You are here
Β
Misc 5 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at August 5, 2026 by Teachoo
Changing of trignometric variables and then applying formula
Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise
- Definition
- Finding principal value
- Formulae based
-
Changing of trignometric variables and then applying formula
- Not clear how to approach
Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 3 questions, selected from your answers, mistakes, and progress.
Transcript
Misc 5 Prove cos-1 4/5 + cos-1 12/13 = cos-1 33/65 Let a = cos-1 4/5 and b = cos-1 12/13 Finding sin a, cos a Let a = cos-1 4/5 cos a = 4/5 We know that sin a = β(1 βcos^2β‘π ) =β(1β(4/5)^2 ) " =" β(9/25) "=" 3/5 Finding sin b, cos b Let b = cos-1 12/13 cos b = 12/13 We know that sin b = β(1 βπππ 2 π) = β(1 β(12/13)^2 ) = β(25/169) = 5/13 Now, cos (a + b) = cos a cos b β sin a sin b cos (a + b) = 4/5 Γ 12/13 β 3/5 Γ 5/13 = 48/65 β 3/13 = (48 β 3(5))/65 = (48 β 15)/65 = ππ/ππ Putting cos a = 4/5 , sin a = 3/5 & cos b = 12/13 , sin b = 5/13 Thus, cos (a + b) = 33/65 a + b = cos-1 (33/65) Putting a = cos-1 4/5 & b = cos-1 12/15 cos-1 π/π + cos-1 ππ/ππ = cos-1 ππ/ππ Which is what we had to prove Hence proved
