Question

If $\alpha ,\beta ,\gamma$ are acute angles and $\cos \theta =\frac{\sin \beta }{\sin \alpha },\cos \varphi =\frac{\sin \gamma }{\sin \alpha }$ and $\cos (\theta -\varphi )-\sin \beta \sin \gamma =0$, then ${\tan }^2\alpha -{\tan }^2\beta -{\tan }^2\gamma$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. none of these

Answer 1

The correct option is B $0$

$\cos \theta =\frac{\sin \beta }{\sin \alpha }$ ...(1)
$\cos \varphi =\frac{\sin \gamma }{\sin \alpha }$ ...(2)
$\cos (\theta -\varphi )=\sin \beta \sin \gamma \Rightarrow \cos \theta \cos \varphi -\sin \theta \sin \varphi =\sin \beta \sin \gamma \Rightarrow \sin \theta \sin \varphi =\cos \theta \cos \varphi -\sin \beta \sin \gamma \Rightarrow {\sin }^2\theta {\sin }^2\varphi ={(\cos \theta \cos \varphi -\sin \beta \sin \gamma )}^2$
From (1) and (2)
$(1-{\cos }^2\theta )(1-{\cos }^2\varphi )={(\frac{\sin \beta }{\sin \alpha }\frac{\sin \gamma }{\sin \alpha }-\sin \beta \sin \gamma )}^2\Rightarrow (1-\frac{{\sin }^2\beta }{{\sin }^2\alpha })(1-\frac{{\sin }^2\gamma }{{\sin }^2\alpha })={\sin }^2\beta {\sin }^2\gamma {(1-{\sin }^2\alpha )}^2\Rightarrow ({\sin }^2\alpha -{\sin }^2\beta )({\sin }^2\alpha -{\sin }^2\gamma )={\sin }^2\beta {\sin }^2\gamma (1+{\sin }^4\alpha -2{\sin }^2\alpha ){\Rightarrow \sin }^4\alpha -{\sin }^2\alpha {\sin }^2\gamma -{\sin }^2\alpha {\sin }^2\beta +{\sin }^2\beta {\sin }^2\gamma ={\sin }^2\beta {\sin }^2\gamma +{\sin }^4\alpha {\sin }^2\beta {\sin }^2\gamma -2{\sin }^2\alpha {\sin }^2\beta {\sin }^2\gamma {\Rightarrow \sin }^4\alpha (1-{\sin }^2\beta {\sin }^2\gamma )-{\sin }^2\alpha ({\sin }^2\beta +{\sin }^2\gamma -2{\sin }^2\beta {\sin }^2\gamma )=0$
${\Rightarrow \sin }^2\alpha =\frac{{\sin }^2\beta +{\sin }^2\gamma -2{\sin }^2\beta {\sin }^2\gamma }{1-{\sin }^2\beta {\sin }^2\gamma }$ ...(3)
$\Rightarrow {\cos }^2\alpha =1-\frac{{\sin }^2\beta +{\sin }^2\gamma -2{\sin }^2\beta {\sin }^2\gamma }{1-{\sin }^2\beta {\sin }^2\gamma }=\frac{1-{\sin }^2\beta -{\sin }^2\gamma +{\sin }^2\beta {\sin }^2\gamma }{1-{\sin }^2\beta {\sin }^2\gamma }$ ...(4)
From (3) and (4)
${\tan }^2\alpha =\frac{{\sin }^2\beta +{\sin }^2\gamma -2{\sin }^2\beta {\sin }^2\gamma }{1-{\sin }^2\beta -{\sin }^2\gamma +{\sin }^2\beta {\sin }^2\gamma }=\frac{{\sin }^2\beta {\cos }^2\gamma +{\sin }^2\gamma {\cos }^2\beta }{{\cos }^2\beta {\cos }^2\gamma }$

Now, ${\tan }^2\alpha -{\tan }^2\beta -{\tan }^2\gamma =0$
Hence, option 'B' is correct.