Question

If $\frac{x}{\tan (\theta +\alpha )}=\frac{y}{\tan (\theta +\beta )}=\frac{z}{\tan (\theta +y)}$, then prove that $\sum \frac{x+y}{x-y}{\sin }^2(\alpha -\beta )=0$

Answer 1

Given,
$\frac{x}{tan(\theta +\alpha )}=\frac{y}{tan(\theta +\beta )}=\frac{z}{tan(\theta +\gamma )}$
$\Rightarrow \frac{(x+y)}{(x-y)}=\frac{\left(tan\right(\theta +\alpha )+tan(\theta +\beta \left)\right)}{\left(tan\right(\theta +\alpha )-tan(\theta +\beta \left)\right)}$
We know that, $\frac{(tanA+tanB)}{(tanA-tanB)}=\frac{sin(A+B)}{sin(A-B)}$
$\Rightarrow \frac{(x+y)}{(x-y)}=\frac{sin\left(\right(\theta +\alpha )+(\theta +\beta \left)\right)}{sin\left(\right(\theta +\alpha )-(\theta +\beta \left)\right)}$
$\Rightarrow \frac{(x+y)}{(x-y)}=\frac{sin(2\theta +\alpha +\beta )}{sin(\alpha -\beta )}$

Consider,$si{n}^2(\alpha -\beta )\times \frac{(x+y)}{(x-y)}=si{n}^2(\alpha -\beta )\times \frac{sin(2\theta +\alpha +\beta )}{sin(\alpha -\beta )}$

$=sin(\alpha -\beta )\times sin(2\theta +\alpha +\beta )$

$=\frac{1}{2}\times (cos(\beta +\theta )-cos(\alpha +\theta )$

Consider,$si{n}^2(\alpha -\beta )\times \frac{(x+y)}{(x-y)}+si{n}^2(\beta -\gamma )\times \frac{(y+z)}{(y-z)}+si{n}^2(\gamma -\alpha )\times \frac{(z+x)}{(z-x)}$

$=\frac{1}{2}\times \left(\right(cos(\beta +\theta )-cos(\alpha +\theta ))+\left(cos\right(\gamma +\theta )-cos(\beta +\theta ))+(cos(\alpha +\theta )-cos(\beta +\theta )\left)\right)=0$