Cot90-thetha/tan thetha+cosec 90- thetha sunglass thetha/tan thetha

1.	Cot90-thetha/tan thetha+cosec 90- thetha sunglass thetha/tan thetha
Answer» Let H =1, therefoe P=(a2-b2)/(a2+b2) and B2 =1^2-(a2-b2)^2/(a2+b2)^2 =1-(a4+b4-2a2b2)/(a4+b4+2a2b2 ) =(a4+b4+2a2b2-a4-b4+2a2b2)/(a4+b4+2a2b2) = 4a2b2/(a2+b2)^2 B=2ab/(a+b) cos =B/H=B/1=B tan=P/H.Put the value of perpendicular and base and find the value of cos ,cosec and cot by reciprocaling these three ratios

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