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Answer 1 · 2017-01-26T13:22:10

#LHS=(1+sinx)/cosx#

Multiplying both numerator and denominator by #1-sinx# we get

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

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

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

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

Verified

Answer 2 · 2017-01-26T13:31:18

See the Explanation

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

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

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