If tan A + sec A = 4, what is CosA?

Mar 23, 2018

Given $\tan A + \sec A = 4$

$\left(\sec A + \tan A\right) = 4. \ldots . \left[1\right]$

So

$\left(\sec A + \tan A\right) \left(\sec A - \tan A\right) = 4 \left(\sec A - \tan A\right)$

$\implies \left({\sec}^{2} A - {\tan}^{2} A\right) = 4 \left(\sec A - \tan A\right)$

$\implies 1 = 4 \left(\sec A - \tan A\right)$

$\implies \sec A - \tan A = \frac{1}{4.} \ldots \ldots \left[2\right]$

Adding  and  we get

$2 \sec A = 4 + \frac{1}{4}$

$\implies \sec A = \frac{17}{8}$

$\implies \cos A = \frac{8}{17}$

Mar 23, 2018

$\cos A = \frac{8}{17}$

Explanation:

Here,

$\sec A + \tan A = 4. \ldots . \to \left(1\right)$

We know that,

color(red)(sec^2A-tan^2A=1

$\left(\sec A + \tan A\right) \left(\sec A - \tan A\right) = 1$

$\implies \left(4\right) \left(\sec A - \tan A\right) = 1. \ldots . \to$From(1)

$\therefore \sec A - \tan A = \frac{1}{4.} \ldots . \to \left(2\right)$

$\sec A + \cancel{\tan} A = 4$
(secA-canceltanA=1/4)/
$2 \sec A = 4 + \frac{1}{4} = \frac{17}{4}$
$\implies \sec A = \frac{17}{8}$
$\implies \cos A = \frac{8}{17}$