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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?

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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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    $\begingroup$ This is really good and simple explanation,well done. $\endgroup$ – Iti Shree Apr 20 '17 at 14:51
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Another method: $$\tan^2A=\sec^2A-1=(4-\tan A)^2-1.$$ Now solve for $\tan A$.

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Rearranging the identity $1 + \tan^2A = \sec^2A$ yields $$\sec^2A - \tan^2A = 1$$
Factoring yields $$(\sec A + \tan A)(\sec A - \tan A) = 1$$ Since $\sec A + \tan A = 4$, we have $$4(\sec A - \tan A) = 1$$ which yields the system of equations \begin{align*} \sec A + \tan A & = 4\\ \sec A - \tan A & = \frac{1}{4} \end{align*} Solve the system for $\sec A$ and $\tan A$, then use the identities \begin{align*} \cos A & = \frac{1}{\sec A}\\ \sin A & = \tan A\cos A \end{align*} to solve for sine and cosine.

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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.

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Together with $$\sin A+1=4\cos A $$ you can use $\sin^2A+\cos^2A=1$ as $$(\sin A+1)(\sin A-1)=\cos^2 A\ .$$ Putting the two together you easily get $$4\cos A(\sin A-1)=\cos^2 A\ ,$$ and hence $$4\sin A-4=\cos A\ .$$ You now just have to solve the linear system in $\sin A$ and $\cos A$: $$\begin{cases} 4\sin A-4=\cos A\\ \sin A+1=4\cos A \end{cases}$$

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