# Thread: Find the exact value of tan -105 degrees

1. ## Find the exact value of tan -105 degrees

tan -105 degrees

Edit 2: Forgot to cancel the 2 before the radical, so it's 2+radical 3...

2. $\displaystyle \tan 105 = \tan (60+45)$

Now apply

$\displaystyle \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$

3. An angle of $\displaystyle 105^{\circ}$ in the clockwise direction is the same as an angle of $\displaystyle 255^{\circ}$ in the anticlockwise direction.

So $\displaystyle \tan{\left(-105^{\circ}\right)} = \tan{\left(255^{\circ}\right)}$

$\displaystyle = \tan{\left(180^{\circ} + 75^{\circ}\right)}$

$\displaystyle = \tan{\left(75^{\circ}\right)}$ since the tangent values are the same in the first and third quadrants...

$\displaystyle = \tan{\left(30^{\circ} + 45^{\circ}\right)}$.

Now make use of the identity $\displaystyle \tan{\left(\alpha + \beta\right)} \equiv \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$.

4. Originally Posted by pickslides
$\displaystyle \tan 105 = \tan (60+45)$

Now apply

$\displaystyle \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$
Yes it is, but the OP is evaluating $\displaystyle \tan{\left(-105^{\circ}\right)}$...

5. Good pick up Hayden, I didn't see that little fella in there, my eyes must be failing me...

I hate being old.

6. Originally Posted by Prove It
Yes it is, but the OP is evaluating $\displaystyle \tan{\left(-105^{\circ}\right)}$...
But $\tan\left(-x\right) = -\tan\left(x\right)$, so $\tan\left(-105^{\circ}\right) = -\tan\left(105^{\circ}\right) = -\cdots$ (Pickslide's post).

7. Originally Posted by TheCoffeeMachine
But $\tan\left(-x\right) = -\tan\left(x\right)$, so $\tan\left(-105^{\circ}\right) = -\tan\left(105^{\circ}\right) = -\cdots$ (Pickslide's post).
I'll pretend I meant that...

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# tan 105=

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