Trigonometric identities are equalities that involves trigonometric functions. Trigonometric identities involves angles, side lengths and other lengths of the triangle. We use these identities for expressions involving trigonometric functions that need to be simplified.  Let’s see what the tangent addition formula looks like and solve an example to make it clear.

The Tangent function in trigonometry is defined by $tan\;x=\frac{sin\;x}{cos\;x}$ . The addition formula for the tangent are achieved from the addition of sine and cosine. To find the addition of a tangent function use below given formula.

The Tangent Addition Formula in term of functions $\alpha$ and \beta is given below,

$\large tan\left(\alpha+\beta\right)=\frac{sin\left(\alpha+\beta\right)}{cos\left(\alpha+\beta\right)}=\frac{tan\; \alpha + tan\; \beta}{1-tan\; \alpha \times tan\; \beta}$

### Solved problems

Question: Find the exact value of tan $285^{\circ}$

Solution:

Let $285^{\circ}$ = $330^{\circ}$ – $45^{\circ}$

Where,
$\alpha=330^{\circ}$
$\beta=45^{\circ}$

$tan\left(\alpha+\beta\right)=\frac{sin\left(\alpha+\beta\right)}{cos\left(\alpha+\beta\right)}=\frac{tan\; \alpha + tan\; \beta}{1-tan\; \alpha \times tan\; \beta}$
$tan\left(330+45\right)=\frac{sin\left(330+45\right)}{cos\left(330+45\right)}=\frac{tan\; 330 + tan\; 45}{1-tan\; 330 \times tan\; 45}$
$tan\;285^{\circ}=\frac{-\frac{\sqrt{3}}{3}-1}{1-\frac{\sqrt{3}}{3}\times \; 1}$
$tan\;285^{\circ}=\frac{-12-6\sqrt{3}}{6}=\frac{-3-\sqrt{3}}{3-\sqrt{3}}\times\frac{3+\sqrt{3}}{3+\sqrt{3}}=\frac{-9-6\sqrt{3}-3}{9-3}$
$tan\;285^{\circ}=-2\;-\sqrt{3}$