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यदि $0 < x, y < \pi$ तथा $\cos x+\cos y-\cos (x+y)=\frac{3}{2}$, है, तो $\sin x+\cos y$ बराबर है
$\frac{1}{2}$
$\frac{1+\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{1-\sqrt{3}}{2}$
Solution
$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$
$\cos ^{2}\left(\frac{x+y}{2}\right)-\cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)$
$+\frac{1}{4} \cdot \cos ^{2}\left(\frac{x-y}{2}\right)+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$
$\Rightarrow\left(\cos \left(\frac{x+y}{2}\right)-\frac{1}{2} \cos \left(\frac{x-y}{2}\right)\right)^{2}+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$
$\Rightarrow \sin \left(\frac{x-y}{2}\right)=0$ and $\cos \left(\frac{x+y}{2}\right)=\frac{1}{2} \cos \left(\frac{x-y}{2}\right)$
$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$
$\therefore \sin x=\frac{\sqrt{3}}{2}$
$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$
