Question

If $\tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3,\tan {\theta }_4$ are the roots of the equation $x^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =0$, then prove that $\tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)=\cot \beta$

Answer 1

$\tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)=\frac{S_1-S_3}{1-S_2+S_4}$
where $S_1=\sin 2\beta ,S_2=\cos 2\beta ,S_3=\cos \beta ,S_4=-\sin \beta$
$\therefore \tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)=\frac{\sin 2\beta -\cos 2\beta }{1-\cos 2\beta -\sin \beta }=\frac{\cos \beta (2\sin \beta -1)}{2{\sin }^2\beta -\sin \beta }=\frac{\cos \beta }{\sin \beta }=\cot \beta$