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

Question

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

Answer 1

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 $\cos(\tan^{-1}(3/4))=4/5$

Answer 2

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,

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