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Last updated at May 12, 2021 by Teachoo

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Example 13 Solve tan–1 2x + tan–1 3x = π/4 Given tan–1 2x + tan 3x = π/4 tan–1 ((2x + 3x)/(1 − 2x × 3x)) = π/4 𝟓𝐱/(𝟏 − 𝟔𝐱𝟐) = tan 𝝅/𝟒 We know that tan–1 x + tan–1 y = tan–1 ((𝐱 + 𝐲)/(𝟏 − 𝐱𝐲)) Replacing x by 2x & y by 3x 5x/(1− 6x2) = 1 5x = 1 × (1 – 6x2) 5x = 1 – 6x2 6x2 + 5x – 1 = 0 6x2 + 6x – x – 1 = 0 6x(x + 1 ) – 1 (x – 1) = 0 (6x – 1) (x + 1) = 0 Thus, x = 1/6 or x = −1 But For x = −1 tan–1 2x + tan–1 3x = π/4 tan–1 (–2) + tan–1 (–3) = π/4 So, L.H.S becomes negative but R.H.S is positive. Thus, x = –1 is not possible. Hence, x = 𝟏/𝟔 is the only solution of the given equation

Examples

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Example 2

Example 3 (i) Important

Example 3 (ii)

Example 4 Deleted for CBSE Board 2022 Exams

Example 5 Important Deleted for CBSE Board 2022 Exams

Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Deleted for CBSE Board 2022 Exams

Example 8 Deleted for CBSE Board 2022 Exams

Example 9 Important

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11 Important Deleted for CBSE Board 2022 Exams

Example 12 Important Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams You are here

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Serial order wise

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.