<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] Let $t=\tan \frac{A}{2}$ then $\tan A = \frac{2t}{1-t^2}$ and $\cos A = \frac{1-t^2}{1+t^2}$ so $$\begin{align*}\frac{1+t^2}{1-t^2} + \frac{2t}{1-t^2}&=\frac{(1+t)^2}{1-t^2} \\ & = \frac{(1+t)(1+t)}{(1-t)(1+t)} \\ & = \frac{1+t}{1-t} = 4\end{align*}$$ 
So, assuming $t\neq 1$ we get $t = \frac{3}{5}$. From this, you can find $\cos A$ and $\sin A$. 
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