Question

TT
X. (5) If sin-' (1-x) - 2 sin ' x =
then find the value of x
2​

Answer 1

$\large\underline{\sf{Correct \: Question- }}$

$\sf \: If \: {sin}^{ - 1} (1 - x) - 2 {sin}^{ - 1} x = \dfrac{\pi}{2} \: then \: find \: value \: of \: x$

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

Given that

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

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

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

$\rm :\longmapsto\:1 - x = cos( {2sin}^{ - 1} x) \: \: \: \: \{ \because \: sin\bigg( \dfrac{\pi}{2} + x \bigg) = cosx \}$

$\rm :\longmapsto\:1 - x = cos\bigg( {cos}^{ - 1}(1 - {2x}^{2}) \bigg) \: \{ \because \: {2sin}^{ - 1} x = {cos}^{ - 1} (1 - {2x}^{2} \}$

$\rm :\longmapsto\: \cancel1 - x = \cancel1 - {2x}^{2} \: \: \: \: \: \: \{ \because \: {cos}^{ - 1} (cosx) = x \}$

$\rm :\longmapsto\: {2x}^{2} - x = 0$

$\rm :\longmapsto\:x(2x - 1) = 0$

$\bf\implies \:x \: = 0 \: \: \: or \: \: \: x = \dfrac{1}{2}$

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

Case :- 1

$\sf \: when \: x \: = \: 0$

$\rm :\longmapsto\:{sin}^{ - 1} (1 - 0) - 2 {sin}^{ - 1} 0 = \dfrac{\pi}{2}$

$\rm :\longmapsto\:{sin}^{ - 1} (1) -0 = \dfrac{\pi}{2}$

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

$\bf\implies \:x \: = \: 0 \: is \: solution$

Case :- 2

$\sf \: when \: x \: = \: \dfrac{1}{2}$

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

$\rm :\longmapsto\:{sin}^{ - 1} \dfrac{1}{2} - 2 {sin}^{ - 1} \dfrac{1}{2} = \dfrac{\pi}{2}$

$\rm :\longmapsto\: - \: {sin}^{ - } \dfrac{1}{2} = \dfrac{\pi}{2}$

$\rm :\longmapsto\: - \: \dfrac{\pi}{6} = \dfrac{\pi}{2}$

$\sf \: which \: is \: absurd$

$\bf\implies \:x \: \ne \: \dfrac{1}{2}$

So,

$\boxed{ \bf{ \: x \: = \: 0 \: is \: solution \: of \: {sin}^{ - 1} (1 - x) - 2 {sin}^{ - 1} x = \dfrac{\pi}{2}}}$

Additional Information :-

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

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

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

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

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

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

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

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

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

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

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

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

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