## Trigonometry (11th Edition) Clone

$sec~\theta-sin~\theta~tan~\theta= cos~\theta$
We can express $sec~\theta-sin~\theta~tan~\theta$ as a single function of $\theta$: $sec~\theta-sin~\theta~tan~\theta$ $= \frac{1}{cos~\theta}-sin~\theta~\frac{sin~\theta}{cos~\theta}$ $= \frac{1}{cos~\theta}-\frac{sin^2~\theta}{cos~\theta}$ $= \frac{1-sin^2~\theta}{cos~\theta}$ $= \frac{cos^2~\theta}{cos~\theta}$ $= cos~\theta$