Question

If $0<\alpha <\pi /2$ and $\sin \alpha +\cos \alpha +\tan \alpha +\cot \alpha +\sec \alpha +cosec\alpha =7$, then $\sin 2\alpha$ is a root of the equation $x^2-44x-36=0$. If this is true enter 1, else enter 0.

Answer 1

$\sin \alpha +\cos \alpha +\tan \alpha +\cot \alpha +\sec \alpha +\csc \alpha =7$
$\Rightarrow (\sin \alpha +\cos \alpha )+(\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha })+(\frac{1}{\cos \alpha }+\frac{1}{\sin \alpha })=7$
$\Rightarrow (\sin \alpha +\cos \alpha )+\left(\frac{1}{\sin \alpha \cos \alpha }\right)+\left(\frac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }\right)=7$
$\Rightarrow (\sin \alpha +\cos \alpha )(1+\frac{2}{{\sin }^2\alpha })=7-\frac{2}{\sin \alpha }$
$\Rightarrow (1+{\sin }^2\alpha )(1+\frac{4}{{\sin }^{2}2\alpha }+\frac{4}{\sin 2\alpha })=49+\frac{4}{{\sin }^{2}2\alpha }-\frac{28}{\sin 2\alpha }$
$\Rightarrow {\sin }^{3}2\alpha -44{\sin }^{2}2\alpha +36\sin 2\alpha =0\Rightarrow {\sin }^{2}2\alpha -44\sin 2\alpha +36=0[\because \sin 2\alpha \ne 0]\Rightarrow \sin 2\alpha =22-8\sqrt{7}$
$x^2-44x-36=0\Rightarrow x=22\pm 8\sqrt{7}$