Question #d720a

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Answer 1 · 2017-11-24T09:08:33

Given #tanx = b/a#

#[(a+b)/(a-b)]^(1/2) + [(a-b)/(a+b)]^(1/2) #

#=[(a/a+b/a)/(a/a-b/a)]^(1/2) + [(a/a-b/a)/(a/a+b/a)]^(1/2) #

#=[(1+tanx)/(1-tanx)]^(1/2) + [(1-tanx)/(1+tanx)]^(1/2) #

#=[(1+tanx)+(1-tanx)]/[(1-tanx)(1+tanx)]^(1/2) #

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

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

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

#=(2cosx)/sqrt(cos2x) #