Examples

Example 1
Important

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 You are here

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

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Serial order wise

Last updated at May 12, 2021 by Teachoo

Example 7 Show that tan-1 π₯ + tan-1 2π₯/(1 βπ₯2) = tan-1 (3π₯ β π₯3)/(1 β 3π₯2) Solving L.H.S tan-1 π₯ + tan-1 2π₯/(1 β π₯2) = tan-1 (π₯ + 2π₯/(1 β π₯2))/(1β π₯ Γ 2π₯/(1 β π₯2)) = tan-1 ((π₯(1 β π₯2) + 2π₯)/(1 β π₯2))/(γ(1 β π₯2) β 2π₯γ^2/(1 β π₯2)) We know that tan-1 x + tan-1 y = tan-1 ((π+π )/(π βππ)) Replacing x by x and y by 2π₯/(1 β π₯2) = tan-1 ((π₯ β π₯3 + 2π₯)/(1 β π₯2))/(γ1 β π₯2β 2π₯γ^2/(1 β π₯2)) = tan-1 ((3π₯ β π₯3)/(1 β π₯2))/(γ1 β 3π₯γ^2/(1 β π₯2)) = tan-1 (3π₯ β π₯3)/(1 β π₯2) Γ (1 β π₯2)/γ1 β 3π₯γ^2 = tan-1 (3π₯ β π₯3)/γ1 β 3π₯γ^2 = R.H.S Thus L.H.S = R.H.S Hence proved