Question

[1/(1 − sin x)]− [1/(1 + sin x)]= 2 tan x sec x

Answer 1

The way I learned how to figure out these problems is you start with one side of the equation of the equation and then solve it for the opposite side. So, you could either start with the left side and do the distributive property or you could start with the right side and go with; (2sinx/cosx)(1/cosx) and do it that way.

Answer 2

LHS
[1/(1 − sin x)]− [1/(1 + sin x)]

Multiply the numerator and denominator with secx ; the rest is obvious. :)

Answer 3

well why not put the left hand side over a common denominator

\[\frac{1 + \sin(x) - (1 - \sin(x)}{(1 - \sin(x))(1 + \sin(x))}\]

the denominator is the difference of 2 squares

simplify the numerator and you get

\[\frac{2\sin(x)}{1 - \sin^2(x)}\]

the denominator can be written as

\[\frac{2\sin(x)}{\cos^2(x))}\]

hopefully you can simplify it from here...

Answer 4

my answer was negative every time.I hate how its always a small mistake. thanks everyone.