Prove that     tan 37 1/2 degree = root6 + root3 - root2 - 2.

Hi  Nikita.,
Please find below the solution to the asked query:

$$\begin{gathered} \tan 2\mathrm{A}=\frac{2.\mathrm{tanA}}{1-{\tan }^2\mathrm{A}} \\ \Rightarrow \tan \left(150\right)=\tan \left(2\times 75\right)=\frac{2\tan 75}{1-{\tan }^{2}75} \\ \mathrm{Now} \tan \left(150\right)=\tan \left(180-30\right)=-\tan 30=-\frac{1}{\sqrt{3}} \\ \Rightarrow -\frac{1}{\sqrt{3}}=\frac{2\tan 75}{1-{\tan }^{2}75} \\ \Rightarrow {\tan }^{2}75-1=2\sqrt{3}\tan 75 \\ \Rightarrow {\tan }^{2}75-2\sqrt{3}\tan 75-1=0 \\ \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{quadratic} \mathrm{equation} \mathrm{in} \tan 75. \\ \Rightarrow \tan 75=\frac{-\left(-2\sqrt{3}\right)\pm \sqrt{{\left(-2\sqrt{3}\right)}^2-4\times 1\times \left(-1\right)}}{2} \\ =\frac{2\sqrt{3}\pm \sqrt{12+4}}{2} \\ =\frac{2\sqrt{3}\pm \sqrt{16}}{2} \\ =\frac{2\sqrt{3}\pm 4}{2} \\ \mathrm{But} \tan 75 \mathrm{cannot} \mathrm{be} \mathrm{negative} \mathrm{as} \mathrm{angle} \mathrm{lies} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{quadrant}. \\ \Rightarrow \tan 75=\frac{2\sqrt{3}+4}{2} \\ \mathbf{\Rightarrow }\mathbf{tan}\mathbf{75}\mathbf{=}\mathbf{2}\mathbf{+}\sqrt{\mathbf{3}} \\ \mathrm{Again} \\ \tan \left(75\right)=\tan \left(2\times 37.5\right)=\frac{2\tan 37.5}{1-{\tan }^{2}37.5} \\ \Rightarrow 2+\sqrt{3}=\frac{2\tan 37.5}{1-{\tan }^{2}37.5} \\ \Rightarrow 2+\sqrt{3}-\left(2+\sqrt{3}\right){\tan }^{2}37.5=2\tan 37.5 \\ \Rightarrow \left(2+\sqrt{3}\right){\tan }^{2}37.5+2\tan 37.5-\left(2+\sqrt{3}\right)=0 \\ \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{quadratic} \mathrm{equation} \mathrm{in} \tan 37.5. \\ \Rightarrow \tan 37.5=\frac{-2\pm \sqrt{{\left(2\right)}^2+4\times \left(2+\sqrt{3}\right)\times \left(2+\sqrt{3}\right)}}{2} \\ =\frac{-2\pm 2\sqrt{1+{\left(2+\sqrt{3}\right)}^2}}{2\left(2+\sqrt{3}\right)} \\ \mathrm{Using} \mathrm{identity} {\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}, \mathrm{we} \mathrm{get}, \\ \tan 37.5=\frac{-1\pm \sqrt{1+4+3+4\sqrt{3}}}{\left(2+\sqrt{3}\right)} \\ =\frac{-1\pm \sqrt{8+4\sqrt{3}}}{\left(2+\sqrt{3}\right)} \\ \mathrm{But} \tan 37.5 \mathrm{cannot} \mathrm{be} \mathrm{negative} \mathrm{as} \mathrm{angle} \mathrm{lies} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{quadrant}. \\ \Rightarrow \tan 37.5=\frac{\sqrt{8+4\sqrt{3}}-1}{2+\sqrt{3}} \\ \mathrm{Rationalizing} \mathrm{the} \mathrm{denominator}, \mathrm{we} \mathrm{get}, \\ \tan 37.5=\frac{\sqrt{\left(8+4\sqrt{3}\right)}.\left(2-\sqrt{3}\right)-\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)} \\ =\frac{\sqrt{\left(8+4\sqrt{3}\right){\left(2-\sqrt{3}\right)}^2}-\left(2-\sqrt{3}\right)}{4-3} \left\{\mathrm{Using} \mathrm{identity} \left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2\right\} \\ \mathrm{Using} \mathrm{identity} {\left(\mathrm{a}-\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}, \mathrm{we} \mathrm{get}, \\ =\frac{\sqrt{\left(8+4\sqrt{3}\right)\left(4+3-4\sqrt{3}\right)}-\left(2-\sqrt{3}\right)}{1} \\ =\frac{\sqrt{\left(8+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}-\left(2-\sqrt{3}\right)}{1} \\ =\sqrt{\left(8+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}-\left(2-\sqrt{3}\right) \\ =\sqrt{56-32\sqrt{3}+28\sqrt{3}-48}-\left(2-\sqrt{3}\right) \\ =\sqrt{8-4\sqrt{3}}-\left(2-\sqrt{3}\right) \\ =\sqrt{6+2-4\sqrt{3}}-\left(2-\sqrt{3}\right) \\ =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(\sqrt{2}\right)}^2-2.\sqrt{6}.\sqrt{2}}-\left(2-\sqrt{3}\right) \\ =\sqrt{{\left(\sqrt{6}-\sqrt{2}\right)}^2}-2+\sqrt{3} \\ =\sqrt{6}-\sqrt{2}-2+\sqrt{3} \\ \mathbf{Hence}\mathbf{ }\mathbf{tan}\mathbf{37}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{tan}\left(\mathbf{37}\frac{\mathbf{1}}{\mathbf{2}}\right)\mathbf{=}\sqrt{\mathbf{6}}\mathbf{+}\sqrt{\mathbf{3}}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{ }\left(\mathbf{Hence}\mathbf{ }\mathbf{proved}\right) \end{gathered}$$

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