How would I simplify #2cos2x-1= (1-tan2x)/(1+tan2x)# ? I understand how to simplify the right side, however, the left side, the 2 in front of the cosine squared x is confusing me Thank you so much!

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Answer 1 · 2017-06-04T15:13:39

Please see below.

#2cos^2x-1#

= #cos^2x+cos^2x-1#

= #cos^2x+1-sin^2x-1#

= #cos^2x-sin^2x#

= #(cos^2x-sin^2x)/(cos^2x-sin^2x)# #-># note that denominator is #1#

= #(1-sin^2x/cos^2x)/(1+sin^2x/cos^2x# #-># dividing each term by #cos^2x#

= #(1-tan^2x)/(1+tan^2x)#