Ex 2.2, 18 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Ex 2.2, 18 tan (sin−1 3/5 + cot−1 3/2 ) Let sin−1 3/5 = a & cot−1 3/2 = b Now a = sin−1 3/5 = a sin a = 3/5 cos a = √(1 −𝑠𝑖𝑛2𝑎) = √(1 −(3/5)^2 ) = √(1 −9/25) = √((25 − 9)/25) = √(16/25) = 4/5 We write sin-1 3/5 & cot-1 3/2 in terms of tan-1 b = cot −1 3/2 cot b = 3/2 tan b = 1/cot⁡𝑏 = 1/(3/2) = 2/3 tan a = sin⁡𝑎/cos⁡𝑎 = (3/5)/(4/5) = 3/5 × 5/4 = 3/4 Now calculating tan ("sin−1 " 3/5 " + cot−1 " 3/2) = tan (a + b) = tan⁡〖𝑎 + tan⁡𝑏 〗/(1 − tan⁡〖𝑎 tan⁡𝑏 〗 ) Putting tan a = 3/4 & tan b = 2/3 = (3/4 + 2/3 )/(1 − 3/4 × 2/3) = ((3(3) + 2(4) )/(4 × 3) )/( (4 × 3 − 3 × 2)/(4 × 3) ) = ((9 + 8 )/(4 × 3) )/( (12 − 6)/(4 × 3) ) = ((17 )/(4 × 3) )/( 6/(4 × 3) ) = (17 )/(4 × 3) × (4 × 3)/( 6) = (17 )/6 Hence, tan ("sin−1 " 3/5 " + cot−1 " 3/2) = 𝟏𝟕/𝟔