Question

If $\frac{x}{\tan (\theta +\alpha )}=\frac{y}{\tan (\theta +\beta )}=\frac{z}{\tan (\theta +\gamma )}$, then find the value of $\left(\frac{x+y}{x-y}\right){\sin }^2(\alpha -\beta )+\left(\frac{y+z}{y-z}\right){\sin }^2(\beta -\gamma )+\left(\frac{z+x}{z-x}\right){\sin }^2(\gamma -\alpha )=$

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (C)

$0$

$\frac{x}{\tan (\theta +\alpha )}=\frac{y}{\tan (\theta -\beta )}=\frac{z}{\tan (\theta -\gamma )}$

$\Rightarrow \frac{(x+y)}{(x-y)}=\frac{\left(\tan \right(\theta -\alpha )+\tan (\theta +\beta \left)\right)}{\left(\tan \right(\theta +\alpha )-\tan (\theta +\beta \left)\right)}$

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

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

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

Consider,

${\sin }^2(\alpha -\beta )\times \frac{(x+y)}{(x-y)}+{\sin }^2(\beta -\gamma )\times \frac{(y+z)}{(y-z)}+{\sin }^2(\gamma -\alpha )\times \frac{(z+z)}{(z-x)}$

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

Option $C$ is correct.