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

Question

cosec[2cot1(5)+cos1(45)]\left[ {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right] is equal to :

Options

Solution

cosec(2cot1(5)+cos1(45))\cos ec\left( {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right) cosec(2tan1(15)+cos1(45))\cos ec\left( {2{{\tan }^{ - 1}}\left( {{1 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right) =cosec(tan1(2(15)1(15)2)+cos1(45)) = \cos ec\left( {{{\tan }^{ - 1}}\left( {{{2\left( {{1 \over 5}} \right)} \over {1 - {{\left( {{1 \over 5}} \right)}^2}}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right) =cosec(tan1(512)+cos1(45)) = \cos ec\left( {{{\tan }^{ - 1}}\left( {{5 \over {12}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right) Let tan1(5/12)=θsinθ=513,cosθ=1213{\tan ^{ - 1}}(5/12) = \theta \Rightarrow \sin \theta = {5 \over {13}},\cos \theta = {{12} \over {13}} and cos1(45)=ϕcosϕ=45{\cos ^{ - 1}}\left( {{4 \over 5}} \right) = \phi \Rightarrow \cos \phi = {4 \over 5} and sinϕ=35\sin \phi = {3 \over 5} =cosec(θ+ϕ) = \cos ec(\theta + \phi ) =1sinθcosϕ+cosθsinϕ = {1 \over {\sin \theta \cos \phi + \cos \theta \sin \phi }} =1513.45+1213.35=6556 = {1 \over {{5 \over {13}}.{4 \over 5} + {{12} \over {13}}.{3 \over 5}}} = {{65} \over {56}}

Practice More Inverse Trigonometric Functions Questions

View All Questions