A2 C3 Trigonometry Proof

Hello, TSR!

Could any A2 Mathematicians please help me to prove the following trig identity:

(1-cosx)/sinx = 1/(cosecx + cotx)

Many thanks!

Split the LHS into two fractions: 1/sinx - cosx/sinx

cosecx - cotx

multiply by [ (cosecx + cotx) / (cosecx + cotx) ]

[ (cosecx -cotx) . (cosecx + cotx) ] / (cosecx + cotx)

(cosec^2x - cot^2x) / (cosecx + cotx)

1 / (cosecx + cotx)

L.H.S = R.H.S

Please don't post full solutions. It's against the rules of this forum.

I'm going to oversimplify; realistically you would do most of this in your head.You could have alternatively split up (1-cosx)/sinx in terms of cosecx & secx then multiply top and bottom by cosecx + cotx .
cosecx + cotx=(1/sinx) +(cosx/sinx) =(1+cosx)/sinx

therefore 1/(cosecx + cotx)= sinx/(1+cosx)
= sinx(1-cosx)/(1+cosx)(1-cosx)
= sinx(1-cosx)/1-(cosx)^2
1-(cosx)^2=1-[1-(sinx)^2)] =(sinx)^2

therefore 1/(cosecx + cotx) =sinx(1-cosx)/(sinx)^2
= (1-cosx)/sinx
L.H.S = R.H.S

Hey - is there any chance you could provide me with a pointer for this question?

Prove this trig identity:

$(\sin \theta - \mathrm{cosec } \theta)^2 \equiv \sin^2 \theta + \mathrm{cot}^2 \theta -1$

I started with the left side, and got up to

$\sin^2 \theta - \frac{2 \sin \theta}{\sin \theta} + \frac{1}{\sin^2 \theta}$

But not really sure where to go from here to end up with

$\sin^2 \theta + \mathrm{cot}^2 \theta -1$

SImplifying the line you're up to:

$\sin^2 \theta -2 + cosec^2 \theta$

Which I'll now write in another way:

$\sin^2 \theta+ cosec^2\theta - 1 - 1$


Does that help?