If y sqrt{1-x^{2}}+x sqrt{1-y^{2}}=1,then fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$. Substitute $x = \sin \alpha$ and $y = \sin \beta$,where $\alpha = \sin^{-1} x$ and $\bet

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$. Substitute $x = \sin \alpha$ and $y = \sin \beta$,where $\alpha = \sin^{-1} x$ and $\beta = \sin^{-1} y$. The equation becomes $\sin \beta \cos \alpha + \sin \alpha \cos \beta = 1$. Using the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,we get $\sin(\alpha + \beta) = 1$. Thus,$\alpha + \beta = \sin^{-1}(1) = \frac{\pi}{2}$. Substituting back,we have $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(\frac{\pi}{2})$. $\frac{1}{\sqrt{1-x^{2}}} + \frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = 0$. Rearranging the terms: $\frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^{2}}}$. Therefore,$\frac{dy}{dx} = -\sqrt{\frac{1-y^{2}}{1-x^{2}}}$.

If $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$,then $\frac{dy}{dx}=$

Similar Questions

If $x^{3}+y^{3}-3 a x y=0$,then $\frac{d y}{d x}$ equals

If $x e^{xy} = y + \sin^2 x$, then the value of $\frac{dy}{dx}$ at $x = 0$ is equal to

If $\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$,then $\frac{dy}{dx} = \dots$

If $\sin x \sqrt{\cos y} - \cos y \sqrt{\sin x} = 0$,then $\frac{dy}{dx} = $

Let $f(x)=x^5+2x^3+3x+1$,$x \in R$,and $g(x)$ be a function such that $g(f(x))=x$ for all $x \in R$. Then $\frac{g(7)}{g^{\prime}(7)}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module