# How do you find the exact value of cos(tan^-1 (5/12)+cot^-1 (5/12))?

Feb 25, 2016

cos(tan^(−1)(5/12)+cot^(−1)(5/12))=cos(pi/2)=0

#### Explanation:

Remember the trigonometric identity

$\tan \theta = \cot \left(\frac{\pi}{2} - \theta\right) = x$ i.e.

${\tan}^{-} 1 x = \theta$ and ${\cot}^{-} 1 x = \left(\frac{\pi}{2} - \theta\right)$

Hence, for any $x$, tan^(−1)x+cot^(−1)x=(theta+pi/2-theta)=pi/2

Hence for any $x$, cos(tan^(−1)x+cot^(−1)x)=cos(pi/2)=0

Note that $x$ is equal to $\tan \theta$, $x$ can take any value $\left\{- \infty , \infty\right\}$ (as this is the range of $\tan \theta$) including $\frac{5}{12}$.