Ex 2.2, 18 - tan (sin-1 3/5 + cot-1 3/2) - Class 12 NCERT - Changing of trignometric variables and then applying formula

 

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Ex 2.2, 18 (Method 1) tan (sin-1 3/5 + cot-1 3/2 ) Let sin-1 3/5 = a & cot-1 3/2 = b Now sin-1 3/5 = a 3/5 = sin a sin a = 3/5 cos a = √(1 βˆ’π‘ π‘–π‘›2π‘Ž) = √(1 βˆ’(3/5)^2 ) = √(1 βˆ’9/25) = √((25 βˆ’ 9)/25) = √(16/25) = 4/5 tan a = sinβ‘π‘Ž/cosβ‘π‘Ž = (3/5)/(4/5) = 3/5 Γ— 5/4 = 3/4 Hence, tan (sin-1 3/5 + cot-1 3/2 ) = 17/6 Hence Proved Ex 2.2, 18 (Method 2) tan (sin-1 3/5 + cot-1 3/2 ) Let sin-1 3/5 = a & cot-1 3/2 = b Now sin-1 3/5 = a 3/5 = sin a sin a = 3/5 cos a = √(1 βˆ’π‘ π‘–π‘›2π‘Ž) = √(1 βˆ’(3/5)^2 ) = √(1 βˆ’9/25) = √((25 βˆ’ 9)/25) = √(16/25) = 4/5 tan a = sinβ‘π‘Ž/cosβ‘π‘Ž = (3/5)/(4/5) = 3/5 Γ— 5/4 = 3/4 tan a = 3/4 a = tan-1 3/4

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