How do you simplify #(1+csc x)/(cos x+ cot x)#?

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Answer 1 · 2015-11-08T22:31:45

#(1+cscx)/(cosx + cotx) = secx#

We will be using the following identities:

#cscx = 1/sinx#

#cotx = cosx/sinx#

#secx = 1/cosx#

#cosx + cotx = cosx(1+1/sinx) = cosx(1+cscx)#

Thus

#(1+cscx)/(cosx + cotx) = (1+cscx)/(cosx(1+cscx)) = 1/cosx = secx#