Find the value of $$\cos\left({\tan^{-1}\left(\dfrac{3}{4}\right)}\right)$$


closed as off-topic by Namaste, Toby Mak, Nosrati, Key Flex, GNUSupporter 8964民主女神 地下教會 Sep 30 '18 at 15:56

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    $\begingroup$ Try to relate it to a 3-4-5 triangle. $\endgroup$ – Gerry Myerson Sep 30 '18 at 12:08
  • $\begingroup$ I didn't understand $\endgroup$ – user187387 Sep 30 '18 at 12:08
  • $\begingroup$ Draw a right triangle that contains an angle with tangent $\frac 34$. $\endgroup$ – lulu Sep 30 '18 at 12:18
  • $\begingroup$ What, precisely, didn't you understand? Do you know what a triangle is? $\endgroup$ – Gerry Myerson Sep 30 '18 at 12:19
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    $\begingroup$ Wrong tags here. This should be tagged "trigonometry". $\endgroup$ – xbh Sep 30 '18 at 12:55

From $\sin^2(x)+ \cos^2(x)=1$ if you devide by $\cos^2$ and rearrange, you get $$ \cos^2(x)=\frac{1}{\tan^2 (x)+1} $$ then set $x=\tan^{-1}(y)$, taking the square root yelds $$ \cos(\tan^{-1}(y))=\frac{1}{\sqrt{y^2+1}} $$ In case $y=3/4$, so $\cos(\tan^{-1}(3/4))=4/5$

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    $\begingroup$ Nice answer! In your first line, a better place to start from might be $\tan^2 x + 1 = \sec^2 x$. $\endgroup$ – Toby Mak Sep 30 '18 at 12:54

Consider this right angled triangle enter image description here

From the figure $$tan \theta=\dfrac{3}{4}$$

$$\implies \theta=\tan ^{-1} \dfrac{3}{4}$$


$$\cos (\tan ^{-1} \dfrac{3}{4})=\cos \theta$$

From the triangle,

$$\cos \theta=\dfrac{4}{5}$$


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