Question

tanA/I+secA- tanA/1-secA= 2 cosec A

Answer 1

Step-by-step explanation:

I hope you understand....

Answer 2

Answer:

as you know

$1 + { \tan(a) }^{2} = { \sec(a) }^{2} \\ = { \tan(a) }^{2} = { \sec(a) }^{2} - 1 \\ - { \tan(a) }^{2} = 1 - { \sec(a) }^{2}$

Step-by-step explanation:

$\frac{tana}{1 + seca} - \frac{ \tan(a) }{1 - \sec(a) } \\ = \frac{ \tan(a)(1 - \sec(a) ) - \tan(a) (1 + \sec(a) )}{(1 + \sec(a) )(1 - \sec(a) } \\ = \frac{ \tan(a) - \tan(a) \sec(a) - \tan(a) - \tan(a) \sec(a) }{1 - { \sec(a) }^{2} } \\ = \frac{ - 2 \tan(a) \sec(a) }{ - { \tan(a) }^{2} } \\ = \frac{2 \sec(a) }{ \tan(a) } \\ = \frac{ 2\frac{1}{ \cos(a) } }{ \frac{ \sin(a) }{ \cos(a) } } \\ = 2 \frac{1}{ \sin(a) } \\ = 2 \csc(a)$