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# Find the value of tan 15 degree

Q (5) Find the value of

$\small (ii) \tan 15\degree$

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$\tan 15\degree = \tan (45\degree - 30\degree)$
We know that,

$\left [ \tan(x-y)= \frac{\tan x - \tan y}{1+\tan x\tan y} \right ]$
By using this we can write

$\tan (45\degree - 30\degree)= \frac{\tan 45\degree - tan30\degree}{1+\tan45\degree\tan30\degree}\\ \\ \Rightarrow \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3} -1\right )}=\frac{3+1-2\sqrt{3}}{\left ( \sqrt{3} \right )^{2}-\left ( 1 \right )^{2}}\\ \\ \Rightarrow \frac {4-2\sqrt{3}}{3-1}=\frac{2\left ( 2-\sqrt{3} \right )}{2}= 2-\sqrt{3}$

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