How do you simplify #costheta/(sin^2theta-1)#?

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Answer 1 · 2016-11-15T04:29:53

#-sectheta#, #theta!=pi/2+-n2pi# where n is a natural number

We have:

#costheta/(sin^2theta-1)#

Remember that there is an identity:

#sin^2theta+cos^2theta=1#

and so

#cos^2theta=-sin^2theta+1# which can be written as:

#cos^2theta=-(sin^2theta-1)# and so

#-cos^2theta=(sin^2theta-1)#

so let's substitute that into our original statement:

#costheta/(sin^2theta-1)#

#costheta/(-cos^2theta)#

and now we can simplify:

#1/(-costheta)=-sectheta#

Ok, so now what are the limitations? It's #sin^2theta!=1# (that would make the denominator in our original question be 0) and so:

#theta!=pi/2+-n2pi# where n is a natural number