Question

Prove that :$\tan {15}^o=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

Answer 1

According to question

$tan{15}^o=tan({45}^0-{30}^o)$

=> $tan({45}^0-{30}^o)=\frac{tan{45}^o-tan{30}^o}{1+tan{45}^{o}tan{30}^o}$ ..........As, $\left[tan\right(A-B)=\frac{tanA-tanB}{1+tanAtanB}]$

=> $tan{15}^o=\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}$

=> $tan{15}^o=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

On rationalising RHS, we get

=> $tan{15}^o=2-\sqrt{3}$

Hence proved.