Question

What is the value of cot 4A?
Also, what is the value of cot 7 1/2 (ie, cot7.5)? need a quick reply

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{can}\mathrm{write} \\ \cot 4\mathrm{A}=\cot 2\left(2\mathrm{A}\right)=\frac{{\cot }^{2}2\mathrm{A}-1}{2\cot 2\mathrm{A}}\left[\mathrm{As},\cot 2\mathrm{x}=\frac{{\cot }^2\mathrm{x}-1}{2\mathrm{cotx}}\right] \\ =\frac{{\left({\displaystyle \frac{{\cot }^2\mathrm{A}-1}{2\mathrm{cotA}}}\right)}^2-1}{2\left({\displaystyle \frac{{\cot }^2\mathrm{A}-1}{2\mathrm{cotA}}}\right)}=\frac{{\left({\cot }^2\mathrm{A}-1\right)}^2-{\left(2\mathrm{cotA}\right)}^2}{4\mathrm{cotA}\left({\cot }^2\mathrm{A}-1\right)} \\ =\frac{{\cot }^4\mathrm{A}+1-2{\cot }^2\mathrm{A}+4{\cot }^2\mathrm{A}}{4\mathrm{cotA}\left({\cot }^2\mathrm{A}-1\right)}=\frac{{\cot }^4\mathrm{A}+1+2{\cot }^2\mathrm{A}}{4\mathrm{cotA}\left({\cot }^2\mathrm{A}-1\right)} \\ =\frac{{\left({\cot }^2\mathrm{A}+1\right)}^2}{4\mathrm{cotA}\left({\cot }^2\mathrm{A}-1\right)}. \\ \mathrm{Now}; \\ \cot {\left(7\frac{1}{2}\right)}^{\circ }=\cot \frac{\mathrm{\pi }}{24}=\frac{\cos \frac{\mathrm{\pi }}{24}}{\sin \frac{\mathrm{\pi }}{24}}=\frac{2\cos \frac{\mathrm{\pi }}{24}\cos \frac{\mathrm{\pi }}{24}}{2\sin \frac{\mathrm{\pi }}{24}\cos \frac{\mathrm{\pi }}{24}}=\frac{2{\cos }^2\frac{\mathrm{\pi }}{24}}{\sin 2\left(\frac{\mathrm{\pi }}{24}\right)}\left[\sin 2\mathrm{x}=2\mathrm{sinx}\mathrm{cosx}\right] \\ =\frac{1+\cos 2\left(\frac{\mathrm{\pi }}{24}\right)}{\sin 2\left(\frac{\mathrm{\pi }}{24}\right)}=\frac{1+\cos \left(\frac{\mathrm{\pi }}{12}\right)}{\sin \left(\frac{\mathrm{\pi }}{12}\right)}\left[\mathrm{As},2{\cos }^2\mathrm{x}=1+\cos 2\mathrm{x}\right] \\ \Rightarrow \cot {\left(7\frac{1}{2}\right)}^{\circ }=\frac{1+\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{\mathrm{\pi }}{6}}\right)}=\frac{1+\left(\cos {\displaystyle \frac{\mathrm{\pi }}{4}}\cos {\displaystyle \frac{\mathrm{\pi }}{6}}+\sin {\displaystyle \frac{\mathrm{\pi }}{4}}\sin {\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{\left(\sin \frac{\mathrm{\pi }}{4}\cos \frac{\mathrm{\pi }}{6}-\cos \frac{\mathrm{\pi }}{4}\sin \frac{\mathrm{\pi }}{6}\right)} \\ =\frac{1+{\displaystyle \frac{\sqrt{3}}{2\sqrt{2}}}+{\displaystyle \frac{1}{2\sqrt{2}}}}{\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}}=\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left(2\sqrt{2}+\sqrt{3}+1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \\ =\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{3-1}=\frac{2\sqrt{6}+2\sqrt{2}+2\sqrt{3}+4}{2}=\sqrt{2}+\sqrt{3}+2+\sqrt{6}. \\ \mathrm{Hence}, \\ \cot {\left(7\frac{1}{2}\right)}^{\circ }=\sqrt{2}+\sqrt{3}+2+\sqrt{6}. \end{gathered}$$