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Ex 2.2, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 2.2, 11 Find the value of tan-1 ["2 cos " (2"sin 1" 1/2)] Solving sin-1 ( / ) Let y = sin-1 (1/2) sin y = 1/2 sin y = sin ( /6) Range of principal value of sin 1 is between /2 and ( )/2 Hence, y = /6 i.e. sin-1 (1/2) = /6 Now solving tan-1 [2 cos ( 2 sin-1 1/2 )] = tan-1 [2 cos (2 /6 )] = tan-1 (2 cos /3 ) = tan-1 (2 cos (180/3 )) = tan-1 (2 cos 60 ) = tan-1 (2 1/2 ) = tan-1(1) = tan-1 (tan /4 ) = /4
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Ex 2.2, 11 You are here
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Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.