Trigonometry problem

**Adikz** · October 1st 2012, 05:11 PM

Can someone help me to simplify this?

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

**Prove It** · October 1st 2012, 05:19 PM

Originally Posted by Adikz

Can someone help me to simplify this?

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

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

$\displaystyle \begin{align*} \frac{1 + \sec{A}}{\sin{A} + \tan{A}} &\equiv \frac{1 + \frac{1}{\cos{A}}}{\sin{A} + \frac{\sin{A}}{\cos{A}}} \\ &\equiv \frac{\frac{\cos{A} + 1}{\cos{A}}}{\frac{\sin{A}\cos{A} + \sin{A}}{\cos{A}}} \\ &\equiv \frac{\cos{A} + 1}{\sin{A}\cos{A} + \sin{A}} \\ &\equiv \frac{\cos{A} + 1}{\sin{A}\left( \cos{A} + 1\right)} \\ &\equiv \frac{1}{\sin{A}} \\ &\equiv \csc{A} \end{align*}$

**Adikz** · October 2nd 2012, 06:01 AM

Thanks.. Can I ask for more?
$-secAcscA ( 1-2sin^2A )$

**skeeter** · October 2nd 2012, 02:26 PM

$-\sec{x}\csc{x}(1-2\sin^2{x})$

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

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

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

$-2\cot(2x)$

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