Question #71ebe

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Answer 1 · 2017-04-07T15:51:11

I have changed the question so that it is an identity. (Without the parentheses it is not.)

Start with

#(1+cosx)/(1-cosx) - (1-cosx)/(1+cosx)#

We want a single quotient, so get a common denominator and simplify.

#(1+cosx)/(1-cosx) - (1-cosx)/(1+cosx) = ((1+cosx)(1+cosx) - (1-cosx)(1-cosx))/((1-cosx)(1+cosx))#

# = ((1+2cosx+cos^2x)-(1-2cosx+cos^2x))/(1-cos^2x)#

It's hard to see whether this helped, so keep simplifying.

# = (1+2cosx+cos^2x-1+2cosx-cos^2x)/(1-cos^2x)#

# = (4cosx)/(1-cos^2x)#

Now, at least I see a #4# in the numerator, so that's good. We want a #sinx# in the denominator, so use the Pythagorean identity to get

# = (4cosx)/sin^2x#

Now, recall that #tanx = sinx/cosx# and rewrite

# = (4cosx)/(sinxsinx) = 4/(sinxsinx/cosx) = 4/(sinxtanx)#