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Q.
If $ \alpha , \beta \left(\alpha\ne\beta\right) $ satisfies the equation a $\cos \theta + b \sin \theta = c$, then the value of $ \left(\frac{\alpha+\beta}{2}\right) $ is
AMUAMU 2019
Solution:
We have,
$a \cos \theta+b \sin \theta=c ..... $(i)
$\because \alpha$ and $\beta(\alpha \neq \beta)$ satisfy the Eq. (i)
$\Rightarrow a \cos \alpha+b \sin \alpha=c .....$ (ii)
and $ a \cos \beta+b \sin \beta=c..... $(iii)
Now, subtracting Eq. (iii) from Eq. (ii), we get
$a \cos \alpha+b \sin \alpha-a \cos \beta-b \sin \beta=0$
$\Rightarrow a(\cos \alpha-\cos \beta)+b(\sin \alpha-\sin \beta)=0$
$\Rightarrow a(\cos \alpha-\cos \beta)=-b(\sin \alpha-\sin \beta)$
$\Rightarrow a \sin \frac{\alpha+\beta}{2}=-b\left[-\cos \left(\frac{\alpha+\beta}{2}\right)\right]$
$\Rightarrow a \sin \frac{(\alpha+\beta)}{2}=b \cos \frac{(\alpha+\beta)}{2}$
$\Rightarrow \tan \left(\frac{\alpha+\beta}{2}\right)=\frac{b}{a}$