# How do you calculate  cos(tan^-1(3/4))?

May 27, 2018

$\cos \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = 0.8$

#### Explanation:

 cos ( tan^-1 (3/4))= ? Let ${\tan}^{-} 1 \left(\frac{3}{4}\right) = \theta$

$\therefore \tan \theta = \frac{3}{4} = \frac{P}{B} , P \mathmr{and} B$ are perpendicular and base

of right triangle , then ${H}^{2} = {P}^{2} + {B}^{2} = {3}^{2} + {4}^{2} = 25$

:.H=5 ; :. cos theta = B/H=4/5 =0.8

$\cos \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = \cos \theta = 0.8$

$\therefore \cos \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = 0.8$ [Ans]

May 27, 2018

$\frac{4}{5}$

#### Explanation:

$\tan \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = \frac{3}{4}$
$\text{Name } y = {\tan}^{-} 1 \left(\frac{3}{4}\right)$
$\text{Then we have}$
$\tan \left(y\right) = \frac{3}{4}$
"Now use "sec²(x) = 1 + tan²(x)
=> sec²(y) = 1 + tan²(y) = 1 + 9/16 = 25/16
$\implies \sec \left(y\right) = \frac{1}{\cos} \left(y\right) = \pm \frac{5}{4}$
$\implies \cos \left(y\right) = \pm \frac{4}{5}$
$\implies \cos \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = \pm \frac{4}{5}$
$\text{We have to take the solution with + sign as}$
$- \frac{\pi}{2} \le \arctan \left(x\right) \le \frac{\pi}{2}$
$\text{and}$
$\cos \left(x\right) > 0 , \mathmr{if} - \frac{\pi}{2} \le x \le \frac{\pi}{2}$
$\implies \cos \left({\tan}^{-} 1 \left(\frac{3}{4}\right)\right) = \frac{4}{5}$

$\text{Note that we could also have used}$
$\tan \left(y\right) = \sin \frac{y}{\cos} \left(y\right)$
$\text{and}$
${\sin}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1$
$\tan \left(y\right) = \sin \frac{y}{\cos} \left(y\right) = \frac{3}{4}$
$\implies \pm \frac{\sqrt{1 - {\cos}^{2} \left(y\right)}}{\cos} \left(y\right) = \frac{3}{4}$
$\implies 1 - {\cos}^{2} \left(y\right) = {\left(\left(\frac{3}{4}\right) \cos \left(y\right)\right)}^{2}$
$\implies \left(1 + \frac{9}{16}\right) {\cos}^{2} \left(y\right) = 1$
$\implies {\cos}^{2} \left(y\right) = \frac{16}{25}$
$\implies \cos \left(y\right) = \frac{4}{5}$