x+sin⁡−1y=π2\sin^{-1}x + \sin^{-1}y = \frac\pi2⁠. Differentiate both sides; the yy term brings dydx\frac{dy}{dx}⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 2, Inverse Trigonometric Functions.

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