If $\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$ then $x^2 - 44x - 36 = 0$ holds for $x=\sin 2A$

If $$\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$$ then prove that $$\sin 2A \quad\text{ is a root of }\quad x^2 - 44x - 36 = 0$$

I have no idea how to solve it. Plz help.

Solution 1:

Let $$\sin{A}+\cos{A}=t$$

$$7=\sin A+\cos A+\frac{1}{\sin A\cos A}+\frac{\sin A+\cos A} {\sin A\cos A}$$

$$\sin 2A =2 \sin A \cos A$$

$$t^2=1+\sin 2A$$

Solution 2:

I think you have the problem wrong, you can show $\sin 2A$ is a root of $x^2-44x+36$ using Awesome's hint.

EDIT:

If $\sin 2A$ is a root of $x^2-44x+36$, then $$\sin 2A = \frac{44 \pm \sqrt{1936 - 144}}{2}=22\pm 8\sqrt{7}$$

Since $|\sin 2A|\leq 1$, we have that $\sin 2A = 22-8\sqrt{7}$.

Using the hint by Awesome,

\begin{align} 7 &=\sin A + \cos A + \frac{1}{\sin A\cos A} + \frac{\sin A + \cos A}{\sin A\cos A} \\ &= t + \frac{1+t}{\frac{1}{2}(t^2-1)} \\ &= t + \frac{2(1+t)}{t^2-1} \\ &= t + \frac{2}{t-1}. \end{align}

This gives the polynomial $t^2-8t+9=0$, which has roots $t=4 \pm \sqrt{7}$. Since $$(\sin A + \cos A)^2 = 1+\sin 2A \leq 1 + 1= 2,$$ we have $$|\sin A + \cos A| \leq \sqrt{2}.$$

This forces $t=4-\sqrt{7}$. You can now compute $\sin 2A = 22-8\sqrt{7}$ as desired.