JEE Main 202226 Jul, Evening shiftMathematicsMediumTrigonometry118× since 2002If 0<x<120 < x < {1 \over {\sqrt 2 }}0<x<21 and sin−1xα=cos−1xβ{{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta }αsin−1x=βcos−1x, then the value of sin(2παα+β)\sin \left( {{{2\pi \alpha } \over {\alpha + \beta }}} \right)sin(α+β2πα) is :AOption A. 4(1−x2)(1−2x2)4 \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)4(1−x2)(1−2x2)BOption B. 4x(1−x2)(1−2x2)4 x \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)4x(1−x2)(1−2x2)COption C. 2x(1−x2)(1−4x2)2 x \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)2x(1−x2)(1−4x2)DOption D. 4(1−x2)(1−4x2)4 \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)4(1−x2)(1−4x2)CheckSkip · show solution
