Question

What is the value of cot inverse of 1/2 + cot inverse of 1/3

Answer 1

$we \: know \: that \\ {cot}^{ - 1} x = \frac{1}{ {tan}^{ - 1} x} \\ \\ so \: converting \: the \: given \: values \: of \: {cot}^{ - 1} \: in \: terms \: of \: {tan}^{ - 1} x \\ we \: will \: get \: the \: question \: as\\ {tan}^{ - 1} 2 + {tan}^{ - 1} 3 \\ we \: also \: know \: that \: \\ { tan}^{ - 1} x + { tan }^{ - 1} y = {tan}^{ - 1} ( \frac{x + y}{1 - xy} ) \\ \\ applying \: this \: formula \: in \: the \: given \: question \\ we \: will \: get \\ \\ {tan}^{ - 1} ( \frac{2 + 3}{1 - 2 \times 3} ) \\ {tan}^{ - 1} ( \frac{5}{ - 5} ) \\ {tan}^{ - 1} ( - 1) \\ \frac{ - \pi}{4}$