# Thread: Derive the given identities

1. ## Derive the given identities

I have a very basic knowledge (or almost none at all) about identities and deriving.

1. (cosbeta)(tanbeta) + (sinbeta) / (tan beta) = 2cosbeta
2. 2csc2beta = secbeta(cscbeta)
3. tanbeta + cotbeta = 2csc2beta
4. sin2beta/sinbeta - cos2beta/cosbeta = secbeta

2. Originally Posted by roflcoptur

1. (cosbeta)(tanbeta) + (sinbeta) / (tan beta) = 2cosbeta
With this type of problems you need to work with only one side of the equation and apply trig identities until you can find the other side of the equation.

$\displaystyle \frac{\cos(\beta)\tan(\beta) + \sin(\beta)}{ \tan (\beta)}$

$\displaystyle \frac{\cos(\beta)\tan(\beta) }{ \tan (\beta)}+ \frac{ \sin(\beta)}{ \tan (\beta)}$

$\displaystyle \cos(\beta)+ \frac{ \sin(\beta)}{ \tan (\beta)}$

$\displaystyle \cos(\beta)+ \cos(\beta)$

$\displaystyle 2 \cos(\beta)$

Have a look at this site:

Table of Trigonometric Identities

3. Originally Posted by roflcoptur
I have a very basic knowledge (or almost none at all) about identities and deriving.

1. (cosbeta)(tanbeta) + (sinbeta) / (tan beta) = 2cosbeta
2. 2csc2beta = secbeta(cscbeta)
3. tanbeta + cotbeta = 2csc2beta
4. sin2beta/sinbeta - cos2beta/cosbeta = secbeta
2. $\displaystyle 2\csc{2\beta} = \frac{2}{\sin{2\beta}}$

$\displaystyle = \frac{2}{2\sin{\beta}\cos{\beta}}$

$\displaystyle = \frac{1}{\sin{\beta}\cos{\beta}}$

$\displaystyle = \sec{\beta}\csc{\beta}$.