What is the value of #tanx + cotx#?

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Answer 1 · 2016-07-01T02:57:26

C is the correct answer.

You need to put on a common denominator. This will be #sinxcosx#.

#=(sin^2x + cos^2x)/(sinxcosx)#

Applying the identity #sin^2x + cos^2x = 1#:

#=1/(sinxcosx)#

Now, recall that #1/sintheta = csctheta# and #1/costheta = sectheta#.

#=cscxsecx#

So, C is the answer that corresponds.

Hopefully this helps!