$\sec\theta + \tan\theta=4$ find $\cos\theta$ given: $\theta\neq90$

Question

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 ?

Answer 1

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 $\cos\theta$?

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Alternatively if $\sec A+\tan A=x\iff x-\sec A=\tan A$

Squaring we get, $$(x-\sec A)^2=\tan^2A\iff x^2-2x\sec A+\sec^2A=\sec^2A-1$$

$$\implies2x\sec A=x^2+1\implies\sec A=?$$

Answer 2

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 $\cos\theta$.

Answer 3

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 $$\cos(\theta)=\frac{1-t^2}{1+t^2}=\frac{1-\frac{9}{25}}{1+\frac{9}{25}}=\frac{8}{17}.$$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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, $$ 0=17\cos^{2}\pars{\theta} - 8\cos\pars{\theta} =\bracks{17\cos\pars{\theta} - 8}\ \underbrace{\cos\pars{\theta}}_{\ds{\color{#c00000}{\not=\ 0}}}\ \imp\ \color{#66f}{\large\cos\pars{\theta} = {8 \over 17}} $$