Find the value of $\frac{a+b}{10}$ 
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.

How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve for it but  it'll be very lengthy. Is there any other short and nicer method.

$\displaystyle \sin x+\cos x+\tan x+\cot x+\sec x+\text{cosec } x=7$
$\iff\displaystyle \sin x+\cos x+\frac{\sin x}{\cos x} +\frac{\cos x}{\sin x} +\frac{1}{\cos x} +\frac{1}{\sin x} =7$
$\displaystyle ( \sin x+\cos x)\left( 1+\frac{1}{\sin x\cos x}\right) +\frac{1}{\sin x \cos x} =7$
$\iff\displaystyle ( \sin x+\cos x)\left( 1+\frac{2}{\sin(2x)}\right) =7-\frac{2}{\sin( 2x)}$
Now square both the sides:
$\displaystyle ( 1+\sin( 2x))\left( 1+\frac{2}{\sin( 2x)}\right)^{2} =\left( 7-\frac{2}{\sin( 2x)}\right)^{2}$
Now this a cubic in $\sin(2x)$
You can simplify this to: $\displaystyle \sin( 2x)\left( \sin^{2}( 2x) -44\sin( 2x) +36\right)$
Now you can find $\sin(2x)$ using Quadratic formula. (assuming $\displaystyle \sin( 2x) \neq \ 0$)
So $\displaystyle \sin( 2x) =22\pm 8\sqrt{7}$ but we ignore the + result as its greater than 1.
So $ a+b = 30$.
