1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
  3. Ex 2.2

Ex 2.2, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at May 29, 2018 by

Ex 2.2, 11 - Find tan-1 [2 cos (2 sin-1 1/2)] - Class 12 - Finding pricipal value

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
    3. Ex 2.2
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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

Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise

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