How do you prove #tan2x + sec2x = (cosx + sinx) / (cosx – sinx)#?

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Answer 1 · 2016-07-01T11:21:59

As below

#RHS = (cosx + sinx) / (cosx – sinx)#

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

#= (cosx + sinx) ^2/ (cos^2x – sin^2x)#

# = (cos^2x + sin^2x+2sinxcosx) / (cos2x)#

#=(1+sin2x)/(cos2x)=1/(cos2x)+(sin2x)/(cos2x)#

#=tan2x+sec2x=LHS#

proved