JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionIf 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 :OptionsA1−y22y{{1 - {y^2}} \over {2y}}2y1−y2B1−y2yy{{1 - {y^2}} \over {y\sqrt y }}yy1−y2C1−y21 - {y^2}1−y2D1−y21+y2{{1 - {y^2}} \over {1 + {y^2}}}1+y21−y2Check AnswerHide SolutionSolutionsin−1xa=cos−1xb=tan−1yc{{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}asin−1x=bcos−1x=ctan−1y sin−1xa=cos−1xb=sin−1x+cos−1xa+b=π2(a+b){{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}asin−1x=bcos−1x=a+bsin−1x+cos−1x=2(a+b)π Now, tan−1yc=π2(a+b){{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}ctan−1y=2(a+b)π 2tan−1y=πca+b2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}2tan−1y=a+bπc ⇒cos(πca+b)=cos(2tan−1y)=1−y21+y2 \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}⇒cos(a+bπc)=cos(2tan−1y)=1+y21−y2