Question

(√3+tan6)(√3+tan12)(√3+tan18)(√3+tan24)

Answer 1

Given:

$\mathsf{(\sqrt{3}+tan6^{\circ})(\sqrt{3}+tan12^{\circ})(\sqrt{3}+tan18^{\circ})(\sqrt{3}+tan24^{\circ})}$

To find: the value

Solution:

• Before we solve the problem, we must know that:

• $\mathsf{tan(A+B)=\frac{tanA+tanB}{1-tanA\:tanB}}$

$\mathsf{\therefore(\sqrt{3}+tan6^{\circ})(\sqrt{3}+tan12^{\circ})(\sqrt{3}+tan18^{\circ})(\sqrt{3}+tan24^{\circ})}$

$\mathsf{=(\sqrt{3}+tan6^{\circ})(\sqrt{3}+tan24^{\circ})(\sqrt{3}+tan12^{\circ})(\sqrt{3}+tan18^{\circ})}$

$\mathsf{=[3+\sqrt{3}(tan6^{\circ}+tan24^{\circ})+tan6^{\circ}\:tan24^{\circ}][3+\sqrt{3}(tan12^{\circ}+tan18^{\circ})+tan12^{\circ}\:tan18^{\circ}]}$

$\mathsf{=[3+1-tan6^{\circ}\:tan24^{\circ}+tan6^{\circ}\:tan24^{\circ}][3+1-tan12^{\circ}\:tan18^{\circ}+tan12^{\circ}\:tan18^{\circ}]}$

$\mathsf{=4\times 4}$

$\mathsf{=16}$

Calculations:

• Here, $\mathsf{tan30^{\circ}=tan(6+24)^{\circ}}$

• $\mathsf{=\frac{tan6^{\circ}+tan24^{\circ}}{1-tan6^{\circ}\:tan24^{\circ}}}$

• $\mathsf{\Rightarrow\frac{1}{\sqrt{3}}=\frac{tan6^{\circ}+tan24^{\circ}}{1-tan6^{\circ}\:tan24^{\circ}}}$

• $\mathsf{\Rightarrow \sqrt{3}(tan6^{\circ}+tan24^{\circ})}$
• $\mathsf{=1-tan6^{\circ}\:tan24^{\circ}}$

• and $\mathsf{tan30^{\circ}=tan(12+18)^{\circ}}$

• $\mathsf{=\frac{tan12^{\circ}+tan18^{\circ}}{1-tan12^{\circ}\:tan18^{\circ}}}$

• $\mathsf{\Rightarrow\frac{1}{\sqrt{3}}=\frac{tan12^{\circ}+tan18^{\circ}}{1-tan12^{\circ}\:tan18^{\circ}}}$

• $\mathsf{\Rightarrow \sqrt{3}(tan12^{\circ}+tan18^{\circ})}$
• $\mathsf{=1-tan12^{\circ}\:tan18^{\circ}}$

Answer:

$\mathsf{(\sqrt{3}+tan6^{\circ})(\sqrt{3}+tan12^{\circ})(\sqrt{3}+tan18^{\circ})(\sqrt{3}+tan24^{\circ})=16}$