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Example 2
Example 3 (i) Important
Example 3 (ii)
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
Last updated at May 12, 2021 by Teachoo
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