Miscellaneous
Misc. 2 Important
Misc. 3 Deleted for CBSE Board 2022 Exams You are here
Misc. 4 Important Deleted for CBSE Board 2022 Exams
Misc. 5 Deleted for CBSE Board 2022 Exams
Misc. 6 Deleted for CBSE Board 2022 Exams
Misc. 7 Important Deleted for CBSE Board 2022 Exams
Misc. 8 Important Deleted for CBSE Board 2022 Exams
Misc. 9 Important Deleted for CBSE Board 2022 Exams
Misc. 10 Important Deleted for CBSE Board 2022 Exams
Misc. 11 Important Deleted for CBSE Board 2022 Exams
Misc 12 Important Deleted for CBSE Board 2022 Exams
Misc. 13 Important Deleted for CBSE Board 2022 Exams
Misc. 14 Deleted for CBSE Board 2022 Exams
Misc 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
Misc 16 (MCQ) Important Deleted for CBSE Board 2022 Exams
Misc 17 (MCQ) Deleted for CBSE Board 2022 Exams
Miscellaneous
Last updated at May 12, 2021 by Teachoo
Misc 3 Prove 2 sin-1 3/5 = tan-1 24/7 We need to convert LHS in form tan-1 Converting sin-1 (π/π) to tan-1 Let x = sin-1 (3/5) sin x = 3/5 Now, cos x = β(1βπ ππ2 π₯) = β(1 β (3/5)^2 ) = β(1 β 9/25) = β((25 β 9)/25) = β(16/25) = 4/5 Thus, tan x = sinβ‘π₯/cosβ‘π₯ tan x = (3/5)/(4/5) tan x = 3/4 x = tanβ1 π/π Solving L.H.S 2 sinβ1 π/π = 2x = 2 tan-1 (3/4) Using 2tan-1 x = tan-1 (ππ/(π β ππ)) = tan-1 (2(3/4)/(1 β (3/4)2)) = tan-1 ((3/2)/(1 β 9/16)) = tan-1 ((3/2)/( (16 β 9)/16)) = tan-1 ((3/2)/( 7/16)) = tan-1 (3/2Γ16/7) = tan-1 (ππ/π) = R.H.S. Hence L.H.S. = R.H.S Hence proved