Question

cot $\left(\frac{\mathrm{\pi }}{4}-2{\cot }^{-1}3\right)$ is equal to ______________________.

Answer 1

Let ${\cot }^{-1}3=\theta \Rightarrow \cot \theta =3$.

$$\begin{gathered} \therefore \cot \left(\frac{\mathrm{\pi }}{4}-2{\cot }^{-1}3\right) \\ =\cot \left(\frac{\mathrm{\pi }}{4}-2\theta \right) \\ =\frac{\cot {\displaystyle \frac{\mathrm{\pi }}{4}}\cot 2\theta +1}{\cot 2\theta -\cot {\displaystyle \frac{\mathrm{\pi }}{4}}}\left[\cot \left(A-B\right)=\frac{\cot A\cot B+1}{\cot B-\cot A}\right] \end{gathered}$$
$$\begin{gathered} =\frac{\cot 2\theta +1}{\cot 2\theta -1} \\ =\frac{{\displaystyle \frac{{\cot }^2\theta -1}{2\cot \theta }}+1}{{\displaystyle \frac{{\cot }^2\theta -1}{2\cot \theta }}-1} \\ =\frac{{\cot }^2\theta -1+2\cot \theta }{{\cot }^2\theta -1-2\cot \theta } \end{gathered}$$
$$\begin{gathered} =\frac{3^2-1+2\times 3}{3^2-1-2\times 3}\left(\cot \theta =3\right) \\ =\frac{9-1+6}{9-1-6} \\ =\frac{14}{2} \\ =7 \end{gathered}$$

cot $\left(\frac{\mathrm{\pi }}{4}-2{\cot }^{-1}3\right)$ is equal to ___7___.