Question

Sin[sin^-1(x/5)+cos^-1(3/5)]=1
find x
​

Answer 1

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

$\sf \: sin\bigg( {sin}^{ - 1}\dfrac{x}{5} + {cos}^{ - 1} \dfrac{3}{5} \bigg) = 1$

$\large\underline{\sf{To\:Find - }}$

$\sf \:\: Value \: of \: x$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf{ \: {sin}^{ - 1} (sinx) = x}}$

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

$\boxed{ \bf{ \: sin\dfrac{\pi}{2} = 1}}$

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

Given that

$\rm :\longmapsto\: \sf \: sin\bigg( {sin}^{ - 1}\dfrac{x}{5} + {cos}^{ - 1} \dfrac{3}{5} \bigg) = 1$

$\rm :\longmapsto\: \sf \: {sin}^{ - 1}\dfrac{x}{5} + {cos}^{ - 1} \dfrac{3}{5} = {sin}^{ - 1}( 1)$

$\rm :\longmapsto\: \sf \: {sin}^{ - 1}\dfrac{x}{5} + {cos}^{ - 1} \dfrac{3}{5} = {sin}^{ - 1}( sin\dfrac{\pi}{2} )$

$\rm :\longmapsto\: \sf \: {sin}^{ - 1}\dfrac{x}{5} + {cos}^{ - 1} \dfrac{3}{5} = \dfrac{\pi}{2}$

$\rm :\longmapsto\: \sf \: {sin}^{ - 1}\dfrac{x}{5} = \dfrac{\pi}{2} - {cos}^{ - 1} \dfrac{3}{5}$

$\rm :\longmapsto\: \sf \: {sin}^{ - 1}\dfrac{x}{5} = {sin}^{ - 1} \dfrac{3}{5}$

$\: \: \: \: \: \: \because \: \bf \{ \: { \bf{ \: {sin}^{ - 1} x = \dfrac{\pi}{2} } \: - \: {cos}^{ - 1} x} \}$

$\rm :\implies\:\dfrac{x}{ \cancel5} = \dfrac{3}{ \cancel5}$

$\bf\implies \:x \: = \: 3$

Additional Information :-

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

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

$\boxed{ \bf{ \: {cos}^{ - 1} (cox) = x}}$

$\boxed{ \bf{ \: {tan}^{ - 1} (tanx) = x}}$

$\boxed{ \bf{ \: {sin}^{ - 1} ( - x) = - \: {sin}^{ - 1} x}}$

$\boxed{ \bf{ \: {cos}^{ - 1} ( - x) =\pi \: - \: {cos}^{ - 1} x}}$

$\boxed{ \bf{ \: {tan}^{ - 1} ( - x) = - \: {tan}^{ - 1} x}}$