Example 7 - Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
Last updated at May 12, 2021 by Teachoo
Examples
Example 1
Important
Example 2
Example 3 (i) Important
Example 3 (ii)
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Example 9 Important
Example 10 Important Deleted for CBSE Board 2022 Exams
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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
Transcript
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
