Question

Prove that

${sin}^{ - 1} x = {cos}^{ - 1} ( \sqrt{1 - {x}^{2} } ) = {tan}^{ - 1} ( \frac{x}{ \sqrt{1 - {x}^{2} } } )$
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Answer 1

heya your answer is here... hope it helps

Answer 2

$\bf \: \huge \: {sin}^{ - 1} x = {cos}^{ - 1} ( \sqrt{1 - {x}^{2} } )$

$\bf \: LHS \: = {sin}^{ - 1} x$

$\bf \: Let \: \: \: \: \: \: \: \: \: \: {sin}^{ - 1} x = \theta ..........(1)\\ \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: x = sin \: \theta$

$\bf \: Now, \: u sing \: the \: formula \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: { \sin }^{2} \theta \: + \cos {}^{2} \theta \: = 1}$

$\bf \: Let's \: put \: the \: value \: of \: sin \: \theta \: \\ \\ \\ \implies \: \: {x}^{2} + {cos}^{2} \theta \: = 1 \\ \\ \\ \implies \: \bf \: \: cos \: \theta \: = 1 - {x}^{2} \\ \\ \\ \implies \bf \: \theta \: = {cos}^{ - 1} \sqrt{1 - {x}^{2} } \\ \\ \\ \bf \: now \: we \: put \: the \: value \: of \: \theta \: from \: the \: first \: equation \: \\ \\ \\ \implies \: \boxed{ \bf \: { \sin}^{ - 1}x = \cos {}^{ - 1} \sqrt{1 - {x}^{2} } }$

$\huge \bf \sin {}^{ - 1} x = \tan {}^{ - 1} ( \frac{x}{ \sqrt{1 - {x}^{2} } } )$

$\bf \: LHS \: = {sin}^{ - 1} x$

$\bf \: Again \: let \: {sin}^{ - 1} x = \theta \: \: \: \: \: \: \: \: ....(1)$

$\bf \: x \: = sin \: \theta$

$\bf \: Now \: using \: the \: formula \\ \\ \\ \bf \: cosec \: \theta \: = \frac{1}{sin \: \theta}$

$\bf \: Now \: put \: the \: value \: of \: sin \: \theta \\ \\ \\ \implies \: \: \bf \: \: cosec \: \theta \: = \frac{1}{x}$

$\bf \: W e \: know \: that \: formula \: \: \\ \\ \\ \boxed{1 + { \cot }^{2} \theta \: = {cosec \: }^{2} \theta \: } \\ \\ \\ let \: put \: the \: value \: of \: cosec \: \theta \\ \\ \\ \implies \: 1 + {cot}^{2} \: \theta \: = ( \frac{1}{x} ) {}^{2} \\ \\ \\ \implies \: \bf \: \: cot \: {}^{2} \theta \: = \frac{1}{ {x}^{2} } - 1 \\ \\ \\ \implies \: \bf \: cot {}^{2} \theta = \frac{1 - {x}^{2} }{ {x}^{2} } \\ \\ \\ \implies \: \bf \: cot \: \theta \: = \sqrt{ \frac{1 - {x}^{2} }{x} } \: \: \\ \\ \\ \therefore \: tan \: \theta = \frac{x}{ \sqrt{1 - {x}^{2} } } \\ \\ \\ \implies \: \bf \: \theta \: = {tan}^{ - 1} \frac{x}{ \sqrt{1 - {x}^{2} } } \\ \\ \\ let \: put \: the \: value \: of \: \theta \: from \: the \: first \: equation \: \\ \\ \\ \boxed{ \bf \: {sin}^{ - 1} = {tan}^{ - 1} \frac{x}{ \sqrt{1 - {x}^{2} } } }$