How do you simplify #(cot θ + tan θ) sec θ#?

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Answer 1 · 2016-06-23T03:52:22

Recall that #tantheta = sintheta/costheta#.

Since #cottheta =1/tantheta#, cottheta = 1/(sintheta/costheta)= costheta/sintheta#.

Also, #sectheta = 1/costheta#.

#=(sintheta/costheta + costheta/sintheta)1/costheta#

#=((sin^2theta + cos^2theta)/(costhetasintheta))1/costheta#

Recall the Pythagorean Identity #sin^2theta + cos^2theta = 1#:

#=1/(costhetasintheta costheta)#

#=1/(cos^2thetasintheta)#

Use the rearranged form of the Pythagorean identity presented above.

#=1/((1 - sin^2theta)sintheta)#

#=1/(sintheta - sin^3theta#

We could have also finished with #sec^2thetacsctheta#, because #1/sintheta = csctheta# and #1/costheta = sectheta#

Hopefully this helps!