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
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
$\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!
$\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.
haha thx, nice trick, never saw it before...again thx a lotNo, 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.
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} $