*of the*

**sum***and*

**left***brackets to be 90° or $\frac{\pi}{2}$*

**right**$\sin(\frac{\pi}{2}-\theta)=\cos(\theta)$

$\sin(\theta)=\cos(\frac{\pi}{2}-\theta)$

$\tan(\frac{\pi}{2}-\theta)=\cot(\theta)$

$\tan(\theta)=\cot(\frac{\pi}{2}-\theta)$

$\sec(\frac{\pi}{2}-\theta)=\csc(\theta)$

$\sec(\theta)=\csc(\frac{\pi}{2}-\theta)$