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

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

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Answer 1 · 2018-03-04T04:02:18

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

Explanation:

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

then $\to tan x = 3$ $rarrtanx=3$

$\to sec x = \sqrt{1 + {tan}^{2} x} = \sqrt{1 + 3^{2}} = \sqrt{10}$ $rarrsecx=sqrt(1+tan^2x)=sqrt(1+3^2)=sqrt(10)$

$\to cos x = \frac{1}{\sqrt{10}}$ $rarrcosx=1/sqrt(10)$

$\to x = {cos}^{- 1} (\frac{1}{\sqrt{10}}) = {tan}^{- 1} (3)$ $rarrx=cos^(-1)(1/sqrt(10))=tan^(-1)(3)$

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

then $\to tan y = 4$ $rarrtany=4$

$\to cot y = \frac{1}{4}$ $rarrcoty=1/4$

$\to csc y = \sqrt{1 + {cot}^{2} y} = \sqrt{1 + {(\frac{1}{4})}^{2}} = \frac{\sqrt{17}}{4}$ $rarrcscy=sqrt(1+cot^2y)=sqrt(1+(1/4)^2)=sqrt(17)/4$

$\to sin y = \frac{4}{\sqrt{17}}$ $rarrsiny=4/sqrt(17)$

$\to y = {sin}^{- 1} (\frac{4}{\sqrt{17}}) = {tan}^{- 1} 4$ $rarry=sin^(-1)(4/sqrt(17))=tan^(-1)4$

Now, $\to cos ({tan}^{- 1} (3)) + sin ({tan}^{- 1} tan (4))$ $rarrcos(tan^(-1)(3))+sin(tan^(-1)tan(4))$

$\to cos ({cos}^{- 1} (\frac{1}{\sqrt{10}})) + sin ({sin}^{- 1} (\frac{4}{\sqrt{17}})) = \frac{1}{\sqrt{10}} + \frac{4}{\sqrt{17}}$ $rarrcos(cos^-1(1/sqrt(10)))+sin(sin^(-1)(4/sqrt(17)))=1/sqrt(10)+4/sqrt(17)$

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

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