Question

Is anybody else here to help me ??​

Answer 1

`\cot(x)+\cot((\pi)/(2)-x)`

`\cot(x)+\tan(x)`

`(\cos(x))/(\sin(x))+(\sin(x))/(\cos(x))`

`(1)/(\sin(x))(\cos(x)+\sin(x)(\sin(x))/(\cos(x)))`

`\csc(x)(\cos(x)+\sin(x)(\sin(x))/(\cos(x)))`

`\csc(x)[(\cos(x)\cos(x))/(\cos(x))+\sin(x)(sin(x))/(\cos(x))]`

`\csc(x)[(\cos(x)\cos(x)+\sin(x)\sin(x))/(\cos(x))]`

`\csc(x)[(\cos^2(x)+\sin^2(x))/(\cos(x))]`

`\csc(x)[(1)/(\cos(x))]`

`\csc(x)[\sec(x)]`

`\csc(x)[\csc((\pi)/(2)-x)]`

`\csc(x)\csc((\pi)/(2)-x)`

Explanation:

I'm going to use
`x` instead of
`\theta` because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

`\cot(x)+\cot((\pi)/(2)-x)`

I'm going to use a cofunction identity for the 2nd term.

This is the identity:
`\tan(x)=\cot((\pi)/(2)-x)` I'm going to use there.

`\cot(x)+\tan(x)`

I'm going to rewrite this in terms of
`\sin(x)` and
`\cos(x)` because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities:
`(\cos(x))/(\sin(x))=\cot(x)` and
`(\sin(x))/(\cos(x))=\tan(x)`

So we have:

`(\cos(x))/(\sin(x))+(\sin(x))/(\cos(x))`

I'm going to factor out
`(1)/(\sin(x))` because if I do that I will have the
`\csc(x)` factor I see on the right by the reciprocal identity:

`\csc(x)=(1)/(\sin(x))`

`(1)/(\sin(x))(\cos(x)+\sin(x)(\sin(x))/(\cos(x)))`

`\csc(x)(\cos(x)+\sin(x)(\sin(x))/(\cos(x)))`

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show
`\cos(x)+\sin(x)(\sin(x))/(\cos(x))` is equal to
`\csc((\pi)/(2)-x)`.

So since I want one term I'm going to write as a single fraction first:

`\cos(x)+\sin(x)(\sin(x))/(\cos(x))`

Find a common denominator which is
`\cos(x)`:

`(\cos(x)\cos(x))/(\cos(x))+\sin(x)(sin(x))/(\cos(x))`

`(\cos(x)\cos(x)+\sin(x)\sin(x))/(\cos(x))`

`(\cos^2(x)+\sin^2(x))/(\cos(x))`

By the Pythagorean Identity
`\cos^2(x)+\sin^2(x)=1` I can rewrite the top as 1:

`(1)/(\cos(x))`

By the quotient identity
`\sec(x)=(1)/(\cos(x))`, I can rewrite this as:

`\sec(x)`

By the cofunction identity
`\sec(x)=\csc(x)=((\pi)/(2)-x)`, we have the second factor of the right hand side:

`\csc((\pi)/(2)-x)`

Let's just do it all together without all the words now:

`\cot(x)+\cot((\pi)/(2)-x)`

`\cot(x)+\tan(x)`

`(\cos(x))/(\sin(x))+(\sin(x))/(\cos(x))`

`(1)/(\sin(x))(\cos(x)+\sin(x)(\sin(x))/(\cos(x)))`

`\csc(x)(\cos(x)+\sin(x)(\sin(x))/(\cos(x)))`

`\csc(x)[(\cos(x)\cos(x))/(\cos(x))+\sin(x)(sin(x))/(\cos(x))]`

`\csc(x)[(\cos(x)\cos(x)+\sin(x)\sin(x))/(\cos(x))]`

`\csc(x)[(\cos^2(x)+\sin^2(x))/(\cos(x))]`

`\csc(x)[(1)/(\cos(x))]`

`\csc(x)[\sec(x)]`

`\csc(x)[\csc((\pi)/(2)-x)]`

`\csc(x)\csc((\pi)/(2)-x)`