# Trigonometry problem

• Oct 1st 2012, 05:11 PM
Trigonometry problem
Can someone help me to simplify this?(Worried)

$\displaystyle \frac{\text{1 + secA}}{\text{sinA+tanA}}$
• Oct 1st 2012, 05:19 PM
Prove It
Re: Trigonometry problem
Quote:

Originally Posted by Adikz
Can someone help me to simplify this?(Worried)

$\displaystyle \frac{\text{1 + secA}}{\text{sinA+tanA}}$

There are MANY equivalent expressions, any of which could be considered "simpler"...

\displaystyle \displaystyle \begin{align*} \frac{1 + \sec{A}}{\sin{A} + \tan{A}} &\equiv \frac{1 + \frac{1}{\cos{A}}}{\sin{A} + \frac{\sin{A}}{\cos{A}}} \\ &\equiv \frac{\frac{\cos{A} + 1}{\cos{A}}}{\frac{\sin{A}\cos{A} + \sin{A}}{\cos{A}}} \\ &\equiv \frac{\cos{A} + 1}{\sin{A}\cos{A} + \sin{A}} \\ &\equiv \frac{\cos{A} + 1}{\sin{A}\left( \cos{A} + 1\right)} \\ &\equiv \frac{1}{\sin{A}} \\ &\equiv \csc{A} \end{align*}
• Oct 2nd 2012, 06:01 AM
Re: Trigonometry problem
Thanks.. Can I ask for more?
$\displaystyle -secAcscA ( 1-2sin^2A )$
• Oct 2nd 2012, 02:26 PM
skeeter
Re: Trigonometry problem
$\displaystyle -\sec{x}\csc{x}(1-2\sin^2{x})$

$\displaystyle -\frac{1}{\cos{x}} \cdot \frac{1}{\sin{x}} \cdot \cos(2x)$

$\displaystyle -\frac{2\cos(2x)}{2\cos{x}\sin{x}}$

$\displaystyle -\frac{2\cos(2x)}{\sin(2x)}$

$\displaystyle -2\cot(2x)$