Question

find the principal value of Sin-' (Sin 2π/3 ) + tan-' (tan 7π/6 )​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: {sin}^{ - 1}\bigg(sin\dfrac{2\pi}{3} \bigg) + {tan}^{ - 1}\bigg(tan\dfrac{7\pi}{6} \bigg)$

Consider,

$\rm :\longmapsto\: {sin}^{ - 1}\bigg(sin\dfrac{2\pi}{3} \bigg)$

We know,

$\rm :\longmapsto\: {sin}^{ - 1}(sinx) = x \: \: if \: x \: \in \: \bigg[ -\dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg]$

Now, here

$\rm :\longmapsto\: \dfrac{2\pi}{3} \: \cancel\in \: \bigg[ -\dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg]$

So,

$\rm :\longmapsto\: {sin}^{ - 1}\bigg(sin\dfrac{2\pi}{3} \bigg)$

$\rm \: = \: \: \: {sin}^{ - 1}\bigg[sin\bigg(\pi - \dfrac{\pi}{3} \bigg) \bigg]$

$\rm \: = \: \: \: {sin}^{ - 1}\bigg[sin\bigg(\dfrac{\pi}{3} \bigg) \bigg]$

$\rm \: = \: \: \:\dfrac{\pi}{3} \: \: \: \: \: \: \: \: \: \: \red{\: as \: \dfrac{\pi}{3} \in \: \bigg[ -\dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg] }$

$\bf :\longmapsto\: {sin}^{ - 1}\bigg(sin\dfrac{2\pi}{3} \bigg) = \dfrac{\pi}{3}$

Now,

Consider,

$\rm :\longmapsto\: {tan}^{ - 1}\bigg(tan\dfrac{7\pi}{6} \bigg)$

Now, here

$\rm :\longmapsto\: \dfrac{7\pi}{6} \: \cancel\in \: \bigg( -\dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg)$

So,

$\rm :\longmapsto\: {tan}^{ - 1}\bigg(tan\dfrac{7\pi}{6} \bigg)$

$\rm \: = \: \: \: {tan}^{ - 1}\bigg[tan\bigg(\pi + \dfrac{\pi}{6} \bigg) \bigg]$

$\rm \: = \: \: \: {tan}^{ - 1}\bigg[tan\bigg(\dfrac{\pi}{6} \bigg) \bigg]$

$\rm \: = \: \: \:\dfrac{\pi}{6} \: \: \: \: \: \: \: \: \: \: \red{\: as \: \dfrac{\pi}{6} \in \: \bigg( -\dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg)}$

$\bf :\longmapsto\: {tan}^{ - 1}\bigg(tan\dfrac{7\pi}{6} \bigg) = \dfrac{\pi}{6}$

Therefore,

$\bf :\longmapsto\: {sin}^{ - 1}\bigg(sin\dfrac{2\pi}{3} \bigg) + {tan}^{ - 1}\bigg(tan\dfrac{7\pi}{6} \bigg)$

$\rm \: = \: \: \:\dfrac{\pi}{3} + \dfrac{\pi}{6}$

$\rm \: = \: \: \:\dfrac{2\pi + \pi}{6}$

$\rm \: = \: \: \:\dfrac{3\pi }{6}$

$\rm \: = \: \: \:\dfrac{\pi }{2}$

Hence,

$\bf :\longmapsto\: {sin}^{ - 1}\bigg(sin\dfrac{2\pi}{3} \bigg) + {tan}^{ - 1}\bigg(tan\dfrac{7\pi}{6} \bigg) = \dfrac{\pi}{2}$