Show that? # (1-tan^2x)/(1+tan^2x) = cos^2-sin^2x #

1 Answer
Steve M
Jan 26, 2018

Consider the LHS of the given expression:

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

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

# \ \ \ \ \ \ \ \ = ((cos^2-sin^2x)/cos^2x)/((cos^2x+sin^2x)/cos^2x) #

# \ \ \ \ \ \ \ \ = ((cos^2-sin^2x)/cos^2x) * (cos^2x/(cos^2x+sin^2x)) #

# \ \ \ \ \ \ \ \ = (cos^2-sin^2x)/(cos^2x+sin^2x) #

# \ \ \ \ \ \ \ \ = (cos^2-sin^2x)/1 \ \ \ # , as #cos^2A+sin^2xA -= 1 #

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

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