Question

If $co{t}^2\theta +cot(\theta -\alpha )cot(\theta -\beta )$ show that $cot2\theta =\frac{1}{2}(cot\alpha +cot\beta )$

Answer 1

From the given relation on changing to sin and cos , we have
$\frac{co{s}^2\theta }{si{n}^2\theta }=\frac{cos(\theta -\alpha )cos(\theta -\beta )}{sin(\theta -\alpha )cos(\theta -\beta )}$
Apply compo, and divi .
$\frac{co{s}^2\theta -si{n}^2\theta }{co{s}^2\theta +si{n}^2\theta }=\frac{cos(\theta -\alpha +\theta -\beta )}{cos(\theta -\alpha +\theta -\beta )}$
Using formula of $cos(A\pm B)$
or $cos2\theta =\frac{cos2\theta -(\alpha +beta)}{cos(\alpha -beta)}[\because cos(-\theta )=cos\theta ]$
$\therefore cos2\theta cos(\alpha -\beta )=cos2\theta cos(\alpha +\beta ),+sin2\theta sin(\alpha +\beta )$
or \, $cos2\theta \{cos(\alpha -\beta )-cos(\alpha +\beta )\}=sin2\theta sin(\alpha +\beta )$
$\therefore \frac{cos2\theta }{sin2\theta }=\frac{sin\alpha cos\beta +cos\alpha sin\beta }{2sin\alpha cos\beta }$
or $cot2\theta =\frac{1}{2}[cot\beta +cot\alpha ]$