Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 10

Find the exact value of $\displaystyle \cos \left( \tan^{-1} 2 + \tan^{-1} 3 \right)$
By applying the sum of angles for cosine,
$\cos (A + B) = \cos A \cos B - \sin A \sin B$
Let the values of right triangle be...




We have,
$\displaystyle \tan A = \frac{\text{opposite}}{\text{adjacent}} = 2$
And,
$\tan B = 3$
So,
$A = \tan^{-1} (2)$ and $B = \tan^{-1} (3)$
We know that $\displaystyle \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}$

and $\displaystyle \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}$
Similarly with $B$,

$\displaystyle \cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{10}}$ and $\displaystyle \sin A = \frac{3}{\sqrt{10}}$
Thus,
$\cos (A + B) = \cos A \cos B - \sin A \sin B$

$\begin{equation} \begin{aligned} \cos \left( \tan^{-1} (2) + \tan^{-1} (3) \right) &= \left( \frac{1}{\sqrt{5}} \right) \left( \frac{1}{\sqrt{10}} \right) - \left( \frac{2}{\sqrt{5}} \right) \left( \frac{3}{\sqrt{10}} \right)\\ \\ &= \frac{-5}{5\sqrt{2}}\\ \\ &= \frac{-1}{\sqrt{2}}\\ \\ &= \frac{-\sqrt{2}}{2} \end{aligned} \end{equation}$