Prove the following identities.............

**sweetG** · July 16th 2008, 06:00 AM

Prove the following identities:

a) (cosec θ + cotθ)(cosecθ - cotθ) = cotθtanθ

b) 2/(1 + sinθ) + 1/(1- sinθ) = (3sec^2)θ - tanθcosecθ

I basically just have to prove Left HandSide = Right Hand Side *

**Soroban** · July 16th 2008, 08:07 AM

Hello, sweetG!

Prove the following identities:

$a)\;\;\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta} \:=\:\cot\theta\tan\theta$ . . . . This is not an identity

$b)\;\;\frac{2}{1 + \sin\theta} + \frac{1}{1-\sin\theta} \:=\:3\sec^2\!\theta - \tan\theta\,{\color{red}\csc\theta\;\;?}$

Combine the fractions:

$\frac{2}{1+\sin\theta}\!\cdot\!{\color{blue}\frac{ 1-\sin\theta}{1-\sin\theta}} + \frac{1}{1-\sin\theta}\!\cdot\!{\color{blue}\frac{1+\sin\thet a}{1+\sin\theta}} \;\;=\;\;\frac{2(1-\sin\theta) + (1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}$

. . $= \;\;\frac{3-\sin\theta}{1-\sin^2\!\theta} \;\;=\;\;\frac{3-\sin\theta}{\cos^2\!\theta} \;\;=\;\;\frac{3}{\cos^2\!\theta} - \frac{\sin\theta}{\cos^2\!\theta}$

. . $= \;\;3\!\cdot\!\frac{1}{\cos^2\!\theta} - \frac{\sin\theta}{\cos\theta}\!\cdot\!\frac{1}{\co s\theta} \;\;=\;\;3\sec^2\!\theta - \tan\theta\,{\color{red}\sec\theta}$

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