Question

Prove? `cscx(1+cosx)(cscx-cotx)=1`

Answer 1

See explanation

Explanation:

We will use the following:

• `(a+b)(a-b) = a^2-b^2`
• `csc(x) = 1/sin(x)`
• `cot(x) = cos(x)/sin(x)`
• `1-cos^2(x) = sin^2(x)`

With those,

`csc(x)(1+cos(x))(csc(x)-cot(x))`

`= (csc(x)+cot(x))(csc(x)-cot(x))`

`=csc^2(x)-cot^2(x)`

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

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

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

`=1`

Answer 2

`LHS=cosectheta(1+costheta)(cosectheta-cottheta)`

`=(cosectheta+cosecthetacostheta)(cosectheta-cottheta)`

`=(cosectheta+1/sinthetaxxcostheta)(cosectheta-cottheta)`

`=(cosectheta+cottheta)(cosectheta-cottheta)`

`=(cosec^2theta-cot^2theta)=1=RHS`

proved

Answer 3

See below:

Explanation:

We have:

`cscx(1+cosx)(cscx-cotx)=1`

distributing out:

`(cscx+cscxcosx)(cscx-cotx)=1`

`csc^2x-cscxcotx+csc^2xcosx-cscxcosxcotx=1`

and now to reorder and simplify:

`1/sin^2x-cancel(cosx/sin^2x)+cancel(cosx/sin^2x)-cos^2x/sin^2x=1`

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

Now we'll use the identity of `sin^2x+cos^2x=1` and so `sin^2x=1-cos^2x`

`(sin^2x)/sin^2x=1`

`1=1`