Question 5 - Examples - Chapter 2 Class 12 Inverse Trigonometric Functions

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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