## Precalculus (6th Edition) Blitzer

The exact value of $\cos \left( {{\tan }^{-1}}\frac{3}{4} \right)$ is $\frac{4}{5}$.
Let $\theta ={{\tan }^{-1}}\frac{3}{4}$ represent the angle in $\left( -\frac{\pi }{2},\frac{\pi }{2} \right)$. \begin{align} & \theta ={{\tan }^{-1}}\frac{3}{4} \\ & \tan \theta =\frac{3}{4} \end{align} As the value of $\tan \theta$ is positive, thus $\theta$ lies in the first quadrant. Hence, the measures of the two sides of the right triangle are $3$ and $4$. The hypotenuse of the triangle is found by applying the Pythagorian Theorem, \begin{align} & r=\sqrt{{{3}^{2}}+{{4}^{2}}} \\ & =\sqrt{25} \\ & =5 \end{align} From above sketch \begin{align} & \theta ={{\tan }^{-1}}\frac{3}{4} \\ & \cos \theta =\cos \left( {{\tan }^{-1}}\frac{3}{4} \right) \end{align} By the basic definition of the cosine function, \begin{align} & \cos \theta =\frac{\text{side adjecent to angle }\theta }{\text{hypotenuse}} \\ & \cos \left( {{\tan }^{-1}}\frac{4}{3} \right)=\frac{4}{5} \end{align} Hence, the value of $\cos \left( {{\tan }^{-1}}\frac{3}{4} \right)$ is $\frac{4}{5}$