If cosθ = n cos(θ+2φ) show that (n+1) tan(θ+φ)=(n-1)cotφ
$\cos \theta$ = $ncos(\theta + 2\phi )$
or, $\frac{\cos \theta }{\cos \left(\theta + 2\phi \right)}$ = $\frac{n}{1}$
or, $\frac{\cos \theta + cos\left(\theta + 2\phi \right)}{\cos \theta – cos\left(\theta + 2\phi \right)}$ = $\frac{n + 1}{n – 1}$
or, $\frac{2cos\frac{\theta + \theta + 2\phi }{2}cos\frac{\theta – \theta – 2\phi }{2}}{2sin\frac{\theta + \theta + 2\phi }{2}sin\frac{\theta + 2\phi – \theta }{2}}$ = $\frac{n + 1}{n – 1}$
or, $\frac{2cos\frac{2\theta + 2\phi }{2}\cos \frac{ – 2\phi }{2}}{2sin\frac{2\theta + 2\phi }{2}sin\frac{2\phi }{2}}$ = $\frac{n + 1}{n – 1}$
or, $\frac{2cos(\theta + \phi )\cos \phi }{2sin(\theta + \phi )sin\phi }$ = $\frac{n + 1}{n – 1}$
or, $\frac{n – 1}{n + 1}$ = $\frac{2sin(\theta + \phi )sin\phi }{2cos(\theta + \phi )\cos \phi }$
or, $\frac{n – 1}{n + 1}$ = $tan\left(\theta + \phi \right)tan\phi$
or, $(n – 1)cot\phi$ = $(n + 1)tan\left(\theta + \phi \right)$
or, $(n + 1)tan\left(\theta + \phi \right)$ = $(n – 1)cot\phi$
$\therefore (n + 1)tan\left(\theta + \phi \right)$ = $(n – 1)cot\phi$ [Proved]