How do you prove ##(tany+coty)/cscy=secy##?

Please see below.

##(tany+coty)/cscy##
= ##(siny/cosy+cosy/siny)xx1/(1/siny)##
= ##(sinyxxsiny+cosyxxcosy)/(sinycosy)xxsiny##
= ##(sin^2y+cos^2y)/(cancelsinycosy)xxcancelsiny##
= ##1/cosy##
= ##secy##

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