(Sinθ - 1)/(1-sin2θ) = (cscθ)/(-cscθ-1)

or

(1-sinθ)/(1+sinθ) = (tanθ-secθ)^2

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- Jun 24th 2008, 08:04 PMrabidsquirrelUrgent Help With Trig Proofs
(Sinθ - 1)/(1

**-**sin2θ) = (cscθ)/(-cscθ-1)

or

(1-sinθ)/(1+sinθ) = (tanθ-secθ)^2 - Jun 24th 2008, 09:33 PMTKHunny
1) Please write what you mean. You have written $\displaystyle \sin(2\theta)$. I am guessing this is not your intent.

2) Are you acquainted with the Pythagorean Identity? $\displaystyle \cos^{2}(x) + \sin^{2}(x) = 1$?

3) $\displaystyle \csc(\theta) = 1/\sin(\theta)$ - Jun 24th 2008, 09:35 PMrabidsquirrel
i intended it to read sin^2 theta

- Jun 24th 2008, 09:40 PMkalagota
- Jun 24th 2008, 09:43 PMrabidsquirrel
- Jun 24th 2008, 09:44 PMrabidsquirrel
- Jun 25th 2008, 01:05 AMkalagota
- Jun 25th 2008, 02:00 AMnikhilHere it is
T=[sin(¤)-1]/[1-sin(¤)^2]=[sin(¤)-1]/[(1-sin(¤))(1+sin(¤)] as [a^2-b^2=(a+b)(a-b)]

=-1/1+sin(¤) now dividing both numerator and denominator by sin(¤) we get

T=-csc(¤)/1+csc(¤)=csc(¤)/-1-csc(¤). Hence proved.

2))A=(1-sin(¤))/(1+sin(¤)) multiplying both numerator and denominator by 1-cos(¤) we get A=(1-sin(¤))^2/cos(¤)^2 or [(1-sin(¤))/cos(¤)]^2 which is=[sec(¤)-tan(¤)]^2 which is= [tan(¤)-sec(¤)]^2 hence proved.

Note (a-b)^2=(b-a)^2