Question

FRIENDS, PLS SOLVE THIS Q FOR ME..I WILL VOTE IT THE BRAINLIEST ANS IF U DO IT RIGHT..

Answer 1

Answer:

$\boxed{\green{\huge{9}}}$

Step-by-step explanation:

$\underline{\bold{Given}} = > {sin}^{ - 1} x + {sin}^{ - 1} y + {sin}^{ - 1} z = \dfrac{3\pi}{2}$

$\underline{\bold{To \: find}} = > value \: of \: {(x + y + z)}^{2}$

$\underline{\bold{Concept \: used}}= > \\range \: of \: {sin}^{ - 1} \: is \: \: | - \dfrac{ \pi}{2} \: \: \: \dfrac{\pi}{2} | \\ so \: maximum \: value \: of \: {sin}^{ - 1} \: is \: \dfrac{\pi}{2}$

$\underline{\bold{Solution }}= > \\ { sin }^{ - 1} x + {sin}^{ - 1} y + {sin}^{ - 1} z = \dfrac{3\pi}{2}$

$= > {sin}^{ - 1} x + {sin}^{ - 1} y + {sin}^{ - 1} z = \dfrac{\pi}{2} + \dfrac{\pi}{2} + \dfrac{\pi}{2}$

$maximum \: value \: of \: {sin}^{ - 1} \: is \: \dfrac{\pi}{2} \: \\ so \: above \: equation \: is \: true \: only \: when$

$= > {sin}^{ - 1} x = \dfrac{\pi}{2} \\ = > {sin}^{ - 1} y = \dfrac{\pi}{2} \\ = > {sin}^{ - 1} z = \dfrac{\pi}{2}$

$now$

${sin}^{ - 1} x = \dfrac{\pi}{2} \\ = > x \: = sin \dfrac{\pi}{2} \\ = > x = 1 \\ simlary \: \\ y = 1 \: \: and \: \: z = 1$

$now$

${(x + y + z)}^{2} = (1 + 1 + 1) ^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {(3)}^{2} \\ = > {(x + y + z)}^{2} = 9$