PLEASE! Help me with this identity....

[Verybad@math]

So I have this identity problem that has me stumped! Could anyone help?

(cscx/sinx)-(cotx/tanx)=1

This one too.....

tan^2xcos^2x+cot^2x=1

 

[chrisr]

C'mon, you can't be giving yourself a name like that,

What is cosec(x)? it's 1/(Sinx), so cosec(x)/(Sinx) = 1/(Sinx) times 1/(Sinx).
This is written 1/(Sin[sup:27uwv76q]2[/sup:27uwv76q]x), which distinguishes it from 1/(Sin{x[sup:27uwv76q]2[/sup:27uwv76q]}).

Similarly, the other part is 1/(tan[sup:27uwv76q]2[/sup:27uwv76q]x).

Since tanx = Sinx/(Cosx),
1/(tanx) = Cosx/(Sinx), then we have 1/(Sin[sup:27uwv76q]2[/sup:27uwv76q]x) - Cos[sup:27uwv76q]2[/sup:27uwv76q]x/(Sin[sup:27uwv76q]2[/sup:27uwv76q]x).

This is (1-Cos[sup:27uwv76q]2[/sup:27uwv76q]x)/(Sin[sup:27uwv76q]2[/sup:27uwv76q]x) which is Sin[sup:27uwv76q]2[/sup:27uwv76q]x/(Sin[sup:27uwv76q]2[/sup:27uwv76q]x).

Practice with that and try to prove the second one.

 

[chrisr]

Possibly you've miswritten the second one...

Did you mean....

Prove that (tan[sup:6pkvk6jl]2[/sup:6pkvk6jl]x)(Cos[sup:6pkvk6jl]2[/sup:6pkvk6jl]x)+(cotan[sup:6pkvk6jl]2[/sup:6pkvk6jl]x)(Sin[sup:6pkvk6jl]2[/sup:6pkvk6jl]x) = 1 ?

 

[soroban]

Hello, Verybad@math!

\(\displaystyle \frac{\csc x}{\sin x} -\frac{\cot x}{\tan x} \:=\:1\)

\(\displaystyle \frac{\csc x}{\sin x} - \frac{\cot x}{\tan x} \;\;=\;\;\frac {\csc x}{\frac{1}{\csc x}} - \frac{\cot x}{\frac{1}{\cot x}} \;\;=\;\;\csc^2\!x - \cot^2\!x \;\;=\;\;1\)

\(\displaystyle \tan^2\!x \cos^2\!x +\cot^2\!x \:=\:1\)

This is not an identity . . .

\(\displaystyle \text{Did you mean: }\;\tan^2\!x\cos^2\!x + \cos^2\!x \:=\:1\;?\)

\(\displaystyle \text{We have: }\;\tan^2\!x\cos^2\!x + \cos^2\!x \;\;=\;\;\cos^2\!x(\tan^2\!x + 1) \;\;=\;\;\cos^2\!x\sec^2\!x \;\;=\;\;1\)