Thread: In terms of tangent beta

1. In terms of tangent beta

Express
(sin BETA + cos BETA)

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in terms of tan Beta how to do???
(cos BETA)

2. Originally Posted by j0nath0n3
Express
(sin BETA + cos BETA)

____________________
in terms of tan Beta how to do???
(cos BETA)

$\frac{\sin \beta + \cos \beta}{\cos \beta}=\frac{\sin \beta}{\cos \beta}+\frac{\cos \beta}{\cos \beta}=\tan \beta + 1$

3. Originally Posted by masters
$\frac{\sin \beta + \cos \beta}{\cos \beta}=\frac{\sin \beta}{\cos \beta}+\frac{\cos \beta}{\cos \beta}=\tan \beta + 1$
Thanked!
sec theta - sin theta tan theta in terms of cos theta

4. Originally Posted by j0nath0n3
sec theta - sin theta tan theta in terms of cos theta
$\sec{\theta} - \sin{\theta}\tan{\theta} = \frac{1}{\cos{\theta}}- \frac{\sin^2{\theta}}{\cos{\theta}}$

Recall that $1 - \sin^2{\theta} = \cos^2{\theta}$. Carry on.

5. Originally Posted by Chop Suey
$\sec{\theta} - \sin{\theta}\tan{\theta} = \frac{1}{\cos{\theta}}- \frac{\sin^2{\theta}}{\cos{\theta}}$

Recall that $1 - \sin^2{\theta} = \cos^2{\theta}$. Carry on.
What Identity is this?

Is $\sin^2{\theta}$ equal to $\sin^2 {\theta}+ \cos^2{\theta} = 1$?
Sorry I am very confused!

6. Originally Posted by j0nath0n3
What Identity is this?
$\sin^2{x}+\cos^2{x}=1 \iff \sin^2{x} = 1-\cos^2{x} \iff \cos^2{x} = 1-\sin^2{x}$

This is known as the Pythagorean identity.

7. Originally Posted by Chop Suey
$\sin^2{x}+\cos^2{x}=1 \iff \sin^2{x} = 1-\cos^2{x} \iff \cos^2{x} = 1-\sin^2{x}$

This is known as the Pythagorean identity.
So in terms of cos thetha sec thetha - sin theta tan theta in terms of cos theta equals
1

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cos squared theta
?

8. Originally Posted by Chop Suey
$\sec{\theta} - \sin{\theta}\tan{\theta} = \frac{1}{\cos{\theta}}- \frac{\sin^2{\theta}}{\cos{\theta}}$

Recall that $1 - \sin^2{\theta} = \cos^2{\theta}$. Carry on.
Look Jonathon,

Chop Suey has taken it to the very last step. There's not much else to do here.

$\sec{\theta} - \sin{\theta}\tan{\theta} = \frac{1}{\cos{\theta}}- \frac{\sin^2{\theta}}{\cos{\theta}}=\frac{1-\sin^2 \theta}{\cos \theta}=\frac{\cos^2 \theta}{\cos \theta}=?$