Question

find the value of (cot inverse 1/2) + (cot inverse 1/3)​

Answer 1

TO FIND :–

• Value of $\bf \: \: \cot ^{ - 1} \left( \dfrac{1}{2} \right) + \cot ^{ - 1} \left( \dfrac{1}{3} \right) = ?$

SOLUTION :–

• Let the function –

$\\ \bf \implies y = \cot ^{ - 1} \left( \dfrac{1}{2} \right) + \cot ^{ - 1} \left( \dfrac{1}{3} \right)$

• Using identity –

$\\ \bf \implies \cot^{ - 1} \left( \dfrac{1}{x} \right) = \tan^{ - 1} (x)$

• So that –

$\\ \bf \implies y = \tan^{ - 1} \left(2\right) + \tan^{ - 1} \left(3\right)$

• Using identity –

$\\ \bf \implies \tan^{ - 1} \left(x\right) + \tan^{ - 1} \left(y\right) = \tan^{ - 1} \left( \dfrac{x + y}{1 - xy} \right)$

• So that –

$\\ \bf \implies y = \tan^{ - 1} \left( \dfrac{2+3}{1 - (2)(3)} \right)$

$\\ \bf \implies y = \tan^{ - 1} \left( \dfrac{2+3}{1 -6} \right)$

$\\ \bf \implies y = \tan^{ - 1} \left( \dfrac{5}{ - 5} \right)$

$\\ \bf \implies y = \tan^{ - 1} \left( - \cancel \dfrac{5}{5} \right)$

$\\ \bf \implies y = \tan^{ - 1} \left( -1 \right)$

$\\ \bf \implies y = \tan^{ - 1} \left \{\tan \left( \dfrac{3\pi}{4} \right) \right \}$

$\\ \bf \implies \large{ \boxed{ \bf y = \dfrac{3\pi}{4}}}$

Answer 2

${\sf{\underline{\underline{\pink{TO \ FIND :–}}}}}$

$Value of \begin{gathered}\bf \: \: \cot ^{ - 1} \left( \dfrac{1}{2} \right) + \cot ^{ - 1} \left( \dfrac{1}{3} \right) = ? \\\end{gathered}$

${\sf{\underline{\underline{\red{SOLUTION :–}}}}}$

${\sf{\underline{\underline{\pink{ Let \ the \ function –}}}}}$

$\begin{gathered}\\ \bf \implies y = \cot ^{ - 1} \left( \dfrac{1}{2} \right) + \cot ^{ - 1} \left( \dfrac{1}{3} \right) \\\end{gathered}$

${\sf{\underline{\underline{\pink{Using \ identity –}}}}}$

$\begin{gathered}\\ \bf \implies \cot^{ - 1} \left( \dfrac{1}{x} \right) = \tan^{ - 1} (x) \\\end{gathered}$

${\sf{\underline{\underline{\pink{ So \ that –}}}}}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left(2\right) + \tan^{ - 1} \left(3\right) \\\end{gathered}$

${\sf{\underline{\underline{\pink{ Using \ identity –}}}}}$

$\begin{gathered}\\ \bf \implies \tan^{ - 1} \left(x\right) + \tan^{ - 1} \left(y\right) = \tan^{ - 1} \left( \dfrac{x + y}{1 - xy} \right) \\\end{gathered}$

${\sf{\underline{\underline{\pink{ So \ that –}}}}}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left( \dfrac{2+3}{1 - (2)(3)} \right) \\\end{gathered}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left( \dfrac{2+3}{1 -6} \right) \\\end{gathered}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left( \dfrac{5}{ - 5} \right) \\\end{gathered}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left( - \cancel \dfrac{5}{5} \right) \\\end{gathered}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left( -1 \right) \\\end{gathered}$

$\begin{gathered}\\ \bf \implies y = \tan^{ - 1} \left \{\tan \left( \dfrac{3\pi}{4} \right) \right \} \\\end{gathered}$

$\begin{gathered}\\ \bf \implies \large{ \boxed{ \bf y = \dfrac{3\pi}{4}}} \\\end{gathered}$