Is anybody else here to help me ??​

Answer:

[tex]\cot(x)+\cot(\frac{\pi}{2}-x)[/tex]

[tex]\cot(x)+\tan(x)[/tex]

[tex]\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}[/tex]

[tex]\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{1}{\cos(x)}][/tex]

[tex]\csc(x)[\sec(x)][/tex]

[tex]\csc(x)[\csc(\frac{\pi}{2}-x)][/tex]

[tex]\csc(x)\csc(\frac{\pi}{2}-x)[/tex]

Step-by-step explanation:

I'm going to use [tex]x[/tex] instead of [tex]\theta[/tex] 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.

[tex]\cot(x)+\cot(\frac{\pi}{2}-x)[/tex]

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

This is the identity: [tex]\tan(x)=\cot(\frac{\pi}{2}-x)[/tex] I'm going to use there.

[tex]\cot(x)+\tan(x)[/tex]

I'm going to rewrite this in terms of [tex]\sin(x)[/tex] and [tex]\cos(x)[/tex] 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: [tex]\frac{\cos(x)}{\sin(x)}=\cot(x)[/tex] and [tex]\frac{\sin(x)}{\cos(x)}=\tan(x)[/tex]

So we have:

[tex]\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}[/tex]

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

[tex]\csc(x)=\frac{1}{\sin(x)}[/tex]

[tex]\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

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 [tex]\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}[/tex] is equal to [tex]\csc(\frac{\pi}{2}-x)[/tex].

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

[tex]\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}[/tex]

Find a common denominator which is [tex]\cos(x)[/tex]:

[tex]\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}[/tex]

[tex]\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}[/tex]

[tex]\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}[/tex]

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

[tex]\frac{1}{\cos(x)}[/tex]

By the quotient identity [tex]\sec(x)=\frac{1}{\cos(x)}[/tex], I can rewrite this as:

[tex]\sec(x)[/tex]

By the cofunction identity [tex]\sec(x)=\csc(x)=(\frac{\pi}{2}-x)[/tex], we have the second factor of the right hand side:

[tex]\csc(\frac{\pi}{2}-x)[/tex]

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

[tex]\cot(x)+\cot(\frac{\pi}{2}-x)[/tex]

[tex]\cot(x)+\tan(x)[/tex]

[tex]\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}[/tex]

[tex]\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{1}{\cos(x)}][/tex]

[tex]\csc(x)[\sec(x)][/tex]

[tex]\csc(x)[\csc(\frac{\pi}{2}-x)][/tex]

[tex]\csc(x)\csc(\frac{\pi}{2}-x)[/tex]

Anyone that answers this right will get 100+ points, any wrong answers or spamming will be reported.

need help i dont know what I am doing

I give brainlest, please help.

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