All the points (x, y) in the plane satisfying | vedclass

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Question

(a)

We have, $x^2+2 x \sin x y+1=0$

$\Rightarrow x^2+2 x \sin x y+\sin ^2 x y+1-\sin ^2 x y=0$

$\Rightarrow \quad(x+\sin x y)^2+\cos ^2 x y=0

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a pair of straight lines

Vedclass · Answer

(a)

We have, $x^2+2 x \sin x y+1=0$

$\Rightarrow x^2+2 x \sin x y+\sin ^2 x y+1-\sin ^2 x y=0$

$\Rightarrow \quad(x+\sin x y)^2+\cos ^2 x y=0$

$	herefore x+\sin x y=0$ and $\cos ^2 x y=0$

$\begin{aligned} \cos ^2 x y &=0 \ \Rightarrow \quad x y &=(2 n+1) \frac{\pi}{2} \end{aligned}$

$\Rightarrow \quad x+1=0$

$\left[\because \cos ^2 x y=0 \Rightarrow \sin x y=1ight]$

$\Rightarrow \quad x=-1$

$	herefore \quad y=-(2 n+1) \frac{\pi}{2}$

All the points $(x, y)$ in the plane satisfying the equation $x^2+2 x \sin (x y)+1=0$ lie on

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