Using the identity find the value of tan15^∘, correct to three decimal places.
[Use √3=1.732]

Correct option is A

Solution:

$$\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}\\\\2\tan \alpha = \tan 2\alpha (1 - \tan^2 \alpha)\\\\2\tan \alpha = \tan 2\alpha - \tan^2 \alpha \tan 2\alpha\\\\2\tan \alpha + \tan 2\alpha \tan^2 \alpha = \tan 2 \alpha\\\text{Let } \alpha = 15^\circ\\2\tan 15^\circ + \tan 30^\circ \tan^2 15^\circ = \tan 30^\circ\\2\tan 15^\circ + \frac{1}{\sqrt{3}} \tan^2 15^\circ = \frac{1}{\sqrt{3}}\\2\sqrt{3} \tan 15^\circ + \tan^2 15^\circ = 1\\\tan^2 15^\circ + 2\sqrt{3} \tan 15^\circ - 1 = 0\\\tan 15^\circ = \frac{-2\sqrt{3} \pm \sqrt{12+4}}{2}\\= \frac{-2\sqrt{3} \pm \sqrt{16}}{2}\\= \frac{-2\sqrt{3} \pm 4}{2}\\= -\sqrt{3} \pm 2\\\tan 15^\circ = 2 - \sqrt{3} or - 2 - \sqrt{3} \\\text {Among given options:}\tan 15^\circ = 2 - 1.732 \\\tan 15^\circ = 0.268\\$$​