Question

If $\frac{x}{\tan (\theta +\alpha )}\simeq \frac{y}{\tan (\theta +\beta )}\simeq \frac{Z}{\tan (\theta +\gamma )}$ , then $\sum \frac{x+y}{x-y}{\sin }^2(\alpha -\beta )$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $0$

Given

$\frac{x}{\tan (\theta +\alpha )}\cong \frac{y}{\tan (\theta +\beta )}\cong \frac{Z}{\tan (\theta +\gamma )}=k$

$⟹x=k\tan (\theta +\alpha )$

$⟹y=k\tan (\theta +\beta )$

$⟹z=k\tan (\theta +\gamma )$

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

$⟹\frac{\frac{\sin (\theta +\alpha )}{\cos (\theta +\alpha )}+\frac{\sin (\theta +\beta )}{\cos (\theta +\beta )}}{\frac{\sin (\theta +\alpha )}{\cos (\theta +\alpha )}-\frac{\sin (\theta +\beta )}{\cos (\theta +\beta )}}$

$⟹\frac{\sin (\theta +\alpha )\cos (\theta +\beta )+\sin (\theta +\beta )\cos (\theta +\alpha )}{\sin (\theta +\alpha )\cos (\theta +\beta )-\sin (\theta +\beta )\cos (\theta +\alpha )}$

$⟹\frac{\sin (\theta +\alpha +\theta +\beta )}{\sin (\theta +\alpha -\theta -\beta )}$

$⟹\frac{\sin (2\theta +\alpha +\beta )}{\sin (\alpha -\beta )}$

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

$\sum \sin (2\theta +\alpha +\beta )\sin (\alpha -\beta )$

$\frac{1}{2}\sum \left(2\sin \right(2\theta +\alpha +\beta \left)\sin \right(\alpha -\beta \left)\right)$

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

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

$⟹\frac{1}{2}\left[\cos 2\right(\theta +\beta )-\cos 2(\theta +\alpha )+\cos 2(\theta +\gamma )-\cos 2(\theta +\beta )+\cos 2(\theta +\alpha )-\cos 2(\theta +\gamma \left)\right]$

$⟹\frac{1}{2}\left(0\right)=0$