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?

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
