Question

The points of intersection of the lines $\frac{\phi }{r}=\cos \theta +\cos (\theta -\alpha$) and $\frac{\phi }{r}=\cos \theta +\cos (\beta -\theta$) is
Note: $\alpha \ne \beta$

• A. $\left(\frac{\phi }{2}\sec \frac{\alpha }{2}\sec \frac{\beta }{2},\frac{\alpha +\beta }{2}\right)$
• B. $\left(\sec \frac{\alpha }{2}\sec \frac{\beta }{2},\frac{\alpha +\beta }{2}\right)$
• C. $\left(\frac{\phi }{2}\cos \frac{\alpha }{2}\cos \frac{\beta }{2},\frac{\alpha +\beta }{2}\right)$
• D. $\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2},\frac{\alpha +\beta }{2}\right)$

Answer 1

The correct option is A $\left(\frac{\phi }{2}\sec \frac{\alpha }{2}\sec \frac{\beta }{2},\frac{\alpha +\beta }{2}\right)$

Solving both equations simultaneously, we get
$\cos \theta +\cos (\theta -\alpha )=\cos \theta +\cos (\beta -\theta )\Rightarrow \cos (\theta -\alpha )=\cos (\beta -\theta )\Rightarrow \theta -\alpha =\beta -\theta \Rightarrow \theta =\frac{\alpha +\beta }{2}$

Substituting $\theta =\frac{\alpha +\beta }{2}$ in the first equation, we get
$\frac{\phi }{r}=\cos \frac{\alpha +\beta }{2}+\cos \frac{\beta -\alpha }{2}\Rightarrow \frac{\phi }{r}=2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\Rightarrow r=\frac{\phi }{2}\sec \frac{\alpha }{2}\sec \frac{\beta }{2}$

Hence, option A is correct.