How do you simplify #csc^2(θ) / (1 + tan^2(θ) )#?

2 Answers
Sep 21, 2015

Answer:

#cot^2 theta#

Explanation:

Let's see, we have:

#(csc^2 theta)/(1+tan^2 theta)#

In this case, I would probably start by rewriting everything in terms of sines and cosines, since we know that
#tan theta = sin theta/cos theta#
and
#csc theta = 1/sin theta#

So we now have:

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

Here, I notice that the denominator can be simplified if I would multiply it by #cos^2 theta#. Here is what I mean. The denominator is:
#1+(sin^2 theta)/(cos^2 theta)#
and if I multiply #cos^2 theta#, we get:
#(1+(sin^2 theta)/(cos^2 theta) )* cos^2 theta#
Opening the parentheses, this becomes:

#cos^2 theta + sin^2 theta#
which is our famous identity and is #=1#.

Yes. The denominator becomes 1 if I would multiply by #cos^2 theta#.
So let's do the following:

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

We can do this since any number over itself (except zero) is equal to 1.
So provided that #theta != pi/2# or #-pi/2#, we can multiply our equation by
#cos^2 theta / cos^2 theta = 1#

Now, we know the denominator becomes 1, and we are left with just the "top":

#= cos^2 theta / sin^2 theta#
#=cot^2 theta#

where #cot theta# is the cotangent of theta.

Sep 22, 2015

Answer:

Simplify: #f(x) = (csc^2 x)/(1 + tan^2 x)#

Ans: #cot^2 x#

Explanation:

Use the trig identity: #1 + tan^2 x = sec^2 x#. We get:

#f(x) = (csc^2 x)/(sec^2 x) = (1/sin^2 x)/(1/cos^2 x) = #
#= (cos^2 x)/(sin^2 x) = cot^2 x#