• # question_answer If $\sin \alpha +cos\beta =\frac{1}{\sqrt{2}}$ and $\cos \alpha +\sin \beta =\sqrt{\frac{2}{3}},$ then the value of $\tan \left( \frac{\alpha -\beta }{2} \right)$ is A) $3\sqrt{3}-6$                B) $\sqrt{3}-2$ C) $4\sqrt{3}-7$    D) $4\sqrt{3}+7$

[c]  Given, $\sin \alpha +\cos \beta =\frac{1}{\sqrt{2}}$                  ? (i) $\cos \alpha +\sin \beta =\frac{\sqrt{2}}{\sqrt{3}}$                     ? (ii) On subtracting Eq. (ii) from Eq. (i), we get $(\sin \alpha -\sin \beta )+(\cos \beta -\cos \alpha )=\frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}}$ $2\cos \frac{\alpha +\beta }{2}\sin \left( \frac{\alpha -\beta }{2} \right)+2\sin \frac{\alpha +\beta }{2}$ $\sin \frac{\alpha -\beta }{2}=\frac{\sqrt{3}-2}{\sqrt{6}}$
 $2\sin \left( \frac{\alpha +\beta }{2} \right)\left[ \cos \frac{\alpha +\beta }{2}+\sin \frac{\alpha +\beta }{2} \right]$ $=\frac{\sqrt{3}-2}{\sqrt{6}}$ ? (iii) On adding Eqs. (i) and (ii), we get $2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ $=\frac{\sqrt{3}+2}{\sqrt{6}}$ $2\cos \left( \frac{\alpha -\beta }{2} \right)\left[ \sin \frac{\alpha +\beta }{2}+\cos \frac{\alpha +\beta }{2} \right]$ $=\frac{\sqrt{3}+2}{\sqrt{6}}$ ? (iv) On dividing Eq. (iii) by Eq. (iv), we get $\tan \left( \frac{\alpha -\beta }{2} \right)=\frac{\sqrt{3}-2}{\sqrt{3}+2}=4\sqrt{3}-7$