Skip to main content
Back to Inverse Trigonometric Functions
JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Easy

Question

If sin1xa=cos1xb=tan1yc{{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}; 0<x<10 < x < 1, then the value of cos(πca+b)\cos \left( {{{\pi c} \over {a + b}}} \right) is :

Options

Solution

sin1xa=cos1xb=tan1yc{{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c} sin1xa=cos1xb=sin1x+cos1xa+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)}} Now, tan1yc=π2(a+b){{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}} 2tan1y=πca+b2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}} cos(πca+b)=cos(2tan1y)=1y21+y2 \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}

Practice More Inverse Trigonometric Functions Questions

View All Questions