# What does cos(arctan(3))+sin(arctan(4)) equal?

Mar 4, 2018

$\cos \left(\arctan \left(3\right)\right) + \sin \left(\arctan \left(4\right)\right) = \frac{1}{\sqrt{10}} + \frac{4}{\sqrt{17}}$

#### Explanation:

Let ${\tan}^{-} 1 \left(3\right) = x$

then $\rightarrow \tan x = 3$

$\rightarrow \sec x = \sqrt{1 + {\tan}^{2} x} = \sqrt{1 + {3}^{2}} = \sqrt{10}$

$\rightarrow \cos x = \frac{1}{\sqrt{10}}$

$\rightarrow x = {\cos}^{- 1} \left(\frac{1}{\sqrt{10}}\right) = {\tan}^{- 1} \left(3\right)$

Also, let ${\tan}^{- 1} \left(4\right) = y$

then $\rightarrow \tan y = 4$

$\rightarrow \cot y = \frac{1}{4}$

$\rightarrow \csc y = \sqrt{1 + {\cot}^{2} y} = \sqrt{1 + {\left(\frac{1}{4}\right)}^{2}} = \frac{\sqrt{17}}{4}$

$\rightarrow \sin y = \frac{4}{\sqrt{17}}$

$\rightarrow y = {\sin}^{- 1} \left(\frac{4}{\sqrt{17}}\right) = {\tan}^{- 1} 4$

Now, $\rightarrow \cos \left({\tan}^{- 1} \left(3\right)\right) + \sin \left({\tan}^{- 1} \tan \left(4\right)\right)$

$\rightarrow \cos \left({\cos}^{-} 1 \left(\frac{1}{\sqrt{10}}\right)\right) + \sin \left({\sin}^{- 1} \left(\frac{4}{\sqrt{17}}\right)\right) = \frac{1}{\sqrt{10}} + \frac{4}{\sqrt{17}}$