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Answer:
ya!
Step-by-step explanation:
Find in the attachment
=[(sinx/cosx)/(1+1/cosx)] + [(1+1/cosx)/(sinx/cosx)]
=[(sinx/cosx)/(cosx+1/cosx)]+[(cosx+1/cosx)/(sinx/cosx)]
= [sinx/(cosx+1)] + [(cosx+1)/sinx]
= [sin^2x+(cosx+1)^2] / [sinx (cosx+1)]
= [2+2cosx] / [sinx(cosx+1)]
=[2(cosx+1)] / [sinx (cosx+1)]
= 2/sinx
= 2 cscx
(I think this will be helpful for you. if you can see the picture, it has more detail in it.)
=[(sinx/cosx)/(cosx+1/cosx)]+[(cosx+1/cosx)/(sinx/cosx)]
= [sinx/(cosx+1)] + [(cosx+1)/sinx]
= [sin^2x+(cosx+1)^2] / [sinx (cosx+1)]
= [2+2cosx] / [sinx(cosx+1)]
=[2(cosx+1)] / [sinx (cosx+1)]
= 2/sinx
= 2 cscx
(I think this will be helpful for you. if you can see the picture, it has more detail in it.)