Ex 2.2, 15 - If tan-1 (x - 1)/(x - 2) + tan-1 (x+1)/(x+2) = pi/4

Ex 2.2, 15 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Ex 2.2, 15 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3

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Question 6 If tanβˆ’1 (x βˆ’ 1)/(x βˆ’ 2) + tanβˆ’1 (x + 1)/(x + 2) = πœ‹/4 , then find the value of x. Given tanβˆ’1 ((𝐱 βˆ’ 𝟏)/(𝐱 βˆ’ 𝟐)) + tanβˆ’1 ((𝐱 + 𝟏)/(𝐱 + 𝟐)) = πœ‹/4We know that tanβˆ’1 x + tanβˆ’1 y = tanβˆ’1 ((𝐱 + 𝐲 )/( 𝟏 βˆ’ 𝐱𝐲)) Replacing x by (π‘₯ βˆ’ 1)/(π‘₯ βˆ’ 2) and y by ((π‘₯ + 1)/(π‘₯ + 2)) tanβˆ’1 [((x βˆ’ 1 )/(x βˆ’ 2) + (x + 1)/(x + 2))/(1βˆ’ (x βˆ’ 1)/(x βˆ’ 2) Γ— (x + 1)/(x + 2))]=" " πœ‹/4 tanβˆ’1 [((x βˆ’ 1 )/(x βˆ’ 2) + (x + 1)/(x + 2))/(1βˆ’ (x βˆ’ 1)/(x βˆ’ 2) Γ— (x + 1)/(x + 2))]="tan " πœ‹/4 = tan-1 [(((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x βˆ’ 2) (x + 2) ))/(((x βˆ’ 2) (x + 2) βˆ’ (x βˆ’ 1) (x + 1))/((x βˆ’ 2) (x + 2) ))]((x βˆ’ 1 )/(x βˆ’ 2) + (x + 1)/(x + 2))/(1βˆ’ (x βˆ’ 1)/(x βˆ’ 2) Γ— (x + 1)/(x + 2)) = "tan " 𝝅/πŸ’ (((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x βˆ’ 2) (x + 2) ))/(((x βˆ’ 2) (x + 2) βˆ’ (x βˆ’ 1) (x + 1))/((x βˆ’ 2) (x + 2) )) = 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)) = 1 ((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x + 2) (x βˆ’ 2) βˆ’ (x βˆ’ 1)(x + 1)) = 1 Using (a + b) (a – b) = a2 – b2 ((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/(π‘₯2 βˆ’ 22 βˆ’[π‘₯2 βˆ’ 12]) = 1 (π‘₯ (π‘₯ + 2) βˆ’ 1 (π‘₯ + 2) + π‘₯ (π‘₯ βˆ’ 2) + 1 (π‘₯ βˆ’ 2))/(π‘₯2 βˆ’ 4 βˆ’ π‘₯2 + 1) = 1 (π‘₯2 + 2π‘₯ βˆ’ π‘₯ βˆ’ 2 + π‘₯2 βˆ’ 2π‘₯ + π‘₯ βˆ’ 2 )/(π‘₯2 βˆ’ π‘₯2 βˆ’ 4 + 1) = 1 (2x2 βˆ’4)/(βˆ’3) = 1 2x2 – 4 = βˆ’3 2x2 = βˆ’3 + 4 2x2 = 1 x2 = 1/2 ∴ x = Β± 𝟏/√𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo