Question

Prove the above equation
(Attached file)​

Answer 1

Step-by-step explanation:

${ \tan( \alpha ) }^{2} - { \tan( \beta ) }^{2} = { \sec( \alpha ) }^{2} - 1 - ( { \sec( \beta ) }^{2} - 1) \\ \\ = { \sec( \alpha ) }^{2} - { \sec( \beta ) }^{2} \\ \\ = \frac{1}{ { \cos( \alpha ) }^{2} } - \frac{1}{ { \cos( \beta ) }^{2} } \\ \\ \\ = \frac{ { \cos( \beta ) }^{2} - { \cos( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} { \cos( \beta ) }^{2} } \\ \\ = \frac{1 - { \sin( \beta ) }^{2} - (1 - { \sin( \alpha )}^{2}) }{ { \cos( \alpha ) }^{2} { \cos( \beta ) }^{2} } \\ \\ \\ \\ = \frac{ { \sin( \alpha ) }^{2} - { \sin( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} { \cos( \beta ) }^{2} }$