Example 11 - Show sin-1 12/13 + cos-1 4/5 + tan-1 63/16 = pi

Example 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Example 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3
Example 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4
Example 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 5

 

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Question 5 Show that sin−1 12/13 + cos−1 4/5 + tan−1 63/16 = π Let a = sin−1 12/13 & b = cos−1 4/5 We convert sin−1 & cos−1 to tan–1 & then use tan (a + b) formula Let a = sin−1 𝟏𝟐/𝟏𝟑 sin a = 12/13 We know that cos a = √(1−sin2⁡𝑎 ) =√(1−(12/13)^2 ) " =" √(25/169) "=" 5/13 Now, tan a = sin⁡𝑎/cos⁡𝑎 = (12/13)/(5/13) = 12/13 × 13/5 = 12/5 Let b = cos−1 𝟒/𝟓 cos b = 4/5 We know that sin b = √("1 – cos2 b " ) = √("1 − " (4/5)^2 ) = √(9/25) = 3/5 Now, tan b = sin⁡𝑏/cos⁡𝑏 = (3/5)/(4/5) = 3/5 × 5/4 = 3/4 We know that tan (a + b) = 𝒕𝒂𝒏⁡〖𝒂 +〖 𝒕𝒂𝒏〗⁡〖𝒃 〗 〗/(𝟏 − 𝒕𝒂𝒏⁡〖𝒂 𝒕𝒂𝒏⁡𝒃 〗 ) Putting tan a = 12/5 and tan b = 3/4 = (12/5 + 3/4)/(1 − 12/5 × 3/4) = ((48 +15 )/20)/((20 − 36)/20) = (63/20)/((−16)/20) = 63/20 × 20/(−16) = (−𝟔𝟑)/( 𝟏𝟔) Hence, tan (a + b) = (−63)/16 a + b = tan-1 (( −63)/16) Putting value of a & b sin−1 𝟏𝟐/𝟏𝟑 + cos−1 𝟒/𝟓 = tan−1 (( −𝟔𝟑)/𝟏𝟔) Solving L.H.S sin−1 12/13 + cos−1 4/5 + tan−1 63/16 Putting values = tan−1 ((−63)/16) + tan−1 (63/16) Using tan−1x + tan−1y = tan−1((𝒙 + 𝒚)/(𝟏 − 𝒙𝒚)) Putting x = (−63)/16 and y by = 63/16 = tan−1(((− 63)/16 + 63/16)/(1 − (− 63)/16 × 63/16)) = tan−1(0/(1+ (( 63)/16)^2 )) = tan−1 0 = π = R.H.S Hence L.H.S = R.H.S Hence proved As tan 180° = 0 tan π = 0 π = tan−1 0 i.e. tan−1 0 = π

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo