Question

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

Answer 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`

Answer 2

Please look at a Proof in Explanation.

Explanation:

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!

Answer 3

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!