what is the above answer in detail
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Step-by-step explanation:
(tana/1-cota +cota/1-tana)+1+secacoseca
=[(sina/cosa)/(sina-cosa/sina)+(cosa/sina)/(cosa-sina/cosa)]+1+secacoseca
=[sin²a/cosa(sina-cosa)-cos²a/sina(sina-cosa)]+1+secacoseca
=[(sin³a-cos³a)/sinacosa(sina-cosa)]+1+secacoseca
=[(sina-cosa)(1+sinacosa)/sinacosa(sina-cosa)]+1+secacoseca
={1+sinacosa/sinacosa}+1+secacoseca
=secacoseca+1+1+secacoseca
=2(1+secacoseca)
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