Question

How do you prove `2tan(x)sec(x) = (1/(1-sin(x))) - (1/(1+sin(x)))`?

Answer 1

See explanation...

Explanation:

Recall these identities involving trig functions:

Quotient Identity: `tan(x) = sin(x)/cos(x)`
Reciprocal Identity: `sec(x) = 1/cos(x)`
Pythagorean Identity: `1-sin^2(x)=cos^2(x)`

Using these, we can rewrite the identity:

`2tan(x)sec(x) = 1/(1-sin(x)) - 1/(1+sin(x))`
`2sin(x)/cos^2(x) = 1/(1-sin(x))-1/(1+sin(x))`

We then make common denominators:

`2sin(x)/cos^2(x) = (1+sin(x))/(1-sin^2(x))-(1-sin(x))/(1-sin^2(x))`
`2sin(x)/cos^2(x) = (1+sin(x)-(1-sin(x)))/(1-sin^2(x))`
`2sin(x)/cos^2(x) = 2sin(x)/(1-sin^2(x))`

And simplify using the Pythagorean Identity from above:

`2sin(x)/cos^2(x) = 2sin(x)/cos^2(x)`