Question

The sum of three positive numbers $\alpha ,\beta ,\gamma$ is equal to $\frac{\pi }{2}.$ If $e^{\cot \alpha },e^{\cot \beta }$ and $e^{\cot \gamma }$ form a geometric progression, then which of the following hold(s) good?

• A. $\sum \cot \alpha =\prod \cot \alpha$
• B. $2\cot \beta =\cot \alpha +\cot \gamma$
• C. $\cot \alpha \cot \gamma =3$
• D. $4\prod \cos \alpha =\sum \sin 2\alpha$

Answer 1

Answer: (A), (B), (C), (D)

$4\prod \cos \alpha =\sum \sin 2\alpha$

$\alpha +\beta =\frac{\pi }{2}-\gamma$
$\Rightarrow \cot (\alpha +\beta )=\cot (\frac{\pi }{2}-\gamma )=\tan \gamma \Rightarrow \frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }=\frac{1}{\cot \gamma }\Rightarrow \cot \alpha \cot \beta \cot \gamma =\cot \alpha +\cot \beta +\cot \gamma \dots \left(1\right)$
Hence, $\sum \cot \alpha =\prod \cot \alpha$

Now, $e^{\cot \alpha },e^{\cot \beta },e^{\cot \gamma }$ are in G.P.
$\therefore e^{2\cot \beta }=e^{\cot \alpha }\cdot e^{\cot \gamma }\Rightarrow \cot \alpha +\cot \gamma =2\cot \beta \dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$\cot \alpha \cot \beta \cot \gamma =3\cot \beta \Rightarrow \cot \alpha \cdot \cot \gamma =3(\because \cot \beta \ne 0)$

Also, $\prod \cos \alpha =\frac{1}{2}\cos \gamma \left(2\cos \alpha \cos \beta \right)$
$=\frac{1}{2}\cos \gamma \left[\cos \right(\alpha +\beta )+\cos (\alpha -\beta \left)\right]=\frac{1}{2}\cos \gamma [\sin \gamma +\cos (\alpha -\beta \left)\right]=\frac{1}{4}\sin 2\gamma +\frac{1}{4}\left[2\cos \gamma \cos \right(\alpha -\beta \left)\right]=\frac{1}{4}\sin 2\gamma +\frac{1}{4}\left[\cos \right(\gamma +\alpha -\beta )+\cos (\gamma +\beta -\alpha \left)\right]=\frac{1}{4}\sin 2\gamma +\frac{1}{4}[\cos (\frac{\pi }{2}-2\beta )+\cos (\frac{\pi }{2}-2\alpha )]=\frac{1}{4}\sin 2\gamma +\frac{1}{4}[\sin 2\beta +\sin 2\alpha ]\Rightarrow \prod \cos \alpha =\frac{1}{4}\sum \sin 2\alpha$
Hence, $4\prod \cos \alpha =\sum \sin 2\alpha$