How do you simplify #(1+costheta)(csctheta-cottheta)#?

Question

5724 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T20:27:42

When at a loss for what to do, it's a good idea to try writing everything using just #sintheta# and #costheta#

#(1+costheta)(csctheta-cottheta) = (1+costheta)(1/sintheta-costheta/sintheta)#

We have a common denominator in the second factor, so let's write it as a single quotient.

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

Now watch and pay attention. This is a good trick to know.

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

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

And we can simplify some more,

# = (sin^2theta)/sintheta#

# = sin theta#

(for #sintheta != 0# which is the same as: for #theta != pi k# with #k# an integer)