How do you verify #cos(x) / (1-sin(x)) = sec(x) + tan(x)#?

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Answer 1 · 2018-07-26T16:36:08

Please see the explanation below

We need

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

#secx=1/cosx#

#tanx=sinx/cosx#

The #LHS=cosx/(1-sinx)#

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

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

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

#=(1+sinx)/cosx#

#=1/cosx+sinx/cosx#

#=secx+tanx#

#=LHS#

#QED#

Answer 2 · 2018-07-26T16:41:00

Please see below.

Here,

#LHS=cosx/(1-sinx)#

#"Multiplynumerator and denominator by cosx" #

#LHS=(cosx*cosx)/(cosx(1-sinx))#

#color(white)(LHS)=cos^2x/(cosx(1-sinx))#

#color(white)(LHS)=(1-sin^2x)/(cosx(1-sinx))to[becausesin^2theta+cos^2theta=1]#

#color(white)(LHS)=((1-sinx)(1+sinx))/(cosx(1-sinx))#

#color(white)(LHS)=(1+sinx)/cosx#

#color(white)(LHS)=1/cosx+sinx/cosx#

#color(white)(LHS)=secx+tanx#

#LHS=RHS#