Question

If $\tan \theta =3$, where $\theta$ is an acute angle, then find the value of $\cos \theta$.

Answer 1

Given : $\tan \theta =3$
In right-angled trianlge $ABC$, we have
$\tan \theta =\frac{\text{Opposite side of angle}\theta }{\text{Adjacent side of angle}\theta }$
$\Rightarrow \frac{3}{1}=\frac{AC}{BC}$
By Pythagoras theorem,
${AB}^2={BC}^2+{AC}^2$
$\Rightarrow {AB}^2={\left(1\right)}^2+{\left(3\right)}^2$
$\Rightarrow {AB}^2=1+9$
$\Rightarrow {AB}^2=10$
$\Rightarrow AB=\sqrt{10}$
We know that,
$\cos \theta =\frac{\text{Adjacent side of angle}\theta }{\text{Hypotenuse}}$
$\Rightarrow \cos \theta =\frac{BC}{AB}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{10}}$