Question

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

Answer 1

According to question,

$\begin{array}{c}\frac{\sin (A+B)}{\sin (A-B)}=\frac{\sin A\cos B+\cos A\sin B÷\cos A\cos B}{\sin A\cos B-\sin A\cos B÷\cos A\cos B}\\ \frac{\sin (A+B)}{\sin (A-B)}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\sin B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\sin B}}\Rightarrow \frac{\tan A+\tan B}{\tan A-\tan B}-----\left(1\right)\\ fromL.H.S.....\\ \frac{x}{tan(\theta +\alpha )}=\frac{y}{tan(\theta +\beta )}=\frac{z}{tan(\theta +\gamma )}\\ \Rightarrow \frac{x}{tan(\theta +\alpha )}=\frac{y}{tan(\theta +\beta )}\\ \Rightarrow \frac{x}{y}=\frac{tan(\theta +\alpha )}{tan(\theta +\beta )}\end{array}$

Using composndo & dividendo,

$\begin{array}{c}\frac{x+y}{x-y}=\frac{\tan (\theta +\alpha )+\tan (\theta +\beta )}{tan(\theta +\alpha )-\tan (\theta +\beta )}-----|usingequ--\left(1\right)\\ \frac{x+y}{x-y}=\frac{\sin (\theta +\alpha +\theta +\beta )}{\sin (\theta +\alpha -\theta -\beta )}=\frac{\sin (2\theta +\alpha +\beta )}{\sin (\alpha -\beta )}\\ \Rightarrow \frac{x+y}{x-y}\left({\sin }^2\right(\alpha -\beta \left)\right)=\frac{\sin (2\theta +\alpha +\beta )si{n}^2(\alpha -\beta )}{\sin (\alpha -\beta )}\\ \Rightarrow \frac{x+y}{x-y}\left({\sin }^2\right(\alpha -\beta \left)\right)=\frac{1}{2}\left[2\sin (2\theta +\alpha +\beta )\sin (\alpha -\beta )\right]\\ [weknow,cosC-\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)\\ \Rightarrow \frac{x+y}{x-y}\left({\sin }^2\right(\alpha -\beta \left)\right)=\frac{1}{2}\left[cos(2\theta +2\beta )-\cos (2\theta +2\alpha \right]-----\left(2\right)\\ similarly........\\ \Rightarrow \frac{y+z}{y-z}\left({\sin }^2\right(\beta -\gamma \left)\right)=\frac{1}{2}\left[cos(2\theta +2\gamma )-\cos (2\theta +2\beta \right]------\left(3\right)\\ \Rightarrow \frac{z+a}{z-a}\left({\sin }^2\right(\gamma -\alpha \left)\right)=\frac{1}{2}\left[cos(2\theta +2\alpha )-\cos (2\theta +2\gamma \right]------\left(4\right)\\ ifaddingequ......\left(2\right),\left(3\right)\&\left(4\right)\\ thenweprovethat,\\ \therefore \frac{x+y}{x-y}{\sin }^2(\alpha -\beta )+\frac{y+z}{y-z}{\sin }^2(\beta -\gamma )+\frac{z+a}{z-a}{\sin }^2(\gamma -\alpha )=0\end{array}$