<s>[INST] Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies 
$$\sin A+1=4\cos A.$$
Then what to do?
 [/INST] Since you have $\sin A + 1 = 4 \cos A$, you can square both sides to yield
$$
\sin^2 A + 2 \sin A + 1 = 16 \cos^2 A = 16 (1 - \sin^2 A).
$$
Rearranging, we obtain
$$
17 \sin^2 A + 2 \sin A - 15 = 0
$$
which is a quadratic equation in $\sin A$ and can be solved to obtain
$$
\sin A = \frac{ -2 \pm 32}{34} = \left\{ \frac{15}{17}, -1 \right\}.
$$
and then we must have
$$
\cos A = \pm \sqrt{ 1 - \sin^2 A} = \left\{ \pm \frac{8}{17}, 0 \right\}.
$$
We now have three candidate solutions for this equation.  However, since we squared the equation and at one point multiplied by $\cos A$, we may have introduced spurious solutions;  so we should check to ensure that these results actually satisfy the original equation.  It turns out that only one of these solutions actually satisfies the original equation;  which one is left as an exercise to the reader.
</s>