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
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Ex 2.2, 11 Find the value of  tan-1 ["2 cos " (2"sin−1" 1/2)] Solving sin-1 (𝟏/𝟐) Let y = sin-1 (1/2) sin y = 1/2 sin y = sin (π/6) Range of principal value of sin −1 is between 𝜋/2 and (−𝜋)/2 Hence, y = π/6 i.e. sin-1 (1/2) = π/6 Now solving tan-1 [2 cos ( 2 sin-1 1/2 )] = tan-1 [2 cos (2 × 𝜋/6 )] = tan-1 (2 cos 𝜋/3 ) = tan-1 (2 cos (180/3 )) = tan-1 (2 × cos 60° ) = tan-1 (2 × 1/2 ) = tan-1(1) = tan-1 (tan 𝜋/4 ) = 𝜋/4

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