1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Concept wise
  3. Formulae based

Ex 2.2, 14 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at Feb. 13, 2020 by

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  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Concept wise

    3. Formulae based

    Changing of trignometric variables and then applying formula →

Transcript

Ex 2.2, 14 If sin ("sin−1 " 1/5 " + cos−1 x" ) = 1 , then find the value of x. sin ("sin−1 " 1/5 " + cos−1 x" ) = 1 We know that sin 90° = 1 i.e. sin π/2 = 1 Comparing (1) & (2) sin ("sin−1 " 1/5 " + cos−1 x" ) = sin π/2 …(1) Comparing angles "sin−1 " 1/5 + "cos−1 x" = 𝜋/2 "sin−1 " 1/5 = 𝝅/𝟐 – "cos−1 x" sin-1 1/5 = sin"−"1 x 1/5 = x x = 𝟏/𝟓 Hence, x = 1/5 We know that sin"−"1 x + cos"−"1 x = 𝜋/2 sin"−"1 x = 𝜋/2 – cos"−"1 x

Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise

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