Question

Prove that: 9π/8-9/4 sin inverse 1/3=9/4 sin inverse 2√2/3.​

Answer 1

Given Question :-

$\rm \: Prove \: that \: \dfrac{9\pi}{8} - \dfrac{9}{4} {sin}^{ - 1}\dfrac{1}{3} = \dfrac{9}{4} {sin}^{ - 1}\dfrac{2 \sqrt{2} }{3}$

Identities Used :-

$\boxed{ \sf \: {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2}}$

$\green{\large\underline{\bf{Solution-}}}$

Consider,

$\rm :\longmapsto\:\dfrac{9\pi}{8} - \dfrac{9}{4} {sin}^{ - 1}\dfrac{1}{3}$

$\: \: \rm = \: \: \dfrac{9}{4} \bigg(\:\dfrac{\pi}{2} - {sin}^{ - 1}\dfrac{1}{3} \bigg)$

$\: \: \rm = \: \: \dfrac{9}{4} \bigg(\:\ {cos}^{ - 1}\dfrac{1}{3} \bigg)$

$\: \: \rm = \: \: \dfrac{9}{4} \bigg(\:\ {sin}^{ - 1}\dfrac{2 \sqrt{2} }{3} \bigg) \: \: \: \{using \: \triangle \: method \}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \sf \: {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2}}$

$\boxed{ \sf \: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2}}$

$\boxed{ \sf \: 2 {tan}^{ - 1}x = {tan}^{ - 1}\bigg(\dfrac{2x}{1 - {x}^{2} } \bigg)}$

$\boxed{ \sf \: 2 {tan}^{ - 1}x = {sin}^{ - 1}\bigg(\dfrac{2x}{1 + {x}^{2} } \bigg)}$

$\boxed{ \sf \: 2 {tan}^{ - 1}x = {cos}^{ - 1}\bigg(\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)}$

$\boxed{ \sf \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x + y}{1 - xy } \bigg) \: \: \: if \: xy \leqslant 1}$

$\boxed{ \sf \: {tan}^{ - 1}x +{tan}^{ - 1}y = \pi + {tan}^{ - 1}\bigg(\dfrac{x + y}{1 - xy } \bigg) \: \: \: if \: xy > 1}$