Misc 2 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Solving by simplifying det.
Question 6 Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 4 Important Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 14 Important Deleted for CBSE Board 2025 Exams
Solving by simplifying det.
Last updated at April 16, 2024 by Teachoo
Misc 2 (Method 1) Evaluate |■8(cos〖α cosβ〗&cos〖α sinβ〗&−sinα@−sinβ&cosβ&0@sinα cosβ&sin〖α sinβ 〗&cosα )| |■8(cos〖α cosβ〗&cos〖α sinβ〗&−sinα@−sinβ&cosβ&0@sinα cosβ&sin〖α sinβ 〗&cosα )| Expanding Determinant along C1 = cos α cos β |■8(cos𝛽&0@sin〖𝛼 sin𝛽 〗&cos𝛼 )| – cos α sin β |■8(〖−sin〗𝛽&0@sin〖𝛼 𝑐𝑜𝑠𝛽 〗&cos𝛼 )| – sin α |■8(sin𝛽&cos𝛽@sin〖𝛼 𝑐𝑜𝑠𝛽 〗&sin〖𝛼 sin𝛽 〗 )| = cos α cos β (cos β cos α – 0) – cos α sin β (– sin β cos α – 0) – sin α (– sin2 β sin α – cos2 β sin α) = cos α cos β ( cos β cos α) – cos α sin β ( – sin β cos α) + sin2 α sin2 β + cos2 β sin2 α = cos2 α cos2 β + cos2 α sin2 β + sin2 α ( sin2 β + cos2 β) = cos2 α (1) + sin2 α (1) = cos 2 α + sin2α = 1 Misc 2 (Method 2) Evaluate |■8(cos〖α cosβ〗&cos〖α sinβ〗&−sinα@−sinβ&cosβ&0@sinα cosβ&sin〖α sinβ 〗&cosα )| Let ∆ = |■8(cos〖α cosβ〗&cos〖α sinβ〗&−sinα@−sinβ&cosβ&0@sinα cosβ&sin〖α sinβ 〗&cosα )| = |■8(𝐜𝐨𝐬〖𝛂 〗 cosβ&𝐜𝐨𝐬〖𝛂 〗 sin β&(−𝒄𝒐𝒔𝜶)/cos𝛼 sinα@−sin β&cos β&0@𝐬𝐢𝐧𝛂 cosβ&𝐬𝐢𝐧𝛂 sinβ&𝒔𝒊𝒏𝜶/sin𝛼 〖.cos〗α )| Taking common cos α from R1 & sin α from R3 = cos α.sin β |■8(cos𝛽&sin𝛽&−tan𝛼@−sinβ&cosβ&0@cosβ&sin𝛽&co𝑡α )| Applying R1→ R1 – R3 = cos α.sin β |■8(cos〖𝛽 〗−𝐜𝒐𝒔𝜷&sin〖𝛽−𝒔𝒊𝒏𝜷 〗&−tan〖𝛼−𝒄𝒐𝒕𝜶 〗@−sinβ&cosβ&0@cosβ&sin𝛽&co𝑡α )| Expanding determinant along R1 = cos α . sin α(0|■8(cos𝛽&0@sin𝛽&cot𝛼 )|−0|■8(〖−sin〗𝛽&0@cos𝛽&cot𝛼 )|−〖(tan〗〖𝛼−cot〖𝛼)|■8(sin𝛽&cos𝛽@cos𝛽&sin𝛽 )|〗 〗 ) = cos α . sin α(0−0−〖(tan〗〖𝛼−cot〖𝛼)|■8(sin𝛽&cos𝛽@cos𝛽&sin𝛽 )|〗 〗 ) = cos α sin α (– (tan α + cotα ) ( – sin2 β – cos2 β )) = cos α sin α ( tan + cot α ) (sin2 β + cos2 β ) = cos α sin α (tan α + cot α ) (1) = cos α sin α (sin𝛼/cos𝛼 + cos𝛼/sin𝛼 ) = cos α sin α ((𝒔𝒊𝒏𝟐 𝜶 + 𝒄𝒐𝒔𝟐 𝜶)/cos〖𝛼 sin𝛼 〗 ) = cos α sin α (𝟏/cos〖𝛼 sin𝛼 〗 ) = 1