# Trig Functions

• Mar 18th 2010, 06:05 PM
Mitch281008
Trig Functions
Hi all,

can someone please check to see if i have this questions correct from a worksheet.
simplify the expression:

(cosec x ) / (tanx + cot x)

= (1/sin x)/[(sin x/cosx) + {(1/sinx) / (sin x / cos x)}]

= (1/sinx)/(1/cosx)

if that makes sense.

how do you guys actually make fractions on here?

thanks,
Mitch
• Mar 18th 2010, 06:32 PM
Soroban
Hello, Mitch281008!

Quote:

Simplify: .$\displaystyle \frac{\csc x}{\tan x + \cot x}$

$\displaystyle = \;\frac{\;\dfrac{1}{\sin x}\;}{\dfrac{\sin x}{\cos x}}+ \frac{\;\dfrac{1}{\sin x}\;}{\dfrac{\sin x}{\cos x}}$ . . . . no!

Does $\displaystyle \frac{1}{2+3} \;=\;\frac{1}{2} + \frac{1}{3}$ ?

You have: .$\displaystyle \frac{\dfrac{1}{\sin x}} {\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}}$

$\displaystyle \text{Multiply by }\frac{\sin x\cos x}{\sin x\cos x}\!:\;\;\frac{\sin x\cos x\left(\dfrac{1}{\sin x}\right)} {\sin x\cos x\left(\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}\right)}$ .$\displaystyle =\; \frac{\cos x}{\sin^2\!x + \cos^2\!x} \;=\;\frac{\cos x}{1} \;=\;\cos x$