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Last updated at Nov. 18, 2020 by Teachoo

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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 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 Rough We know that sin 30° = 1/2 sin (𝜋/6) = 1/2 𝜋/6 = sin–1 1/2 ∴ x = 0 is the only solution Option C is correct Answer

Formulae based

Inverse Trigonometry Formulas

Example 3 Important

Ex 2.2,1 Not in Syllabus - CBSE Exams 2021

Ex 2.2, 2 Important Not in Syllabus - CBSE Exams 2021

Example 8 Not in Syllabus - CBSE Exams 2021

Ex 2.2, 12 Important Not in Syllabus - CBSE Exams 2021

Ex 2.2, 14 Important Not in Syllabus - CBSE Exams 2021

Example 4

Ex 2.2, 3 Important Not in Syllabus - CBSE Exams 2021

Misc. 8 Important Not in Syllabus - CBSE Exams 2021

Ex 2.2, 4 Important Not in Syllabus - CBSE Exams 2021

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Example 13 Important Not in Syllabus - CBSE Exams 2021

Misc. 14 Not in Syllabus - CBSE Exams 2021

Misc. 17 Important Not in Syllabus - CBSE Exams 2021

Misc. 13 Important Not in Syllabus - CBSE Exams 2021

Example 10 Important Not in Syllabus - CBSE Exams 2021

Misc. 3 Not in Syllabus - CBSE Exams 2021

Misc. 4 Important Not in Syllabus - CBSE Exams 2021

Misc 12 Important Not in Syllabus - CBSE Exams 2021

Misc 16 Important Not in Syllabus - CBSE Exams 2021 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 9 years. He provides courses for Maths and Science at Teachoo.