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  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise

Transcript

We know that sin"βˆ’"1 x + cos"βˆ’"1 x = πœ‹/2 sin"βˆ’"1 x = πœ‹/2 – cos"βˆ’"1 x Replacing x by (π‘₯ βˆ’ 1)/(π‘₯ βˆ’ 2) and y by ((π‘₯ + 1)/(π‘₯ + 2)) = tanβˆ’1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x βˆ’ 2) (x + 2) ) Γ— ((x βˆ’ 2) (x + 2))/((x + 2) (x βˆ’ 2) βˆ’ (x βˆ’ 1)(x + 1))] = tanβˆ’1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x + 2) (x βˆ’ 2) βˆ’ (x βˆ’ 1)(x + 1)) ] Using (a + b) (a – b) = a2 – b2 = tanβˆ’1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/(π‘₯2 βˆ’ 22 βˆ’[π‘₯2 βˆ’ 12]) ] = tanβˆ’1 [(π‘₯ (x + 2) βˆ’ 1 (x + 2) + x (x βˆ’ 2) + 1 (x βˆ’ 2))/(π‘₯2 βˆ’ 4 βˆ’ π‘₯2 + 1) ] = tanβˆ’1 [(π‘₯2 + 2π‘₯ βˆ’ π‘₯ βˆ’ 2 + π‘₯2 βˆ’ 2π‘₯ + π‘₯ βˆ’ 2 )/(π‘₯2 βˆ’ π‘₯2 βˆ’ 4 + 1) ] = tanβˆ’1 [(2x2 βˆ’4)/(βˆ’3)] Therefore, tanβˆ’1 ((𝐱 βˆ’ 𝟏)/(𝐱 βˆ’ 𝟐)) + tanβˆ’1 ((𝐱 + 𝟏)/(𝐱 + 𝟐)) = tanβˆ’1 [(𝟐𝐱𝟐 βˆ’πŸ’)/(βˆ’πŸ‘)] Given tanβˆ’1 (x βˆ’ 1)/(x βˆ’ 2) + tanβˆ’1 (x + 1)/(x + 2) = πœ‹/4 Putting values tanβˆ’1 [(2x2 βˆ’4)/(βˆ’3)] = Ο€/4 (2x2 βˆ’4)/(βˆ’3) = tan Ο€/4 (2x2 βˆ’4)/(βˆ’3) = 1 2x2 – 4 = βˆ’3 2x2 = βˆ’3 + 4 2x2 = 1 x2 = 1/2 x = Β± 1/√2 Thus, x = Β± 𝟏/√𝟐

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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.