How to prove #cot^4x(sec^4x-1)-2cot^2x = 1# ?

#LHS=cot^4x(sec^4x-1)-2cot^2x#

#=cot^4x(sec^2x-1)(sec^2x+1)-2cot^2x#

#=cot^4xtan^2x(sec^2x+1)-2cot^2x#

#=cot^2x(sec^2x+1)-2cot^2x#

#=cot^2xsec^2x+cot^2x-2cot^2x#

#=cos^2x/sin^2x xx1/cos^2x-cot^2x#

#=csc^2x-cot^2x#

# = 1=RHS#

We will use, #sec^2x=1+tan^2x#.

#cot^4x(sec^4x-1)-2cot^2x#,

#=cot^2x*cot^2x*(sec^2x-1)(sec^2x+1)-2cot^2x#,

#=cot^2x{(cot^2x)(tan^2x)}{(1+tan^2x)+1}-2cot^2x#,

#=cot^2x{1}(2+tan^2x)-2cot^2x#,

#=cancel(2cot^2x)+cot^2x*tan^2x-cancel(2cot^2x)#,

#=1#, as desired!

Use these two trig identities:

#cottheta=costheta/sintheta#

#sectheta=1/costheta#

And the Pythagorean identity (and its variations):

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

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

Now, rewrite the problem in terms of sine and cosine, then simplify it down using the Pythagorean identity:

#color(white)=cot^4x(sec^4x-1)-2cot^2x#

#=cos^4x/sin^4x(1/cos^4x-1)-2*cos^2x/sin^2x#

#=(cos^4x)/sin^4x*1/(cos^4x)-cos^4x/sin^4x*1-2*cos^2x/sin^2x#

#=color(red)cancelcolor(black)(cos^4x)/sin^4x*1/color(red)cancelcolor(black)(cos^4x)-cos^4x/sin^4x-2*cos^2x/sin^2x#

#=1/sin^4x-cos^4x/sin^4x-2*cos^2x/sin^2x#

Getting a common denominator:

#=(1-cos^4x-2cos^2xsin^2x)/sin^4x#

#=(1-2cos^2xsin^2x-cos^4x)/sin^4x#

#=(1-cos^2xsin^2x-cos^2xsin^2x-cos^4x)/sin^4x#

#=(1-cos^2xsin^2x-cos^2x(sin^2x+cos^2x))/sin^4x#

#=(1-cos^2xsin^2x-cos^2x)/sin^4x#

Rewrite #1# as #sin^2x+cos^2x#:

#=(sin^2x+cos^2x-cos^2xsin^2x-cos^2x)/sin^4x#

#=(sin^2x+color(red)cancelcolor(black)(cos^2x)-cos^2xsin^2xcolor(red)cancelcolor(black)(color(black)-cos^2x))/sin^4x#

#=(sin^2x-cos^2xsin^2x)/sin^4x#

#=(sin^2x(1-cos^2x))/sin^4x#

#=(sin^2x*sin^2x)/sin^4x#

#=(sin^4x)/sin^4x#

#=1#

That's the proof. Hope this helped!