Question

Use the identity tan(theta) = sin(theta) / cos(theta) to show that tan(???? + ????) = tan(????)+tan(????) / 1−tan(????) tan(????)
.

Answer 1

Answer:

See the proof below.

Step-by-step explanation:

For this case we need to proof the following indentity:

$tan(x+y) = \frac{tan (x) + tan(y)}{1- tan(x) tan(y)}$

So we need to begin with the definition of tangent, we know that $tan (x) =\frac{sin(x)}{cos(x)}$ and we can do this:

$tan (x+y) = \frac{sin (x+y)}{cos(x+y)}$   (1)

We also have the following identities:

$sin (a+b) = sin (a) cos(b) + sin (b) cos(a)$

$cos(a+b)= cos(a) cos(b) - sin(a) sin(b)$

Now we can apply those identities into equation (1) like this:

$tan (x+y) =\frac{sin (x) cos(y) + sin (y) cos(x)}{cos(x) cos(y) - sin(x) sin(y)}$   (2)

We can divide numerator and denominator from expression (2) by $\frac{1}{cos(x) cos(y)}$ we got this:

$tan (x+y) = \frac{\frac{sin (x) cos(y)}{cos (x) cos(y)} + \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{cos(x) cos(y)}{cos(x) cos(y)} -\frac{sin(x)sin(y)}{cos(x) cos(y)}}$

And simplifying we got:

$tan (x+y) = \frac{tan(x) + tan(y)}{1-tan(x) tan(y)}$

And that complete the proof.