http://img.skitch.com/20081203-mmu4b...4nj8q5pi51.jpg

Thanks a lot!

Actually the third one i got, it was relatively easy. may i please have help with the others? thanks

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- Dec 3rd 2008, 10:01 AMgobbajeezalusVerifying trig identities, picture included! thanks a lot!
http://img.skitch.com/20081203-mmu4b...4nj8q5pi51.jpg

Thanks a lot!

Actually the third one i got, it was relatively easy. may i please have help with the others? thanks - Dec 3rd 2008, 10:17 AMChop Suey
1. Multiplying the LHS by $\displaystyle \frac{1+\sin{\theta}}{1+\sin{\theta}}$ gives:

$\displaystyle \frac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}} $

Recall that $\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1$

$\displaystyle \frac{\cos{\theta}(1+\sin{\theta})}{\cos^2{\theta} } = \frac{1+\sin{\theta}}{\cos{\theta}}$

Something similar can be done to the RHS.

2. $\displaystyle \cos{\theta}\cot{\theta} $

$\displaystyle = \cos{\theta} \frac{\cos{\theta}}{\sin{\theta}} = \frac{\cos^2{\theta}}{\sin{\theta}} = \frac{1-\sin^2{\theta}}{\sin{\theta}}$

Now simply split and simplify, and you're done.

3 is too easy. Recall that $\displaystyle \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$