Question

Find tan75, hence evaluate tan75 + cot 75

Answer 1

Dear student
$$\begin{gathered} \tan 75°=\tan \left(30°+45°\right) \\ =\frac{\tan 30°+\tan 45°}{1-\tan 30°\tan 45°} \\ =\frac{{\displaystyle \frac{1}{\sqrt{3}}}+1}{1-\left({\displaystyle \frac{1}{\sqrt{3}}}\times 1\right)} \\ =\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ =\frac{{\left(\sqrt{3}+1\right)}^2}{3-1} \\ =\frac{3+1+2\sqrt{3}}{2} \\ =\frac{4+2\sqrt{3}}{2} \\ =2+\sqrt{3} \\ \mathrm{So},\tan 75°=2+\sqrt{3} \\ \mathrm{and}\mathrm{we}\mathrm{know}\mathrm{that}\cot 75°=\frac{1}{\tan 75°}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3} \\ \mathrm{So},\tan 75°+\cot 75° \\ =2+\sqrt{3}+2-\sqrt{3} \\ =4 \\ \mathrm{Note}: \\ \tan \left(\mathrm{x}+\mathrm{y}\right)=\frac{\mathrm{tanx}+\mathrm{tany}}{1-\mathrm{tanx}\mathrm{tany}} \\ {\mathrm{x}}^2-{\mathrm{y}}^2=\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}-\mathrm{y}\right) \end{gathered}$$
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