a stands for alpha b stands for beta

a)cos a= b) 1+cot squared a=

sin a

c) tan(-a)= d) csc(90degrees - theta)=

e)tan a - tan b=

1+tan a tan b

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- Jul 14th 2008, 11:30 AMDave19complete the following identities.
a stands for alpha b stands for beta

a)__cos a__= b) 1+cot squared a=

sin a

c) tan(-a)= d) csc(90degrees - theta)=

e)__tan a - tan b__=

1+tan a tan b - Jul 14th 2008, 01:44 PMtopsquark
In order to prove an identity one must have both sides of the identity...

If a) is

$\displaystyle \frac{cos(a)}{sin(a)}$

then you should know this. Hint: What are the definitions of tan(x), cot(x), sec(x), and csc(x)?

b) This is one side of a "standard" identity. Look at the link in your other thread and you should find it.

c) This is not so much of an identity as it is a simplification.

$\displaystyle tan(-a) = \frac{sin(-a)}{cos(-a)}$

So what are sin(-a) and cos(-a)?

d) $\displaystyle csc(90 - \theta) = \frac{1}{sin(90 - \theta)}$

How do you find $\displaystyle sin(90 - \theta)$?

e) Hint: What is the form for tan(a + b)? (See that link again.)

-Dan - Jul 14th 2008, 02:24 PMmasters
a)$\displaystyle \frac{\cos\alpha}{\sin\alpha}=\cot\alpha$

b)$\displaystyle 1+\cot^2\alpha=\csc^2\alpha$

c)$\displaystyle \tan(-\alpha)=-\tan\alpha$

d)$\displaystyle \csc(90-\theta)=\sec\theta$

e)$\displaystyle \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\tan(\alpha-\beta)$