Misc 7 - Prove tan-1 63/16 = sin-1 5/13 + cos-1 3/5 - Miscellaneous

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Misc 7 (Method 1) Prove tan-1 63/16 = sin-1 5/13 + cos-1 3/5 Let a = sin-1 5/13 , b = cos-1 3/5 Let a = sin-1 5/13 sin a = 5/13 We know that cos2 a = 1 – sin2 a cos a = √(1 – sin2 𝑎) = √("1 – " 25/169) = √((169 − 25)/169) = √(144/169) = 12/13 Let b = cos-1 3/5 cos b = 3/5 We know that sin2 b = 1 – cos2 b sin b = √("1 – cos2 b" ) = √("1 –" (3/5)"2" )= √("1 –" 9/25) =√((25 − 9)/25) =√(16/25) = 4/5 Now we know that tan (a + b) = 𝑡𝑎𝑛⁡〖𝑎 +〖 𝑡𝑎𝑛〗⁡〖𝑏 〗 〗/(1 − 𝑡𝑎𝑛⁡〖𝑎 𝑡𝑎𝑛⁡𝑏 〗 ) Putting tan a = 5/12 and tan b = 4/3 tan (a + b) = (5/12 + 4/3)/(1 − 5/12 × 4/3) = ((5 × 3 + 4 × 12)/36)/(1 − 20/36) = ((15 + 48)/36)/((36 − 20)/36) = (63/36)/(16/36) = 63/36 × 36/16 = 63/16 Hence, tan (a + b) = 63/16 a + b = tan-1 (63/16) Putting values of a & b sin-1 5/13 + cos-1 3/5 = tan-1 (63/16) Hence L.H.S = R.H.S Hence Proved

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