1. ## Verifying Identities

1. a. Use a calculator to determine which one of the following equations is an identity.

1/ sin x + cos x = csc x + sec x

1/ 1 - sin^2 x = 1 + tan^2x

b. Verify the identity equation.

2. Verify the following identity.

tan x-tan y = sin(x-y) / cos x cos y

2. just give a value to x and see if both sides are equal..

3. Originally Posted by accordry
1. a. Use a calculator to determine which one of the following equations is an identity.

1/ sin x + cos x = csc x + sec x

1/ 1 - sin^2 x = 1 + tan^2x

b. Verify the identity equation.
Graph these in your calculator and see if the graphs are the same. If not, then the identities are false.

2. Verify the following identity.

tan x-tan y = sin(x-y) / cos x cos y
$\displaystyle \tan x=\frac{\sin x}{\cos x}$ and $\displaystyle \tan y=\frac{\sin y}{\cos y}$.

Thus, $\displaystyle \tan x-\tan y=\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}$

Now find a common denominator. In this case, it would be $\displaystyle \cos x\cos y$

Thus, $\displaystyle \frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}=\frac{\sin x}{\cos x}\cdot\frac{\cos y}{\cos y}-\frac{\sin y}{\cos y}\cdot\frac{\cos x}{\cos x}=\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}$

Note that $\displaystyle \sin x\cos y-\cos x\sin y=\sin(x-y)$

Thus, $\displaystyle \frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}=\frac{\sin(x-y)}{\cos x\cos y}$

We have established the identity.

Does this make sense?

--Chris

4. Hello, accordry!

2. Verify: .$\displaystyle \tan x - \tan y \:=\: \frac{\sin(x-y)}{\cos x\cos y}$
On the right side, we have:

$\displaystyle \frac{\sin(x-y)}{\sin x\cos y} \;=\;\frac{\sin x\cos y - \cos x\sin y}{\cos x\cos y} \;=\;\frac{\sin x\,{\color{red}\rlap{/////}}\cos y}{\cos x\,{\color{red}\rlap{/////}}\,\cos y} - \frac{{\color{red}\rlap{/////}}\,\cos x\sin y}{{\color{red}\rlap{/////}}\cos x\,\cos y}$

. . . . . $\displaystyle = \;\frac{\sin x}{\cos x} - \frac{\sin y}{\cos y} \;=\;\tan x - \tan y$

5. how would you put the 1st ones in your calculator