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

1 Answer
Kevin G.
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)#

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