If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt. 
$$ 0<\theta<\pi/2$$
  and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$
  then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +36=0$$

I tried to use the above given equation of all the trigonometric ratios, though I ended up with an expression of $\sin\theta\cos\theta$. 
But this too is in the form of $\sin\theta$ and $\cos\theta$ which was $\sin\theta\cos\theta=\frac{1}{6-\sin\theta-\cos\theta}$.
When I put that in the quadratic equation, it would again transform into the form of $\sin\theta$ and $\cos\theta$, and hence at last I couldn't prove the thing.
Even hints would work, as I would like to solve the question myself.
 A: With an obvious short notation and setting $x=2cs$, we reduce to the common denominator
$$s+c+\frac sc+\frac cs+\frac 1s+\frac 1c=\frac{sx+cx+2+2c+2s}{x}=\frac{(x+2)(c+s)+2}x=7.$$
As we know that $x\ne0$ and with $(c+s)^2=x+1$, we rewrite
$$(x+2)^2(x+1)=(7x-2)^2,$$
$$x^3-44x^2+36x=0.$$
A: $$\sin(\theta)+\cos(\theta)+\tan(\theta)+\cot(\theta)+\sec(\theta)+\csc(\theta)=7$$
$$\sin(\theta)+\cos(\theta)+\frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}+\frac{1}{\cos(\theta)}+\frac{1}{\sin(\theta)}=7$$
$$\sin^2\theta\cos\theta+\sin\theta\cos^2\theta+\sin^2\theta+\cos^2\theta+\sin\theta+\cos\theta=7\sin\theta\cos\theta$$
Let $\sin\theta+\cos\theta=u; \sin\theta\cos\theta=v$
$$uv+1+u=7v$$
$$u(1+v)=7v-1$$
$$u=\frac{7v-1}{v+1}$$
$$u^2=\left(\frac{7v-1}{v+1}\right)^2$$
$u^2=(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta=1+2v$
$$1+2v=\left(\frac{7v-1}{v+1}\right)^2$$
where $v=\sin\theta\cos\theta=\frac12 \sin2\theta$
Let $\sin2\theta=x$. Then $$1+x=\left(\frac{\frac72x-1}{\frac12x+1}\right)^2$$
$$1+x=\left(\frac{7x-2}{x+2}\right)^2$$
$$\color{red}{x^2-44x+36=0}$$
