$\sec\theta + \tan\theta=4$ find $\cos\theta$ given: $\theta\neq90$ I tried the following :
\begin{align}\sec\theta + \tan\theta&=4\\
\frac1{\cos\theta} + \frac{\sin\theta}{\cos\theta}&=4\\
\frac{1+\sin\theta}{\cos\theta}&=4\\
\frac{1+\sin\theta}4&=\cos\theta\end{align}
now don't know how to evaluate further ?
 A: HINT:
As $\sec^2\theta-\tan^2\theta=1$
$$\sec\theta+\tan\theta=4\iff\sec\theta-\tan\theta=\frac14$$
Can you find $\sec\theta$ and then $\cos\theta$?

Alternatively if $\sec A+\tan A=x\iff x-\sec A=\tan A$
Squaring we get, $$(x-\sec A)^2=\tan^2A\iff x^2-2x\sec A+\sec^2A=\sec^2A-1$$
$$\implies2x\sec A=x^2+1\implies\sec A=?$$
A: Hint: since $1+\cos\theta >0$ and $\frac{1+\cos\theta}{\sin\theta} =4$,
$\sin\theta >0$ so $\sin\theta = \sqrt{1-\cos^2\theta}$. This leads to a quadratic equation on $\cos\theta$.
A: one way is to use the half angle formule like @Mark Bennet suggested. then you have
$$\sec(\theta)=\frac{1+t^2}{1-t^2},\qquad\tan(\theta)=\frac{2t}{1-t^2}$$ where $t=\tan\left(\frac{\theta}{2}\right)$. This results in the following equationt
$$\frac{1+2t+t^2}{1-t^2}=\frac{(1+t)^2}{(1-t)(1+t)}=\frac{1+t}{1-t}=4\\t=\frac{3}{5}\rightarrow\theta=2\arctan\left(\frac{3}{5}\right)$$
and $$\cos(\theta)=\frac{1-t^2}{1+t^2}=\frac{1-\frac{9}{25}}{1+\frac{9}{25}}=\frac{8}{17}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\sec\pars{\theta} + \tan\pars{\theta}=4\ \imp\ 1 + \sin\pars{\theta} =4\cos\pars{\theta}
\\[5mm]&\imp\ 1 - \cos^{2}\pars{\theta}=\sin^{2}\pars{\theta}
=\bracks{4\cos\pars{\theta} - 1}^{2}
\end{align}

Then,
  $$
0=17\cos^{2}\pars{\theta} - 8\cos\pars{\theta}
=\bracks{17\cos\pars{\theta} - 8}\
\underbrace{\cos\pars{\theta}}_{\ds{\color{#c00000}{\not=\ 0}}}\ \imp\
\color{#66f}{\large\cos\pars{\theta} = {8 \over 17}}
$$

