Identities : Theta?

BombBomb

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Oct 15, 2009
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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.
 
\(\displaystyle \frac{\tan x(1+\cot^2x)}{\cot x}\)

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.
 
BombBomb said:
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.
 
Subhotosh Khan said:
BombBomb said:
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.
 
BombBomb said:
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?
 
Hello, BombBomb!

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


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

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




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

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

 
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