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

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

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

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If this question can be reworded to fit the rules in the help center, please edit the question.

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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.

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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.

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

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.

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  • $\begingroup$ 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. $\endgroup$ – th ie Jul 20 '15 at 20:16

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