Question

Tan(A+45°) - tan(A-45°) =2sec2A​

Answer 1

Answer:

$\tan(A + 45 \degree) - \tan(A - 45 \degree) \\ \\ = \frac{ \tan(A) + \tan(45 \degree) }{1 - \tan(A) \tan(45 \degree) } - \frac{ \tan(A) - \tan(45 \degree) }{1 + \tan(A) \tan(45 \degree) }$

but tan 45° is 1;

$= \frac{ \tan(A) }{1 - \tan(A) } - \frac{ \tan(A) }{1 + \tan(A) } \\ \\ \frac{\tan(A) + { \tan }^{3} A + \tan(A) - { \tan}^{3} A}{1 - { \tan }^{2} A } \\ \\ = (\frac{2 \tan(A) }{1 - { \tan }^{2} A}) \\ \\ = \frac{2 \sin(A) }{ \cos(A) } \times { \sec }^{2} A \\ \\ = \frac{2}{ \cos(2A) } \\ \\ = 2 \sec(2A)$