JEE Main 202126 Feb, Morning shiftMathematicsMediumTrigonometry118× since 2002If sin1xa=cos−1xb=tan−1yc{{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}asin1x=bcos−1x=ctan−1y; 0<x<10 < x < 10<x<1, then the value of cos(πca+b)\cos \left( {{{\pi c} \over {a + b}}} \right)cos(a+bπc) is :AOption A. 1−y22y{{1 - {y^2}} \over {2y}}2y1−y2BOption B. 1−y2yy{{1 - {y^2}} \over {y\sqrt y }}yy1−y2COption C. 1−y21 - {y^2}1−y2DOption D. 1−y21+y2{{1 - {y^2}} \over {1 + {y^2}}}1+y21−y2CheckSkip · show solution
