Question

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

• A. b / a
• B. c / a
• C. a / b
• D. c / b

Answer 1

Answer: (A)

We have,
$a\cos \theta +b\sin \theta =c.....$(i)
$\because \alpha$ and $\beta (\alpha \ne \beta )$ 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
$a\cos \alpha +b\sin \alpha -a\cos \beta -b\sin \beta =0$
$\Rightarrow a(\cos \alpha -\cos \beta )+b(\sin \alpha -\sin \beta )=0$
$\Rightarrow a(\cos \alpha -\cos \beta )=-b(\sin \alpha -\sin \beta )$
$\Rightarrow a\sin \frac{\alpha +\beta }{2}=-b\left[-\cos \left(\frac{\alpha +\beta }{2}\right)\right]$
$\Rightarrow a\sin \frac{(\alpha +\beta )}{2}=b\cos \frac{(\alpha +\beta )}{2}$
$\Rightarrow \tan \left(\frac{\alpha +\beta }{2}\right)=\frac{b}{a}$