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Gennadij [26K]
2 years ago
8

Show tan(???? − ????) = tan(????)−tan(????) / 1+tan(????) tan(????) .

Mathematics
1 answer:
anyanavicka [17]2 years ago
8 0

Answer:

See the proof below

Step-by-step explanation:

For this case we need to proof the following identity:

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

We need to begin with the definition of tangent:

tan (x) =\frac{sin(x)}{cos(x)}

So we can replace into our formula and we got:

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

We have the following identities useful for this case:

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

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

If we apply the identities into our equation (1) we got:

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

Now we can divide the numerator and denominato from expression (2) by \frac{1}{cos(x) cos(y)} and 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{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(y)}{cos(x) cos(y)}}

And simplifying we got:

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

And this identity is satisfied for all:

(x-y) \neq \frac{\pi}{2} +n\pi

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