Question

Question #cbd96

Answer 1

Look at the steps shown in the description...

Explanation:

`(cot^2x)(1-cos^2x)=1-sin^2x`

So using the Left Hand Side:

LHS`=(cot^2x)(1-cos^2x)`

Expand the brackets:

`=cot^2x-cot^2x*cos^2x`

Using this: `cot^2x=1/(tan^2x)`

`=1/(tan^2x)-1/(tan^2x)cos^2x`

`=1/(tan^2x)-(cos^2x)/(tan^2x)`

Turn into single fraction:

`=(1-cos^2x)/tan^2x`

Using this: `1-cos^2x=sin^2x` (derived from `sin^2x+cos^2x=1`)

`=(sin^2x)/tan^2x`

Using this: `tan^2x=sin^2x/(cos^2x)`

`=(sin^2x)/(sin^2x/cos^2x)`

flip and multiply (dividing multiple fractions):

`=(sin^2x)*(cos^2x/sin^2x)`

`=(sin^2x*cos^2x)/sin^2x`

`=(cancel(sin^2x)*cos^2x)/cancel(sin^2x)`

`=cos^2x`

Using this: `cos^2x=1-sin^2x` (derived from `sin^2x+cos^2x=1`)

`=1-sin^2x`

Hence Left Hand Side (LHS) `=` Right Hand Side (RHS)

`=>(cot^2x)(1-cos^2x)=1-sin^2x`

Answer 2

See the proof below

Explanation:

We need

`cotx=cosx/sinx`

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

`1-cos^2x=sin^2x`

`cos^2x=1-sin^2x`

Therefore,

`LHS=(cot^2x)(1-cos^2x)`

`=cos^2x/cancel(sin^2x)*cancel(sin^2x)`

`=cos^2x`

`=1-sin^2x`

`=RHS`

`QED`