# $\sec\theta + \tan\theta=4$ 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 ?

• The substitution $t=\tan \frac {\theta}2$ with $\sin \theta =\frac {2t}{1+t^2}$ and $\cos \theta =\frac {1-t^2}{1+t^2}$ might help. – Mark Bennet Nov 7 '14 at 17:22

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

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

• Very elegant... – Simon S Nov 7 '14 at 17:35
• @SimonS, Added an alternative method – lab bhattacharjee Nov 8 '14 at 4:35

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 $\cos\theta$.

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


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