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JEE Main 2019
Inverse Trigonometric Functions
Inverse Trigonometric Functions
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Question

If the inverse trigonometric functions take principal values then cos1(310cos(tan1(43))+25sin(tan1(43))){\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right) is equal to :

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Solution

cos1(310cos(tan1(43))+25sin(tan1(43))){\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right) =cos1(310.35+25.45) = {\cos ^{ - 1}}\left( {{3 \over {10}}\,.\,{3 \over 5} + {2 \over 5}\,.\,{4 \over 5}} \right) =cos1(12)=π3 = {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}

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