Show that #(2cos^2(x/2)tanx)/(tanx) = (tanx+cosxtanx)/(tanx)#?

1 Answer
Oct 24, 2015

Okay, let's consider dividing both sides by #tanx# to begin with:

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

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

Notice how we can use the identity #sin^2(u) + cos^2(u) = 1#, where #u = x/2#:

#2cos^2(x/2) - [sin^2(x/2) + cos^2(x/2)] = cosx#

#2cos^2(x/2) - sin^2(x/2) - cos^2(x/2) = cosx#

#cos^2(x/2) - sin^2(x/2) = cosx#

Then, we can use the identity #cos(u + v) = cosucosv - sinusinv#, where #u = v = x/2# and work backwards:

#cos(x/2)cos(x/2) - sin(x/2)sin(x/2) = cosx#

#cos(x/2 + x/2) = cosx#

#color(blue)(cosx = cosx)#