Question

tan(A+B)/tan(A-B)=sin^2A-sin^2B/cos^2A-sin^2B

Answer 1

Answer:

$\frac{ \tan( \alpha + \beta ) }{ \tan( \alpha - \beta ) } \\ \\ = \frac{ \sin( \alpha + \beta ) \cos( \alpha - \beta ) }{ \sin( \alpha - \beta ) \cos( \alpha + \beta ) } \\ \\ \frac{( \sin( \alpha ) \cos( \beta ) + \cos( \alpha ) \sin( \beta ) ) ( \cos( \alpha ) \cos( \beta ) + \sin( \alpha ) \sin( \beta ) ) }{( \sin( \alpha ) \cos( \beta ) - \cos( \alpha ) \sin( \beta ) ) ( \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta )) } \\ \\ = \frac{ \sin( \alpha ) \cos( \alpha ) ( \cos {}^{2} ( \beta ) + \sin {}^{2} ( \beta )) + \sin( \beta ) \cos( \beta )( \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha )) }{\sin( \alpha ) \cos( \alpha ) ( \cos {}^{2} ( \beta ) - \sin {}^{2} ( \beta )) + \sin( \beta ) \cos( \beta )( \sin {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha )) } \\ \\ = \frac{ \sin( \alpha ) \cos( \alpha ) + \sin( \beta ) \cos( \beta ) }{ \sin( \alpha ) \cos( \alpha ) \cos(2 \beta ) - \sin( \beta ) \cos( \beta ) \cos(2 \alpha ) }$

Multiply and divide by 2 , use sin2x formula ,