Verifying trig identities, picture included! thanks a lot!

**gobbajeezalus** · December 3rd 2008, 10:01 AM

Thanks a lot!

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

**Chop Suey** · December 3rd 2008, 10:17 AM

Originally Posted by gobbajeezalus

Thanks a lot!

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

1. Multiplying the LHS by $\frac{1+\sin{\theta}}{1+\sin{\theta}}$ gives:

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

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

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

Something similar can be done to the RHS.

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

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

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

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