Correct Answer - Option 1 :
\(\sqrt{\frac{3}{2}}\)
Concept:
- sin A - sin B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\sin \left ( \frac{A-B}{2} \right )\)
- cos A + cos B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)
Calculation:
Given,
cos x + cos y = \(\rm \frac{1}{\sqrt{2}}\) and sin x - sin y = \(\rm \frac{\sqrt{3}}{2}\)
We know that ,
sin A - sin B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\sin \left ( \frac{A-B}{2} \right )\)
cos A + cos B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)
∴ cos x + cos y = \(\rm \frac{1}{\sqrt{2}}\) ⇒ \(\rm 2cos\left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )\)= \(\rm \frac{1}{\sqrt{2}}\) .... (i)
and sin x - sin y = \(\rm \frac{\sqrt{3}}{2}\) ⇒ \(\rm 2cos\left ( \frac{x+y}{2} \right )\sin \left ( \frac{x-y}{2} \right )\) = \(\rm \frac{\sqrt{3}}{2}\) .... (ii)
Dividing (ii) by (i) , we get
tan \(\rm \left ( \frac{x - y}{2} \right )\) = \(\sqrt{\frac{3}{2}}\) .
The correct option is 1.