Correct Answer - Option 3 :
\(i\tan \left( {\frac{{\alpha - \beta }}{2}} \right)\)
Calculation:
Given, x = cos α + i sin β and y = cos β + i sin β
take, x + y = (cos α + i sin β) + (cos β + i sin β)
\(= 2cos \dfrac{α + β}{2} . cos \dfrac{α - β}{2} + i 2 sin \dfrac{α+ β}{2}. cos \dfrac{α -β}{2}\)
\(= 2cos \dfrac{α - β}{2}\left[cos \dfrac{α+β}{2}+ i sin \dfrac{α + β}{2}\right].....(1)\)
again, x - y = (cos α + i sin α) - (cos β + i sin β)
= (cos α - cos β) + i (sin α + sin β)
\(= 2sin \dfrac{α + β}{2} . sin \dfrac{ β-\alpha }{2} + 2i cos \dfrac{α+ β}{2}. sin \dfrac{α -β}{2}\)
\(= -2sin \dfrac{α + β}{2} . sin \dfrac{α - β}{2} + 2i\; cos \dfrac{α+ β}{2}. sin \dfrac{α -β}{2}\)
\(= 2i^2 sin \dfrac{α + β}{2} . sin \dfrac{α - β}{2} + 2i\; cos \dfrac{α+ β}{2}. sin \dfrac{α -β}{2}\)
\( 2 i sin \dfrac{\alpha- \beta }{2} \left[ cos \dfrac{\alpha + \beta}{2} + i sin \dfrac{ \alpha + \beta}{2} \right].....(2)\)
From equation (1) & (2)
\(\dfrac{x - y}{x + y} = \dfrac{2 i sin \left(\dfrac{\alpha + \beta}{2}\right)}{\left(2 cos\dfrac{\alpha + \beta}{2}\right)}= i \;tan \left(\dfrac{\alpha - \beta}{2}\right)\)
