# If  tanA+secA=2  then find cosA ?

Oct 27, 2016

$\cos A = \frac{4}{5}$

#### Explanation:

$\tan A + \sec A = 2$

Substituting the definition of $\tan$ and $\sec$ gives us:
$\sin \frac{A}{\cos} A + \frac{1}{\cos} A = 2$

Multiplying by $\cos A$:
$\sin A + 1 = 2 \cos A$

$\therefore \sin A = 2 \cos A - 1$
$\therefore {\sin}^{2} A = {\left(2 \cos A - 1\right)}^{2}$

Using the fundamental identity ${\sin}^{2} X + {\cos}^{2} X \equiv 1$
$1 - {\cos}^{2} A = {\left(2 \cos A - 1\right)}^{2}$
$\therefore 1 - {\cos}^{2} A = 4 {\cos}^{2} A - 4 \cos A + 1$
$\therefore 5 {\cos}^{2} A - 4 \cos A = 0$
$\therefore \cos A \left(5 \cos A - 4\right) = 0$
$\cos A = 0$ or $5 \cos A - 4 = 0 \implies \cos A = \frac{4}{5}$

We were told that A is an acute angle, so we can eliminate the solution $\cos A = 0$

Hence, $\cos A = \frac{4}{5}$