# Find the value of $\cos\left({\tan^{-1}\left(\frac{3}{4}\right)}\right)$. [closed]

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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• Try to relate it to a 3-4-5 triangle. – Gerry Myerson Sep 30 '18 at 12:08
• I didn't understand – user187387 Sep 30 '18 at 12:08
• Draw a right triangle that contains an angle with tangent $\frac 34$. – lulu Sep 30 '18 at 12:18
• What, precisely, didn't you understand? Do you know what a triangle is? – Gerry Myerson Sep 30 '18 at 12:19
• Wrong tags here. This should be tagged "trigonometry". – 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 $$\cos(\tan^{-1}(3/4))=4/5$$

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

Consider this right angled triangle

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

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

So

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

From the triangle,

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