Last updated at May 12, 2021 by Teachoo

Transcript

Example 11 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−sin2𝑎 ) =√(1−(12/13)^2 ) " =" √(25/169) "=" 5/13 Now, tan a = sin𝑎/cos𝑎 = (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 = sin𝑏/cos𝑏 = (3/5)/(4/5) = 3/5 × 5/4 = 3/4 We know that tan (a + b) = 𝒕𝒂𝒏〖𝒂 +〖 𝒕𝒂𝒏〗〖𝒃 〗 〗/(𝟏 − 𝒕𝒂𝒏〖𝒂 𝒕𝒂𝒏𝒃 〗 ) 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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Example 4 Deleted for CBSE Board 2022 Exams

Example 5 Important Deleted for CBSE Board 2022 Exams

Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Deleted for CBSE Board 2022 Exams

Example 8 Deleted for CBSE Board 2022 Exams

Example 9 Important

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11 Important Deleted for CBSE Board 2022 Exams You are here

Example 12 Important Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.