Question #20ee7

1 Answer
Narad T.
Nov 28, 2017

See the proof below

Explanation:

We need

#secx=1/cosx#

#tanx=sinx/cosx#

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

#a^2-b^2=(a+b)(a-b)#

Therefore,

#LHS=(secx+tanx)^2=(1/cosx+sinx/cosx)^2#

#=(1+sinx)^2/(cos^2x)#

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

#=(1+sinx)^2/((1+sinx)(1-sinx))#

#=(1+sinx)/(1-sinx)#

#=RHS#

#QED#

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