# Identities

• March 9th 2010, 06:19 PM
purplec16
Identities
Can someone help me with these two problems
$sec \theta + csc \theta-cos\theta-sin\theta= sin\theta tan\theta+ cos\theta cot\theta$

and

$\frac{cot(-t)+tan(-t)}{cot t}= -sec^2 t$
• March 9th 2010, 06:36 PM
Stroodle
For the first one, you can take the RHS:

$sin\theta tan\theta+cos\theta cot\theta = sin\theta\times\frac{sin\theta}{cos\theta}+cos\the ta\times\frac{cos\theta}{sin\theta}$

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

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

$=\frac{1}{cos\theta}-cos\theta+\frac{1}{sin\theta}-sin\theta$

$=sec\theta+csc\theta-cos\theta-sin\theta = LHS$
• March 9th 2010, 06:44 PM
purplec16
Do you know how to do the other one?
• March 9th 2010, 06:55 PM
Stroodle
For the second one you can take the LHS:

$\frac{cot(-t)+tan(-t)}{cot(t)}=\frac{-cot(t)-tan(t)}{cot(t)}$

$=-1-\frac{\frac{sin(t)}{cos(t)}}{\frac{cos(t)}{sin(t)} }$

$=-1-\frac{sin^2(t)}{cos^2(t)}$

$=-1-tan^2(t)$

$=-sec^2(t)=RHS$
• March 9th 2010, 07:01 PM
Prove It
Quote:

Originally Posted by purplec16
Can someone help me with these two problems
$sec \theta + csc \theta-cos\theta-sin\theta= sin\theta tan\theta+ cos\theta cot\theta$

and

$\frac{cot(-t)+tan(-t)}{cot t}= -sec^2 t$

$\frac{\cot{(-t)} + \tan{(-t)}}{\cot{t}} = \frac{\frac{\cos{(-t)}}{\sin{(-t)}} + \frac{\sin{(-t)}}{\cos{(-t)}}}{\frac{\cos{t}}{\sin{t}}}$

$= \frac{\frac{\cos{t}}{-\sin{t}} + \frac{-\sin{t}}{\cos{t}}}{\frac{\cos{t}}{\sin{t}}}$

$= \frac{-\cot{t} - \tan{t}}{\cot{t}}$

$= -1 - \frac{\tan{t}}{\cot{t}}$

$= -1 - \frac{\tan{t}}{\frac{1}{\tan{t}}}$

$= -1 - \tan^2{t}$

$= -(1 + \tan^2{t})$

$= -\sec^2{t}$.
• March 9th 2010, 07:24 PM
purplec16
THANK YOU SO MUCH!!!!!!!!!!!!!!!!!(Rofl)