# Find exact values of $\tan(105^\circ)$ and $\tan(11\pi/12)$ without calculator [closed]

How do you find the exact values of the following without using a calculator?

$$\tan(105^\circ) \qquad \tan(11\pi/12)$$

## closed as off-topic by Zev Chonoles, Zain Patel, user223391, A.P., MicahJul 20 '15 at 20:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Zev Chonoles, Zain Patel, Community, A.P., Micah
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Hint: Use the sum-angle formula for $\tan$:

$$\tan(\alpha+\beta)=\frac{\tan \alpha + \tan \beta}{1-\tan \alpha\tan \beta}$$

for some nice values of $\alpha,\beta$ of which you know the tangent.

Hint: $$\tan(105^\circ) = \tan(60^\circ+45^\circ)$$

$$\tan \theta = \frac{\sin\theta}{\cos\theta}$$ $$\sin(A+B) = \sin A\cos B + \cos A \sin B$$ $$\cos(A+B) = \cos A\cos B - \sin A \sin B$$

Alternatively, you can use the double angle formula for $\tan (A + B)$ as wythagoras suggested.

$$\tan 105^\circ = \tan (90+15)^\circ = -\cot 15^\circ = -1/\tan 15^\circ$$

Now, $\tan 2x = 2\tan x / (1-\tan^2 x)$

Put $x=15^\circ$ in the above equation and calculate $\tan 15^\circ$ from here and put in $$\tan 105^\circ = -1/\tan15^\circ$$

Do the same thing for your other question.

• Thanks @Micah, you have edited my response in such a beautiful fashion. I am new here, so, i dont know how to write mathematical equations here. – th ie Jul 20 '15 at 20:16