Misc. 4 - Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Transcript

Misc 4 Prove sin–1 8/17 + sin–1 3/5 = tan–1 77/36 Let a = sin–1 8/17 & b = sin–1 3/5 We convert sin–1 to tan–1 & then use tan (a – b) formula Let a = sin–1 𝟖/𝟏𝟕 sin a = 8/17 Now, cos a = √(1 −𝑠𝑖𝑛2 𝑎) = √(1 −(8/17)^2 ) = √(225/289) = 15/17 Thus, tan a = (sin 𝑎)/(cos a) = (8/17)/(15/17) = 8/17 × 17/15 = 𝟖/𝟏𝟓 Let b = sin–1 𝟑/𝟓 sin b = 3/5 Now, cos b = √(1 −𝑠𝑖𝑛2 𝑏) = √(1 −(3/5)^2 )= √(16/25) = 4/5 Thus, tan b = sin⁡𝑏/cos⁡𝑏 = (3/5)/(4/5) = 3/5 × 5/4 = 𝟑/𝟒 Now tan (a + b) = tan⁡〖𝑎 +〖 tan〗⁡〖𝑏 〗 〗/(1 − tan⁡〖𝑎 tan⁡𝑏 〗 ) = (8/15 + 3/4)/(1 − 8/15 × 3/4) = ((8(4) + 3(15) )/(15 × 4) )/( (15 × 4 − 8 × 3)/(15 × 4) ) = ((32 + 45 )/(15 × 4) )/( (60 − 24)/(15 × 4) ) = (32 + 45)/(15 × 4) × (15 × 4)/(60 −24) = 77/36 Hence, tan (a + b) = 𝟕𝟕/𝟑𝟔 a + b = tan–1 77/36 Putting value of a & b sin–1 8/17 + sin–1 3/5 = tan–1 77/36 Hence Proved