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

1 Answer
Apr 20, 2018

The expression simplifies to cot^2xcot2x.

Explanation:

First, we need these basic trig identities:

cscx=1/sinxcscx=1sinx

secx=1/cosxsecx=1cosx

cotx=cosx/sinxcotx=cosxsinx

Now, using the Pythagorean identity,

sin^2x+cos^2x=1sin2x+cos2x=1

we can derive a new identity if we divide everything by sin^2xsin2x:

sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2xsin2xsin2x+cos2xsin2x=1sin2x

1+cot^2x=csc^2x1+cot2x=csc2x

Use all these to simplify our expression:

color(white)=(1+cot^2x)/sec^2x=1+cot2xsec2x

=csc^2x/sec^2x=csc2xsec2x

=(1/sin^2x)/(1/cos^2x)=1sin2x1cos2x

=1/sin^2x*cos^2x/1=1sin2xcos2x1

=cos^2x/sin^2x=cos2xsin2x

=cot^2x=cot2x

That's it. Hope this helped!