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If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1}(a)$ then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$y / x$
Given, $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1} a$
$$
\Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\cos \left(\sin ^{-1} a\right)=c
$$
On applying componendo and dividendo law, we get
$$
\frac{2 x^2}{2 y^2}=\frac{c+1}{1-c} \Rightarrow \frac{y^2}{x^2}=\frac{1-c}{1+c}
$$
On differentiating both sides w.r.t ' $x$ ', we get
$$
2 x^2 y \frac{d y}{d x}-2 y^2 x=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}
$$
$$
\Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\cos \left(\sin ^{-1} a\right)=c
$$
On applying componendo and dividendo law, we get
$$
\frac{2 x^2}{2 y^2}=\frac{c+1}{1-c} \Rightarrow \frac{y^2}{x^2}=\frac{1-c}{1+c}
$$
On differentiating both sides w.r.t ' $x$ ', we get
$$
2 x^2 y \frac{d y}{d x}-2 y^2 x=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}
$$
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