How do you verify #cosy/(1-siny)=(1+siny)/cosy#?

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Answer 1 · 2016-11-21T19:31:54

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#cos y/(1-siny)=(1+siny)/cosy#

Left Side : #=cos y/(1-siny)#

#=cos y/(1-siny) * (1+siny)/(1+siny)# --> multiply by conjugate

#= (cos y(1+siny))/(1-sin^2y)#

#= (cos y(1+siny))/cos^2y#

#= (cancelcos y(1+siny))/cos^cancel 2y#

#=(1+siny)/cosy#

#:.=# Right Side