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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Misc 7 Prove tan–1 63/16 = sin–1 5/13 + cos–1 3/5 Let a = sin–1 5/13 , b = cos–1 3/5 Finding tan a & tan b We convert sin–1 & cos–1 to tan–1 & then use tan (a + b) formula Let a = sin–1 𝟓/𝟏𝟑 sin a = 5/13 We know that cos a = √(1 –sin2 𝑎) = √("1 – " (5/13)^2 ) = √(144/169) = 12/13 Now, tan a = (sin 𝑎)/(cos a) = (5/13)/(12/13) = 5/13×13/12 = 5/12 Let b = cos–1 𝟑/𝟓 cos b = 3/5 We know that sin b = √("1 – cos2 b" ) = √("1 –" (3/5)^2 ) =√(16/25) = 4/5 Now, tan b = sin⁡𝑏/cos⁡𝑏 = (4/5)/(3/5) = 4/5×5/3 = 4/3 Now we know that tan (a + b) = 𝑡𝑎𝑛⁡〖𝑎 +〖 𝑡𝑎𝑛〗⁡〖𝑏 〗 〗/(1 − 𝑡𝑎𝑛⁡〖𝑎 𝑡𝑎𝑛⁡𝑏 〗 ) Putting tan a = 5/12 & 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 Thus, 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.