# Thread: Trigonometry problem

1. ## Trigonometry problem

Can someone help me to simplify this?

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

2. ## Re: Trigonometry problem

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*}

3. ## Re: Trigonometry problem

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

4. ## Re: Trigonometry problem

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

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

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

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

$-2\cot(2x)$