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# Find the value

Question:

If $y=\frac{\cos x-\sin x}{\cos x+\sin x}$, show that $\frac{d y}{d x}+y^{2}+1=0$

Solution:

$=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$

Formula:

$\frac{d(\sin x)}{d x}=\cos x$ and $\frac{d(\cos x)}{d x}=-\sin x$

According to the quotient rule of differentiation

If $y=u / v$

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$

$=\frac{(\cos x+\sin x) \times(-\sin x-\cos x)-(\cos x-\sin x) \times(-\sin x+\cos x)}{(\cos x+\sin x)^{2}}$

$=\frac{-(\cos x+\sin x)^{2}-(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$

$=-\frac{(\cos x+\sin x)^{2}}{(\cos x+\sin x)^{2}}-\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)^{2}}$

$=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$

$\frac{d y}{d x}+y^{2}+1=0$

HENCE PROVED.