Question

Prove this (tanx)/(1+cosx) + (sinx)/(1-cosx)= cotx+secxcscx

Answer 1

HI!!

Answer 2

probably if you add you get it

Answer 3

\[\frac{ tanx }{ 1+cosx }+\frac{ sinx }{ 1-cosx }=cotx+secxcscx\]

Answer 4

Hi Misty

Answer 5

wanna try it?

Answer 6

Yes. I'm so confused

Answer 7

it is loads easier if you replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\) since 98% of this is pure algebra

Answer 8

so the left hand side, the side we can add is
\[\frac{ \tan(x) }{ 1+\cos(x) }+\frac{ \sin(x) }{ 1-\cos(x)}\]

Answer 9

if we use the letters \(a\) and \(b\) it is easier to see what to do
\[\frac{\frac{b}{a}}{1+a}+\frac{b}{1-a}\]

Answer 10

you see what i did there?

Answer 11

Yes i'm gonna write it down and then send you my answer

Answer 12

ok go ahead
i am hoping you recognize that the denominator will be \((1+a)(1-a)=1-a^2\)

Answer 13

how you doing?

Answer 14

I got it but it wasn't the way you told me cause I couldnt get the concept but I got down to the answer with a whole bunch of work

Answer 15

tan(x)/[1 + cos(x)] + sin(x)/[1 - cos(x)]
= tan(x)[1 - cos(x)]/{[1 + cos(x)][1 - cos(x)]} + sin(x)[1 + cos(x)]/{[1 - cos(x)][1 + cos(x)]}
= tan(x)[1 - cos(x)]/[1 - cos^2(x)] + sin(x)[1 + cos(x)]/[1 - cos^2(x)]
= tan(x)[1 - cos(x)]/sin^2(x) + sin(x)[1 + cos(x)]/sin^2(x)
= {tan(x)[1 - cos(x) + sin(x)[1 + cos(x)]}/sin^2(x)
= [tan(x) - tan(x)cos(x) + sin(x) + sin(x)cos(x)]/sin^2(x)
= [tan(x) - {sin(x)/cos(x)}cos(x) + sin(x) + sin(x)cos(x)]/sin^2(x)
= [tan(x) - sin(x) + sin(x) + sin(x)cos(x)]/sin^2(x)
= [tan(x) + sin(x)cos(x)]/sin^2(x)
= [{sin(x)/cos(x)} + sin(x)cos(x)]/sin^2(x)
= [{1/cos(x)} + cos(x)]/sin(x)
= [sec(x) + cos(x)]/sin(x)
= sec(x)/sin(x) + cos(x)/sin(x)
= sec(x)csc(x) + cot(x)
= cot(x) + sec(x)csc(x)