## Trigonometry (11th Edition) Clone

$\tan(-\frac{2\pi}{5})$ is the answer to this exercise.
$$\cot\frac{9\pi}{10}$$ As tangent and cotangent are cofunctions, in order to write $\cot\frac{9\pi}{10}$ in terms of cofunction, we now need to find $\theta$, which must satisfy $$\tan\theta=\cot\frac{9\pi}{10}\hspace{1cm}(1)$$ According to Cofunction Identity: $\tan\theta=\cot(\frac{\pi}{2}-\theta)$ Apply this to the equation $(1)$: $$\cot(\frac{\pi}{2}-\theta)=\cot(\frac{9\pi}{10})$$ $$\frac{\pi}{2}-\theta=\frac{9\pi}{10}$$ $$\theta=\frac{\pi}{2}-\frac{9\pi}{10}=\frac{-4\pi}{10}=-\frac{2\pi}{5}$$ $\tan(-\frac{2\pi}{5})$ is the answer to this exercise.