If $\frac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$. If $\dfrac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$.
My Attempt:
$$\dfrac {\sin A+\tan A}{\cos A}=9$$
$$\dfrac {\sin A+ \dfrac {\sin A}{\cos A}}{\cos A}=9$$
$$\dfrac {\sin A.\cos A+\sin A}{\cos^2 A}=9$$
$$\dfrac {\sin A(1+\cos A)}{\cos^2 A}=9$$
$$\tan A.\sec A(1+\cos A)=9$$
$$\tan A(1+\sec A)=9$$
How do I go further?
 A: Hint:   using the tangent half-angle formulas, let $\,t=\tan(A/2)\,$, then the equation becomes:
$$
\frac{2t}{1+t^2} + \frac{2t}{1-t^2}=9 \,\frac{1-t^2}{1+t^2} \;\;\iff\;\; 9 t^4 - 18 t^2 - 4 t + 9 = 0
$$
The quartic has $2$ real roots which can be solved in radicals, but the calculations are not pretty.

[ EDIT ]  Once $\,t\,$ is determined, $\,\sin A = 2t/(1+t^2)\,$ follows. Or, to determine $\,x = \sin A\,$ directly, one can eliminate $t$ between the equation above and $\,(1+t^2)x-2t=0\,$ using resultants:

$$
1312 x^4 + 288 x^3 - 2592 x^2 - 288 x + 1296 = 0 \;\;\iff\;\; 82 x^4 + 18 x^3 - 162 x^2 - 18 x + 81 = 0
$$
A: By the Fundamental Theorem of Trigonometry we have:
$$\tag {*} \sin^2x+\cos^2x=1 \Rightarrow \cos x=\sqrt {1-\sin^2x}$$
Now:
$$\frac {\sin A + \tan A}{\cos A}=9$$
$$\frac {\sin A + \frac {\sin A}{cos A}}{\sqrt {1-\sin^2A}}=9$$
$$\frac {\sin A + \frac {\sin A}{\sqrt {1-\sin^2A}}}{\sqrt {1-\sin^2A}}=9$$
Let $\sin A=x$. Thus:
$$\frac {x+ \frac {x}{\sqrt {1-x^2}}}{\sqrt {1-x^2}}=9$$
$$x+ \frac {x}{\sqrt {1-x^2}}=9 \sqrt {1-x^2}$$
$$x\sqrt {1-x^2}+x=9(1-x^2)$$
$$x\sqrt {1-x^2}=9(1-x^2)-x$$
$$x^2(1-x^2)=81(1-x^2)^2-18x(1-x^2)+x^2$$
$$x^2-x^4=81x^4-162x^2+81-18x+18x^3+x^2$$
$$82x^4+18x^3-162x^2-18x+81=0$$
Solve the equation and we are done
