# Question #62402

Jul 23, 2016

see explanation.

#### Explanation:

There are 3 possible approaches that may be taken.

(1) Manipulate left side to obtain the right side.

(2) Manipulate the right side to obtain the left side.

(3) Manipulate both sides together until they are the same.

Using (1) along with the following $\textcolor{b l u e}{\text{trig. identities}}$

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

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

left side =$\frac{\sin A \cos B - \cos A \sin B}{\sin A \cos B}$

$= \frac{{\cancel{\sin A \cos B}}^{1}}{{\cancel{\sin A \cos B}}^{1}} - \frac{\cos A \sin B}{\sin A \cos B}$

$= 1 - \frac{\cos A}{\sin A} \times \frac{\sin B}{\cos B}$

$= 1 - \cot A \tan B = \text{ right side"rArr"proved}$