Question

Tan(A+45°) - tan(A-45°) =2sec2A​

Answer 1

`\tan(A + 45 \degree) - \tan(A - 45 \degree) \\ \\ = ( \tan(A) + \tan(45 \degree) )/(1 - \tan(A) \tan(45 \degree) ) - ( \tan(A) - \tan(45 \degree) )/(1 + \tan(A) \tan(45 \degree) ) \\ \\`

but tan 45° is 1;

`= ( \tan(A) )/(1 - \tan(A) ) - ( \tan(A) )/(1 + \tan(A) ) \\ \\ \frac{\tan(A) + { \tan }^(3) A + \tan(A) - { \tan}^(3) A}{1 - { \tan }^(2) A } \\ \\ = (\frac{2 \tan(A) }{1 - { \tan }^(2) A}) \\ \\ = (2 \sin(A) )/( \cos(A) ) * { \sec }^(2) A \\ \\ = (2)/( \cos(2A) ) \\ \\ = 2 \sec(2A)`