# Prove that tan(A+B)*tan(A-B)=cos²B-cos²A/cos²B-sin²A?

Dec 18, 2017

It is proved below:

#### Explanation:

$L . H . S = \tan \left(A + B\right) . \tan \left(A - B\right)$

=(sin(A+B)sin(A-B))/(cos(A+B)cos(A-B))color(brown)[[[As,tantheta=sintheta/costheta]]

$= \frac{\left(\sin A \cos B + \cos A \sin B\right) \left(\sin A \cos B - \cos A \sin B\right)}{\left(\cos A \cos B - \sin A \sin B\right) \left(\cos A \cos B + \sin A \sin B\right)} \textcolor{\mathmr{and} a n \ge}{\left[A s . \sin \left(a + b\right) = \sin a \cos b + \cos a \sin b \mathmr{and} \cos \left(a + b\right) = \cos a \cos b - \sin a \sin b\right]}$

$= \frac{{\sin}^{2} A {\cos}^{2} B - {\cos}^{2} A {\sin}^{2} B}{{\cos}^{2} A {\cos}^{2} B - {\sin}^{2} A {\sin}^{2} B}$$\textcolor{b l u e}{\left[A s . \left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}\right]}$

$= \frac{\left(1 - {\cos}^{2} A\right) {\cos}^{2} B - {\cos}^{2} A \left(1 - {\cos}^{2} B\right)}{\left(1 - {\sin}^{2} A\right) {\cos}^{2} B - {\sin}^{2} A \left(1 - {\cos}^{2} B\right)}$$\textcolor{g r e e n}{\left[A s . {\sin}^{2} \theta = 1 - {\cos}^{2} \theta \mathmr{and} {\cos}^{2} \theta = 1 - {\sin}^{2} \theta\right]}$

$= \frac{{\cos}^{2} B - \cancel{{\cos}^{2} A {\cos}^{2} B} - {\cos}^{2} A + \cancel{{\cos}^{2} A {\cos}^{2} B}}{{\cos}^{2} B - \cancel{{\sin}^{2} A {\cos}^{2} B} - {\sin}^{2} A + \cancel{{\sin}^{2} A {\cos}^{2} B}}$

$= \frac{{\cos}^{2} B - {\cos}^{2} A}{{\cos}^{2} B - {\sin}^{2} A} = R . H . S$$\textcolor{b l u e}{\left[P r o v e d\right]}$