If $$sin$$$$-1x2-y2x2+y2=log$$ a then dydxis

Given $$sin$$$$-1x2-y2x2+y2=log$$ aApply sine function on both $$sidesx2-y2x2+y2=$$$$sin$$ $$(log$$ $$a)Apply$$ Componendo and Dividendo $$rule2x2-2y2=$$$$sin$$$$(log$$ $$a)+1sin(log$$ $$a)-1x2y2=1+sinloga1-sinlogaObserving$$ that the right hand side of the above equation is $$constantHencey2x2=kDifferentiate$$ both sides with respect to $$xx22ydydx-2xy2x4=0xdydx-y=0dydx=yx$$

Questions

If $\sin^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a then \frac{dy}{dx}$ is

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\frac{x}{y}

\frac{y}{x^{2}}

\frac{x^{2} - y^{2}}{x^{2} + y^{2}}

\frac{y}{x}

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detailed solution

Correct option is D

Given $\sin^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a$

Apply sine function on both sides

$\frac{x^{2} - y^{2}}{x^{2} + y^{2}} = \sin (\log a)$

Apply Componendo and Dividendo rule

$\begin{array}{rcl} \frac{2 x^{2}}{- 2 y^{2}} & = & \frac{\sin (\log a) + 1}{\sin (\log a) - 1} \\ \frac{x^{2}}{y^{2}} & = & \frac{1 + \sin (\log a)}{1 - \sin (\log a)} \end{array}$

Observing that the right hand side of the above equation is constant

Hence

$\frac{y^{2}}{x^{2}} = k$

Differentiate both sides with respect to $x$

$\begin{array}{rcl} \frac{x^{2} (2 y \frac{d y}{d x}) - 2 x (y^{2})}{x^{4}} & = & 0 \\ x \frac{d y}{d x} - y & = & 0 \\ \frac{d y}{d x} & = & \frac{y}{x} \end{array}$