Question

simplify the expression
csc(sin^2x+cos^2x tanx)/sinx+cosx

Answer 1

I know the answer is 1, I just need to know how to get it

Answer 2

I've just tried an example to verify it is 1. Lets take x=0.
csc(sin^2(0)+cos^2(0)tan(0)) / sin(0) +cos(0) =
csc( 0 + 1x0 ) / 0 + 1 =
1/ (0 x 0) + 1 = infinity.
Are you sure you havent left out a few brackets ?

Answer 3

This is the equation \[\frac{ cscx(\sin^2x+\cos^2x tanx) }{ sinx+cosx }\]

Answer 4

The expression in parentheses can be written:
(sin^2 x*cos x + cos^2 x*sin x), by replacing tan x = sin x/cos x.
Factor out sin x*cos x:
sin x*cos x(sin x + cos x).
Finally: y= (1/sin x)(sin x*cos x) (sin x + cos x)/(sin x + cos x)
After simplification y = cos x.

Answer 5

Here's one way to do it


\[\Large \frac{ \csc(x)\left(\sin^2(x) + \cos^2(x)*\tan(x)\right) }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \csc(x)\left(\sin^2(x) + \cos^2(x)*\frac{\sin(x)}{\cos(x)}\right) }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \csc(x)\left(\sin^2(x) + \cos(x)*\sin(x)\right) }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \csc(x)*\sin^2(x) + \csc(x)*\cos(x)*\sin(x) }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \csc(x)*\sin^2(x) + \cos(x)*\csc(x)*\sin(x) }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \frac{1}{\sin(x)}*\sin^2(x) + \cos(x)*\frac{1}{\sin(x)}*\sin(x) }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \frac{\sin^2(x)}{\sin(x)} + \cos(x)*\frac{\sin(x)}{\sin(x)} }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \sin(x) + \cos(x)*1 }{ \sin(x)+\cos(x) }\]


\[\Large \frac{ \sin(x) + \cos(x) }{ \sin(x)+\cos(x) }\]


\[\Large 1\]

Answer 6

Nice one (the x was missing from the csc in the original post)

Answer 7

In the third step why did cos^2x change to cosx?

Answer 8

because cos^2 divided by cos turns into cos

it's like saying x^2 divided by x = x

Answer 9

Got it! Thank you!

Answer 10

you're welcome

Answer 11

Where does cos x (tan x = sin x/cos x) go? From line (2) to Line (3)