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If sin α + sin β = a and cos α - cos β = b, then the value of \(\tan \left( {\frac{{\alpha - \beta }}{2}} \right)\) will be
1. a + b
2. a - b
3. \(- \frac{a}{b}\)
4. \( - \frac{b}{a}\)

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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, \(\frac{{ - \sin \left( {\frac{{\alpha - \beta }}{2}} \right)}}{{\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}} = \frac{b}{a}\)

\(\tan \frac{{\left( {\alpha - \beta } \right)}}{2} = \frac{{ - b}}{a}\)

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