Question

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha - \cos \beta = b \text{ then }  \tan \frac{\alpha - \beta}{2} =\]

 

विकल्प

• \[- \frac{a}{b}\]

   

• \[- \frac{b}{a}\]

   

• \[\sqrt{a^2 + b^2}\]

   

• none of these

Answer 1

\[- \frac{b}{a}\]

\[\text{ Given } : \]
\[sin\alpha + sin\beta = a\]
\[ \Rightarrow 2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = a . . . (1)\]
\[\text{ Also } , \]
\[cos\alpha + cos\beta = b\]
\[ \Rightarrow - 2\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2} = b . . . (2)\]
\[\text{ On dividing (1) by (2), we get} \]
\[\frac{- \cos\frac{\alpha - \beta}{2}}{\sin\frac{\alpha - \beta}{2}} = \frac{a}{b}\]
\[ \Rightarrow \frac{- \sin\frac{\alpha - \beta}{2}}{\cos\frac{\alpha - \beta}{2}} = \frac{b}{a}\]
\[ \Rightarrow \tan\frac{\alpha - \beta}{2} = - \frac{b}{a}\]