Correct Answer - Option 4 :
\( - \frac{b}{a}\)
Concept:
\(\sin c + \sin D = 2\sin \left( {\frac{{C + D}}{2}} \right).\cos \frac{{C - D}}{2} \to \left( 1 \right)\)
\(\cos C - \cos D = 2\sin \left( {\frac{{C + D}}{2}} \right).\sin \left( {\frac{{D - C}}{2}} \right) \to \left( 2 \right)\)
Calculation:
Given,
sin α + sin β = a and cos α – cos β = b
Using (1) & (2)
sin α + sin β = a
\(2\sin \left( {\frac{{\alpha + \beta }}{2}} \right).\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = a\) -----(1)
cos α – cos β = b
\(2\sin \left( {\frac{{\alpha + \beta }}{2}} \right).\sin \left( {\frac{{\beta - \alpha }}{2}} \right) = b\) ----- (2)
On dividing (2) by (1) we get-
\(\frac{{2\sin \left( {\frac{{\alpha + \beta }}{2}} \right).\sin \left( {\frac{{\beta - \alpha }}{2}} \right)}}{{2\sin \left( {\frac{{\alpha + \beta }}{2}} \right).\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}} = \frac{b}{a}\)
\(\frac{{\sin \left( {\frac{{\beta - \alpha }}{2}} \right)}}{{\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}} = \frac{b}{a}\) ----- (3)
As we know, sin (β – α) = -sin (α – β)
Then, equation (A) becomes, \(\frac{{ - \sin \left( {\frac{{\alpha - \beta }}{2}} \right)}}{{\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}} = \frac{b}{a}\)
\(\tan \frac{{\left( {\alpha - \beta } \right)}}{2} = \frac{{ - b}}{a}\)
