Hi could you please help me with 1+cot^2x/sec^2x ?

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Answer 1 · 2018-04-20T01:01:41

The expression simplifies to #cot^2x#.

First, we need these basic trig identities:

#cscx=1/sinx#

#secx=1/cosx#

#cotx=cosx/sinx#

Now, using the Pythagorean identity,

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

we can derive a new identity if we divide everything by #sin^2x#:

#sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2x#

#1+cot^2x=csc^2x#

Use all these to simplify our expression:

#color(white)=(1+cot^2x)/sec^2x#

#=csc^2x/sec^2x#

#=(1/sin^2x)/(1/cos^2x)#

#=1/sin^2x*cos^2x/1#

#=cos^2x/sin^2x#

#=cot^2x#

That's it. Hope this helped!