Trig. Identities; Tricky one :S

**2shoes** · April 27th 2009, 04:34 PM

Prove:

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

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

**Referos** · April 27th 2009, 05:40 PM

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

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

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

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

**2shoes** · April 27th 2009, 06:28 PM

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

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

Originally Posted by Referos

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

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.

**Prove It** · April 27th 2009, 07:01 PM

Originally Posted by 2shoes

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

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

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 $\frac{1 + \cos{x}}{1 + \cos{x}} = 1$ , 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.

**2shoes** · April 28th 2009, 05:22 PM

No, because $\frac{1 + \cos{x}}{1 + \cos{x}} = 1$ , 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

**Soroban** · April 28th 2009, 05:58 PM

Hello, 2shoes!

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

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

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

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woah...what just happned?

Thx

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