Question

Let $L={\sin }^2(\alpha +\beta )+3\sin (\alpha +\beta )\cos (\alpha +\beta )+4{\cos }^2(\alpha +\beta )$ where $\tan \alpha$ and $\tan \beta$ are the roots of the quadratic equation $t^2+3t+4=0,$ $M={\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma -2\cos \alpha \cos \beta \cos \gamma$ where $\alpha +\beta +\gamma =\pi$ and $N=\cot \alpha \cot \gamma$ where $\cot \alpha ,\cot \beta$ and $\cot \gamma$ are in arithmetic progression and $\alpha +\beta +\gamma =\frac{\pi }{2}.$ Then the value of $(L+M+N)$ is

Answer 1

$\tan \alpha$ and $\tan \beta$ are the roots of the quadratic equation $t^2+3t+4=0$
$\therefore \tan \alpha +\tan \beta =-3$ and $\tan \alpha \tan \beta =4$
We know,
$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{-3}{1-4}$
$\Rightarrow \tan (\alpha +\beta )=1$

Now, $L={\sin }^2(\alpha +\beta )+3\sin (\alpha +\beta )\cos (\alpha +\beta )+4{\cos }^2(\alpha +\beta )$
$\Rightarrow L=\frac{{\tan }^2(\alpha +\beta )+3\tan (\alpha +\beta )+4}{1+{\tan }^2(\alpha +\beta )}$
$\Rightarrow L=\frac{1+3+4}{1+1}=4$
$\therefore L=4$

$M={\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma -2\cos \alpha \cos \beta \cos \gamma$
We can write ${\sin }^2\alpha +{\sin }^2\beta$
$=\frac{1}{2}[2{\sin }^2\alpha +2{\sin }^2\beta ]$
$=\frac{1}{2}[1-\cos 2\alpha +1-\cos 2\beta ]$
$=\frac{1}{2}[2-2\cos (\alpha +\beta \left)\cos \right(\alpha -\beta \left)\right]$
$=1-\cos (\alpha +\beta )\cos (\alpha -\beta )$
$=1+\cos \left(\gamma \right)\cos (\alpha -\beta )(\because \alpha +\beta +\gamma =\pi )$
$\therefore {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$
$=1+\cos \left(\gamma \right)\cos (\alpha -\beta )+{\sin }^2\gamma$
$=2-\cos \left(\gamma \right)\left[\cos \right(\gamma )-\cos (\alpha -\beta \left)\right]$
$=2+\cos \left(\gamma \right)\left[2\cos \right(\alpha \left)\cos \right(\beta \left)\right]$
$=2+2\cos \alpha \cos \beta \cos \gamma$
$\therefore \sum {\sin }^2\alpha =2+2\prod \cos \alpha$
$\Rightarrow \sum {\sin }^2\alpha -2\prod \cos \alpha =2$
$\therefore M=2$

$N=\cot \alpha \cot \gamma$
Since $\cot \alpha ,\cot \beta$ and $\cot \gamma$ are in A.P.
$\therefore 2\cot \beta =\cot \alpha +\cot \gamma$
and $\alpha +\beta +\gamma =\frac{\pi }{2}$
$\Rightarrow \alpha +\gamma =\frac{\pi }{2}-\beta$
$\Rightarrow \cot (\alpha +\gamma )=\cot (\frac{\pi }{2}-\beta )=\tan \beta$
$\Rightarrow \frac{\cot \alpha \cdot \cot \gamma -1}{\cot \alpha +\cot \gamma }=\tan \beta$
$\Rightarrow \frac{\cot \alpha \cdot \cot \gamma -1}{2\cot \beta }=\tan \beta$
$\Rightarrow \cot \alpha \cdot \cot \gamma =3$
$\therefore N=3$

Hence, $L+M+N=4+2+3=9$