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Answer to Question #20858 in Trigonometry for Prattyush

Question #20858
prove that under root sec square a plus under root cosec square a=cot a +tan a
Expert's answer
Sqrt(sec^2(a)+cosec^2(a))==Sqrt(1/sin^2a+1/cos^2a)=Sqrt((sin^2a+cos^2a)/sin^2acos^2a=1/sinacosa=(sin^2a+cos^2a)/sinacosa=sina/cosa+cosa/sina=tga+cota

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