Question

the value of tan^6 15 - 15 tan^4 15 + 15 tan^2 15 - 3 is

Answer 1

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{following}\mathrm{expression}. \\ {\tan }^{6}15-15{\tan }^{4}15+15{\tan }^{2}15-3 \\ \mathrm{First}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\tan 15. \\ \mathrm{Use}\mathrm{the}\mathrm{formula} \\ \tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{\mathrm{tanA}-\mathrm{tanB}}{1+\mathrm{tanA}\mathrm{tanB}} \\ \mathrm{Take}\mathrm{A}=45,\mathrm{B}=30\mathrm{to}\mathrm{get}, \\ \tan 15=\tan \left(45-30\right) \\ =\frac{\tan 45-\tan 30}{1+\tan 45\tan 30} \\ =\frac{1-{\displaystyle \frac{1}{\sqrt{3}}}}{1+{\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ =\frac{{\left(\sqrt{3}-1\right)}^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)} \\ =\frac{3+1-2\sqrt{3}}{3-1} \\ =\frac{4-2\sqrt{3}}{2} \\ =\mathbf{2}\mathbf{-}\sqrt{\mathbf{3}} \\ \mathrm{This}\mathrm{gives}, \\ {\tan }^{2}15={\left(2-\sqrt{3}\right)}^2 \\ =4+3-4\sqrt{3} \\ =7-4\sqrt{3} \\ {\tan }^{4}15={\left(7-4\sqrt{3}\right)}^2 \\ =49+48-56\sqrt{3} \\ =97-56\sqrt{3} \\ {\tan }^{6}15={\left(97-56\sqrt{3}\right)}^2 \\ =9409+9408-10864\sqrt{3} \\ =18817-10864\sqrt{3} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{given}\mathrm{expression}. \\ {\tan }^{6}15-15{\tan }^{4}15+15{\tan }^{2}15-3=\left(18817-10864\sqrt{3}\right)-15\left(97-56\sqrt{3}\right) \\ +15\left(7-4\sqrt{3}\right)-3 \\ =18817-10864\sqrt{3}-1455+840\sqrt{3} \\ +105-60\sqrt{3}-3 \\ =\left(18817-1455+105-3\right) \\ +\sqrt{3}\left(-10864+840-60\right) \\ =\mathbf{}\mathbf{17464}\mathbf{}\mathbf{-}\mathbf{10084}\sqrt{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$