Question

Does (1/Sin^2x) = Csc^2x ?

Answer 1

yes

Answer 2

alright so if i have (1/Sin^2x) + (1/Cos^2x) = (Csc^2x)(Sec^2x) What would the proof look like?

Answer 3

let me do it on paper real quick

Answer 4

ok

Answer 5

no you have the wrong equation

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

which can be simplified

Answer 6

This is a problem on my trig homework though

Answer 7

get common denominator on the left

Answer 8

which means its

\[\frac{1}{\sin^2(x)\cos^2(x)} = \frac{1}{\sin^2(x)} \times \frac{1}{\cos^2(x)}\]

you should be able to finish it from here

Answer 9

thanks

Answer 10

Heres another proof

Answer 11

Secx - Cosx = SinxTanx

Answer 12

well sec is 1/cos

so you need a common denominator

so the left hand side becomes

\[\frac{1 - \cos^2(x)}{\cos(x)}\]

make the obvious substitution to the numerator and then split it to

sin(x) x sin(x)/cos(x)

which give the solution

Answer 13

But that gives Tanx x Tanx

Answer 14

nevermind

Answer 15

no make the substitution and you get

\[\frac{\sin^2(x)}{\cos(x)} = \frac{\sin(x) \times \sin(x)}{\cos(x)}\]

hope this helps

Answer 16

then if you split the numerator you have

\[\sin(x) \times \frac{\sin(x)}{\cos(x)}\]

Answer 17

Another Question, ... Tanx + Cotx = SecxCscx

Answer 18

well tan = sin/cos and cot= cos/sin

to you have

sin/cos + cos/sin

you'll need a common denominator... and cross multiply...

then simplify

Answer 19

So (Cos^2x + Sin^2x) / CosxSinx

Answer 20

yes or you could simplify the top because (cos^2x+sin^2x)=1

Answer 21

as long as it matches it doesnt really matter

Answer 22

Oh okay thank you so much

Answer 23

i gave you a medal gotta go bye