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

1 Answer
Dec 14, 2015

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)