# complete the following identities.

• Jul 14th 2008, 11:30 AM
Dave19
complete 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 PM
topsquark
Quote:

Originally Posted by Dave19
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

In order to prove an identity one must have both sides of the identity...

If a) is
$\frac{cos(a)}{sin(a)}$
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.
$tan(-a) = \frac{sin(-a)}{cos(-a)}$
So what are sin(-a) and cos(-a)?

d) $csc(90 - \theta) = \frac{1}{sin(90 - \theta)}$
How do you find $sin(90 - \theta)$?

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

-Dan
• Jul 14th 2008, 02:24 PM
masters
Quote:

Originally Posted by Dave19
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

a) $\frac{\cos\alpha}{\sin\alpha}=\cot\alpha$

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

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

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

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