"If ${({\\sin ^{ - 1}}x)^2} - {({\\cos ^{ - 1}}x)^2} = a$; 0 < x < 1, a $\\ne$ 0, then the value of 2x2 $-$ 1 is :"

"Given $a = {({\\sin ^{ - 1}}x)^2} - {({\\cos ^{ - 1}}x)^2}$ $= ({\\sin ^{ - 1}}x + {\\cos ^{ - 1}}x)({\\sin ^{ - 1}}x - {\\cos ^{ - 1}}x)$ $= {\\pi \\over 2}\\left( {{\\pi \\over 2} - 2{{\\cos }^{ - 1}}x} \\right)$ $\\Rightarrow 2{\\cos ^{ - 1}}x = {\\pi \\over 2} - {{2a} \\over \\pi }$ $\\Rightarrow {\\cos ^{ - 1}}(2{x^2} - 1) = {\\pi \\over 2} - {{2a} \\over \\pi }$ $$ \\Rightarrow 2{x^2} - 1 = \\cos \\left( {{\\pi \\over 2} - {{2a} \\over \\pi }} \\right)$$"

Medium MCQ +4 / -1 PYQ · JEE Mains 2021

If ${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$; 0 < x < 1, a $\ne$ 0, then the value of 2x² $-$ 1 is :

A $\cos \left( {{{4a} \over \pi }} \right)$
B $\sin \left( {{{2a} \over \pi }} \right)$ Correct answer
C $\cos \left( {{{2a} \over \pi }} \right)$
D $\sin \left( {{{4a} \over \pi }} \right)$

Solution

Given $a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$ $= ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$ $= {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$ $\Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$ $\Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$ $$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$$

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Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range

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If ${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$; 0 < x < 1, a $\ne$ 0, then the value of 2x2 $-$ 1 is :

Solution

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If ${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$; 0 < x < 1, a $\ne$ 0, then the value of 2x² $-$ 1 is :