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Last updated at Aug. 9, 2021 by Teachoo

Transcript

Misc 16 Solve sin−1(1 – x) – 2sin−1 x = π/2 , then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2 sin−1 (1 – x) – 2sin−1 x = π/2 –2sin−1 x = 𝝅/𝟐 – sin−1 (1 – x) − 2sin−1 x = cos−1 (1 – x) We know that sin−1 x + cos−1x = 𝝅/𝟐 Replace x by (1 − x) sin-1 (1 − x) + cos−1 (1 − x) = 𝜋/2 cos-1 (1 − x) = 𝜋/2 – sin−1 (1 − x) Let sin−1 x = a, Hence our equation becomes −2a = cos−1 (1 – x) cos (−2a) = 1 – x cos (2a) = (1 – x) 1 – 2 sin2 a = 1 – x We assumed that sin−1 x = a 1 – 2 [sin(sin−1 x)]2 = 1 – x 1 – 2x2 = 1 – x 1 – 2x2 – 1 + x = 0 1 – 1 – 2x2 + x = 0 –2x2 + x = 0 0 = 2x2 – x 2x2 – x = 0 x (2x – 1) = 0 So, x = 0 and x = 1/2 But x = 𝟏/𝟐 does not satisfy the equation Taking equation sin−1(1 – x) – 2sin−1 x = π/2 Putting x = 𝟏/𝟐 in L.H.S sin−1(1− 1/2) – 2 sin−1 (1/2) = sin−1(1/2) – 2 sin−1 (1/2) = 𝜋/6 – 2 × 𝜋/6 = (𝜋 − 2𝜋)/6 = (− 𝜋)/6 ≠ 𝝅/𝟐 Hence x = 1/2 not possible ∴ x = 0 is the only solution Option C is correct Answer

Formulae based

Inverse Trigonometry Formulas

Example 3 (i) Important

Ex 2.2,1 Deleted for CBSE Board 2022 Exams

Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams

Example 8 Deleted for CBSE Board 2022 Exams

Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams

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Example 13 Important Deleted for CBSE Board 2022 Exams

Misc. 14 Deleted for CBSE Board 2022 Exams

Misc 17 (MCQ) Deleted for CBSE Board 2022 Exams

Misc. 13 Important Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Misc. 3 Deleted for CBSE Board 2022 Exams

Misc. 4 Important Deleted for CBSE Board 2022 Exams

Misc 12 Important Deleted for CBSE Board 2022 Exams

Misc 16 (MCQ) Important Deleted for CBSE Board 2022 Exams You are here

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.