Question

Find the value of
tan⁻¹(√3) - sec⁻¹(-2)

Answer 1

We know that the principal value branch of

${tan}^{ - 1} \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ]$
${sec}^{ - 1} \: \: [0 ,\: \pi] - (\frac{\pi}{2})$

now

${tan}^{ - 1} ( \sqrt{3} ) = \alpha \\ \\ then \\ \\ tan( \alpha ) = \sqrt{3} = tan( \frac{\pi}{3} ) \\ \\ where \: \frac{\pi}{3} \: belongs \: to \: [\frac{ - \pi}{2} ,\frac{\pi}{2}]$

So as

${sec}^{ - 1} ( - 2) = \beta \\ \\ then \\ \\ \sec( \beta ) = - 2= - sec( \frac{\pi}{3} ) \\ \\ = sec(\pi - \frac{\pi}{3} ) \\ \\ = sec (\frac{2\pi}{3}) \\ \\ where \: \frac{2\pi}{3} \: belongs \: to \: \: \: [0 ,\: \pi] - (\frac{\pi}{2} )$

now

tan⁻¹(√3) - sec⁻¹(-2) =
$= \frac{\pi}{3} - \frac{2\pi}{3} \\ \\ = \frac{ - \pi}{3}$