If cos x + sin x = frac{1}{2} and 0

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(D) Given $\cos x + \sin x = \frac{1}{2}$.Squaring both sides,we get $(\cos x + \sin x)^2 = (\frac{1}{2})^2$.$\cos^2 x + \sin^2 x + 2 \sin x \cos x =

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$\frac{-4-\sqrt{7}}{3}$

Vedclass · Answer

(D) Given $\cos x + \sin x = \frac{1}{2}$.Squaring both sides,we get $(\cos x + \sin x)^2 = (\frac{1}{2})^2$.$\cos^2 x + \sin^2 x + 2 \sin x \cos x = \frac{1}{4}$.Since $\cos^2 x + \sin^2 x = 1$,we have $1 + 2 \sin x \cos x = \frac{1}{4}$.$2 \sin x \cos x = \frac{1}{4} - 1 = -\frac{3}{4}$.Now,$(\cos x - \sin x)^2 = \cos^2 x + \sin^2 x - 2 \sin x \cos x = 1 - (-\frac{3}{4}) = 1 + \frac{3}{4} = \frac{7}{4}$.So,$\cos x - \sin x = \pm \frac{\sqrt{7}}{2}$.Case $1$: $\cos x + \sin x = \frac{1}{2}$ and $\cos x - \sin x = \frac{\sqrt{7}}{2}$.Adding gives $2 \cos x = \frac{1+\sqrt{7}}{2} \implies \cos x = \frac{1+\sqrt{7}}{4}$.Subtracting gives $2 \sin x = \frac{1-\sqrt{7}}{2} \implies \sin x = \frac{1-\sqrt{7}}{4}$.Then $	an x = \frac{\sin x}{\cos x} = \frac{1-\sqrt{7}}{1+\sqrt{7}} = \frac{(1-\sqrt{7})^2}{1-7} = \frac{1+7-2\sqrt{7}}{-6} = \frac{8-2\sqrt{7}}{-6} = \frac{-4+\sqrt{7}}{3}$.Case $2$: $\cos x + \sin x = \frac{1}{2}$ and $\cos x - \sin x = -\frac{\sqrt{7}}{2}$.Adding gives $2 \cos x = \frac{1-\sqrt{7}}{2} \implies \cos x = \frac{1-\sqrt{7}}{4}$.Subtracting gives $2 \sin x = \frac{1+\sqrt{7}}{2} \implies \sin x = \frac{1+\sqrt{7}}{4}$.Then $	an x = \frac{\sin x}{\cos x} = \frac{1+\sqrt{7}}{1-\sqrt{7}} = \frac{(1+\sqrt{7})^2}{1-7} = \frac{1+7+2\sqrt{7}}{-6} = \frac{8+2\sqrt{7}}{-6} = \frac{-4-\sqrt{7}}{3}$.Since $0  0$. In Case $1$,$\sin x = \frac{1-\sqrt{7}}{4}  0$,which is accepted. Thus,$	an x = \frac{-4-\sqrt{7}}{3}$.

If $\cos x + \sin x = \frac{1}{2}$ and $0 < x < \pi$,then $\tan x =$

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