Question

Prove
(sinsquaredx+4sinx+3)/cossquaredx=(sinx+3)/1-sinx

Answer 1

is the left side all over cos^2x ?

Answer 2

or just the 3

Answer 3

all of it

Answer 4

there i edited it

Answer 5

\[\frac{\sin^2x+4sinx+3}{\cos^2x}=\frac{sinx+3}{1-sinx}?\]

Answer 6

What's the question?

Answer 7

yeah thats how it is, i dont know how to start

Answer 8

Prove? Simplify? Solve?

Answer 9

prove

Answer 10

First you gotta factor the numerator

Answer 11

what do you get?

Answer 12

um
(sinx+1)(sinx+3)?

Answer 13

Good, you noticed that the RHS got all the terms in sine, so we're gonna change the denominator to sine, by using \(cos^2x=1-sin^2x\)
\[\frac{(sinx+1)(sinx+3)}{1-\sin^2x}\]
The denominator can be factored as well, try factoring it

Answer 14

how would i factor that?

Answer 15

oh wait

Answer 16

obviously

Answer 17

(1+sin)(1-sin)

Answer 18

Yes

Answer 19

Then just cancel and you got it

Answer 20

aha thanks !

Answer 21

yw :)