Question #16fc9

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Answer 1 · 2017-08-23T06:13:15

See Below

LHS:
#=(1+sin x)/cos x * ( 1-sin x)/(1-sin x)#

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

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

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

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

#:.=#RHS

Answer 2 · 2017-08-23T06:13:39

See the proof below

We need

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

Therefore,

#LHS=(1+sinx)/cosx#

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

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

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

#=cosx/(1-sinx)#

#=RHS#

#QED#