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

Transcript

Misc 4 (Method 1) Prove sin-1 8/17 + sin-1 3/5 = tan-1 77/36 Let a = sin-1 8/17 & b = sin-1 3/5 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 Misc 4 (Method 2) Prove sin-1 8/17 + sin-1 3/5 = tan-1 77/36 Let a = sin-1 8/17 & b = sin-1 3/5 Solving L.H.S sin-1 8/17 + sin-1 3/5 = a + b Putting values from (1) & (2) = tan-1 8/15 + tan-1 3/4 = tan-1((8/15 + 3/4)/(1 − 8/15 × 3/4 )) = tan-1(((8(4) + 3(15) )/(15 × 4) )/( (15 × 4 − 8 × 3)/(15 × 4) )) = tan-1(((32 + 45 )/(15 × 4) )/( (60 − 24)/(15 × 4) )) = tan-1((32 + 45)/(15 × 4) × (15 × 4)/(60 −24)) = tan-1(77/36) = R.H.S Hence, sin-1 8/17 + sin-1 3/5 = tan-1 77/36 Hence proved

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