# Thread: Trig. Identities; Tricky one :S

1. ## Trig. Identities; Tricky one :S

Prove:

$\displaystyle tanU/(secU+1) = (secU-1)/tanU$

i tried this many times, but i cant seem to prove them.
I really need help with this, plz if can prove it, plz show your steps b/c I want to learn to do it myself for later questions.

Thx

2. $\displaystyle LHS = \frac{tanx}{secx+1} = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+1} = \frac{sinx}{1+cosx} = LHS$

$\displaystyle RHS = \frac{secx-1}{tanx} = \frac{\frac{1}{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=$

$\displaystyle = \frac{1-cosx}{sinx} * \color{red}\frac{(1+cosx)}{(1+cosx)} \color{black}= \frac{1-cos^2x}{sinx+sinxcosx} = \frac{sin^2x}{sinx(1+cosx)} = \frac{sinx}{1+cosx} = LHS$

The key for solving this and other similar identities is the red part. Hope that helps!

3. ## woah...what just happned?

$\displaystyle LHS = \frac{tanx}{secx+1} = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+1} = \frac{sinx}{1+cosx} = LHS$

$\displaystyle RHS = \frac{secx-1}{tanx} = \frac{\frac{1}{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=$

Originally Posted by Referos
$\displaystyle = \frac{1-cosx}{sinx} * \color{red}\frac{(1+cosx)}{(1+cosx)} \color{black}= \frac{1-cos^2x}{sinx+sinxcosx} = \frac{sin^2x}{sinx(1+cosx)} = \frac{sinx}{1+cosx} = LHS$

The key for solving this and other similar identities is the red part. Hope that helps!
why would u just times it by 1+cosx? and would that not change the whole thing, and make it a different answer? so they will not equal each other.
could you plz expain why you did that.

4. Originally Posted by 2shoes
$\displaystyle LHS = \frac{tanx}{secx+1} = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+1} = \frac{sinx}{1+cosx} = LHS$

$\displaystyle RHS = \frac{secx-1}{tanx} = \frac{\frac{1}{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=$

why would u just times it by 1+cosx? and would that not change the whole thing, and make it a different answer? so they will not equal each other.
could you plz expain why you did that.
No, because $\displaystyle \frac{1 + \cos{x}}{1 + \cos{x}} = 1$, and multiplying anything by 1 leaves it unchanged.

This is the trick that's used in mathematics all the time - multiplying by a cleverly disguised 1 to change how something looks.

5. ## Thx

No, because $\displaystyle \frac{1 + \cos{x}}{1 + \cos{x}} = 1$, and multiplying anything by 1 leaves it unchanged.

This is the trick that's used in mathematics all the time - multiplying by a cleverly disguised 1 to change how something looks.
haha thx, nice trick, never saw it before...again thx a lot

6. Hello, 2shoes!

Prove: .$\displaystyle \frac{\tan x}{\sec x+1} \:=\:\frac{\sec x-1}{\tan x}$

Multiply by $\displaystyle \frac{\sec x - 1}{\sec x - 1}$

. . $\displaystyle \frac{\tan x}{\sec x + 1}\cdot\frac{\sec x - 1}{\sec x - 1} \;=\;\frac{\tan x(\sec x - 1)}{\underbrace{\sec^2\!x - 1}_{\text{This is }\tan^2\!x}} \;=\;\frac{\tan x(\sec x-1)}{\tan^2\!x} \;=\;\frac{\sec x - 1}{\tan x}$