# Thread: Simplify expression and Identity

1. ## Simplify expression and Identity

I still cant get the right answer for:

(15tanXcscX-3cscX)/(9tanXcscX-3cscX)
the choices are
A)5/3 B)5/3 sinX+cosX C) (5sinX-cosX)/(3sinX-cosX) D) 5/3 sinX-cosX

Also this problem

1-(cos^2X/1+sinX)=sinX

And finally
(1-sin^2X)/(sinX-cscX)= -sinX

2. Hello, Amanda!

$\frac{15\tan x\csc x-3\csc x}{9\tan x\csc x-3\csc x}$

. . $(A)\;\tfrac{5}{3}\qquad(B)\;\tfrac{5}{3}\sin x + \cos x \qquad(C)\;\frac{5\sin x-\cos x}{3\sin x-\cos x}\qquad(D)\;\tfrac{5}{3}\sin x-\cos x$

Factor and reduce: . $\frac{3\csc x(5\tan x - 1)}{3\csc x(3\tan x - 1)} \;=\;\frac{5\tan x - 1}{3\tan x - 1} \;=\;\frac{5\,\dfrac{\sin x}{\cos x} - 1}{3\,\dfrac{\sin x}{\cos x} -1}$

Multiply by $\frac{\cos x}{\cos x}\!:\;\;\frac{\cos x\left(5\,\frac{\sin x}{\cos x} - 1\right)} {\cos x\left(3\,\frac{\sin x}{\cos x} - 1\right)} \;=\;\frac{5\sin x - \cos x}{3\sin x - \cos x}$ . . . answer (C)

$1 - \frac{\cos^2\!x}{1+\sin x} \:=\:\sin x$

The left side is: . $1 - \frac{1-\sin^2\!x}{1+\sin x} \;\;=\;\;1 - \frac{(1-\sin x)(1+\sin x)}{1 + \sin x}$

. . $= \;\;1 - (1 - \sin x) \;\;=\;\;1 - 1 + \sin x \;\;=\;\;\sin x$

$\frac{1-\sin^2\!x}{\sin x-\csc x}\:=\: -\sin x$

The left side is: . $\frac{1-\sin^2\!x}{\sin x - \csc x} \;=\; \frac{\cos^2\!x}{\sin x - \frac{1}{\sin x}}$

Multiply by $\frac{\sin x}{\sin x}\!:\;\;\frac{\sin x}{\sin x}\cdot\frac{\cos^2\!x}{\sin x - \frac{1}{\sin x}} \;=\;\frac{\sin x\cos^2\!x}{\sin^2\!x - 1} \;=\;\frac{\sin x\cos^2\!x}{-(1-\sin^2\!x)}$

. . . . $= \;\frac{\sin x\cos^2\!x}{-\cos^2\!x} \;=\;\frac{\sin x}{-1} \;=\;-\sin x$

3. THANK YOU SOOOOOOOOOOOO MUCH!!