Find the value of tan 15°

Asked by Abhisek | 1 year ago |  103

##### Solution :-

It can be written as

tan15° = tan (45° – 30°)

Using formula

$$\dfrac{tan45°-tan30°}{1+tan45°+tan 30°}$$

$$tan(x-y)= \dfrac{tanx-tany}{1+tanxtany}$$

Since we know,

Therefore we have,

$$tan15°=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+1\dfrac{1}{\sqrt{3}}}$$

$$\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

$$\dfrac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}$$

Further computing we have,

tan15° = $$\dfrac{3+1-2\sqrt{3}}{(\sqrt{3})^2-(1)^2}$$

$$\dfrac{4-2\sqrt{3}}{3-1}$$

$$2-\sqrt{3}$$

Answered by Pragya Singh | 1 year ago

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