How do you verify the identify tantheta+cottheta=secthetacsctheta?

Sep 28, 2016

see explanation.

Explanation:

We attempt to show by manipulation that the left side has the same form as the right side.

Using the $\textcolor{b l u e}{\text{trigonometric identities}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\tan \theta = \frac{\sin \theta}{\cos \theta} \text{ and } \cot \theta = \frac{\cos \theta}{\sin \theta}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

left side $= \tan \theta + \cot \theta$

$= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin} \theta$

To combine these fractions we require a common denominator of $\cos \theta \sin \theta .$

$\Rightarrow \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta} + \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta}$

$= \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos \theta \sin \theta}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} \theta + {\cos}^{2} \theta = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sec \theta = \frac{1}{\cos \theta} \text{ and } \csc \theta = \frac{1}{\sin \theta}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow \frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos \theta \sin \theta} = \frac{1}{\cos \theta} \times \frac{1}{\sin \theta}$

$= \sec \theta \csc \theta = \text{ right side"rArr" verified}$