Question

Find the value of
cosec⁻¹(-√2) + cot⁻¹(√3)

Answer 1

we know that the principal values of cosec⁻¹x is [-π/2,π/2]-{0}

let

${cosec}^{ - 1} ( - \sqrt{2} ) = \alpha\\ \\ cosec \: \alpha= - \sqrt{2} = cosec( \frac{ - \pi}{4}) \\ \\ = >\alpha= \frac{ - \pi}{4}....eq1 belongs \: to \: [\frac{ - \pi}{2}, \frac{\pi}{2}] - {0}$

So,

we know that the principal values of cot⁻¹x is [0,π]

${cot}^{ - 1} ( \sqrt{3} ) = \beta \\ \\ then \\ \\ cot \: \beta = \sqrt{3} = cot \: ( \frac{\pi}{6} ) \\ \\ = > \beta= \frac{\pi}{6} ...eq2\: \: belongs \: to \:[ 0 ,\: \pi]$

add both equations

$- \frac{\pi}{4} + \frac{\pi}{6} \\ \\ = \frac{ - 3\pi +2\pi}{12} \\ \\ = - \frac{\pi}{12}$