Identities : Theta?

[BombBomb]

Help please. There is theta and it's confusing.

(tan[theta])(1 + cot^2 theta)
cot[theta]


Also, I don't know how to put the symbols. So, if you can help with that I will be happy. =p


Edit:

I need help with this too.

(csc x)(sec x - cos x)

I got this so far..

(1/sin x)(1/cos x - cos x)

or is it wrong? Should I multiply the 2? Maybe put it in pythagorean identities? :/ I'll try these...but I need help so I can be sure.

 

[Loren]

$\displaystyle \frac{\tan x(1+\cot^2x)}{\cot x}$

1+cot[sup:22vn2rtq]2[/sup:22vn2rtq]x = csc[sup:22vn2rtq]2[/sup:22vn2rtq]x

And if you get everything in terms of sin and cosine, I think you will see some cancellation and end up with sec[sup:22vn2rtq]2[/sup:22vn2rtq]x.

 

[Deleted member 4993]

Help please. There is theta and it's confusing.

(tan[theta])(1 + cot^2 theta)
cot[theta]

$\displaystyle \frac{tan(\theta)\cdot \left [1 + cot^2(\theta)\right ]}{cot(\theta)}$

$\displaystyle = \frac{tan(\theta) + cot(\theta)}{cot(\theta)}$

$\displaystyle = tan^2(\theta) + 1$

$\displaystyle = sec^2(\theta)$


Also, I don't know how to put the symbols. So, if you can help with that I will be happy. =p


Edit:

I need help with this too.

(csc x)(sec x - cos x)

I got this so far..

(1/sin x)(1/cos x - cos x)

or is it wrong? Should I multiply the 2? Maybe put it in pythagorean identities? :/ I'll try these...but I need help so I can be sure.

 

[Deleted member 4993]

Help please. There is theta and it's confusing.

(tan[theta])(1 + cot^2 theta)
cot[theta]

$\displaystyle \frac{tan(\theta)\cdot \left [1 + cot^2(\theta)\right ]}{cot(\theta)}$

$\displaystyle = \frac{tan(\theta) + cot(\theta)}{cot(\theta)}$

$\displaystyle = tan^2(\theta) + 1$

$\displaystyle = sec^2(\theta)$


Also, I don't know how to put the symbols. So, if you can help with that I will be happy. =p


Edit:

I need help with this too.

(csc x)(sec x - cos x)

I got this so far..

(1/sin x)(1/cos x - cos x) = 1/sin(x) * [1-cos[sup:3u8z75r7]2[/sup:3u8z75r7](x)]/cos(x) = 1/sin(x) * [sin[sup:3u8z75r7]2[/sup:3u8z75r7](x)]/cos(x) ... continue

or is it wrong? Should I multiply the 2? Maybe put it in pythagorean identities? :/ I'll try these...but I need help so I can be sure.

 

[Denis]

Help please. There is theta and it's confusing.

(tan[theta])(1 + cot^2 theta)
cot[theta]

Well, if it's confusing, simply change "theta" to "x"; kapish?

 

[soroban]

Hello, BombBomb!

You don't like $\displaystyle \theta$? . . . Change it to $\displaystyle x.$

$\displaystyle \frac{\tan x(1 + \cot^2\!x)}{\cot x}$

$\displaystyle \text{We have: }\;\frac{\tan x\csc^2\!x}{\frac{1}{\tan x}} \;=\;\tan^2\!x\csc^2\!x \;=\;\left(\frac{\sin x}{\cos x}\right)^2\left(\frac{1}{\sin x}\right)^2 \; = \;\frac{\sin^2\!x}{\cos^2\!x}\cdot\frac{1}{\sin^2\!x} \;=\;\frac{1}{\cos^2\!x} \;=\;\sec^2\!x$

$\displaystyle \csc x (\sec x - \cos x)$

$\displaystyle \text{We have: }\;\frac{1}{\sin x}\left(\frac{1}{\cos x} - \cos x\right) \;=\;\frac{1}{\sin x}\left(\frac{1-\cos^2\!x}{\cos x}\right) \;=\;\frac{1}{\sin x}\cdot\frac{\sin^2\!x}{\cos x} \;=\;\frac{\sin x}{\cos xc} \;=\;\tan x$