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Misc 15 - Using properties of determinants, prove |sin cos

Misc. 15 - Chapter 4 Class 12 Determinants - Part 2
Misc. 15 - Chapter 4 Class 12 Determinants - Part 3

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Misc 15 Using properties of determinants, prove that: |■8(sin⁡α&cos⁡α&cos⁡〖(α+δ)〗@sin⁡β&cos⁡β&cos⁡〖(β+δ)〗@sin⁡γ&cos⁡γ&cos⁡〖(γ+δ)〗 )| = 0 Let ∆ = |■8(sin⁡α&cos⁡α&cos⁡〖(α+δ)〗@sin⁡β&cos⁡β&cos⁡〖(β+δ)〗@sin⁡γ&cos⁡γ&cos⁡〖(γ+δ)〗 )| Using cos (x + y) = cos x cos y – sin x sin y = |■8(sin⁡α&cos⁡α&cos𝛼 cos⁡〖δ −sin⁡〖𝛼 sin⁡𝛿 〗 〗@sin⁡β&cos⁡β&cos⁡𝛽 cos⁡〖𝛿−sin⁡〖𝛽 sin⁡𝛿 〗 〗@sin⁡γ&cos⁡γ&cosγcos 𝛿 −sin⁡〖γ sin⁡𝛿 〗 )| Expressing elements of 2nd row as sum of two elements = |■8(sin⁡α&cos⁡α&cos 𝛼 cos⁡〖δ 〗@sin⁡β&cos⁡β&cos⁡𝛽 cos⁡𝛿@sin⁡γ&cos⁡γ&cos γ cos 𝛿 )| + |■8(sin⁡α&cos⁡α&−sin⁡〖𝛼 sin⁡𝛿 〗@sin⁡β&cos⁡β&−sin⁡〖𝛽 sin⁡𝛿 〗@sin⁡γ&cos⁡γ&−sin⁡〖γ sin⁡𝛿 〗 )| Using Property : If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms ,then the determinant is expressed as a sum of two (or more) determinants. Taking cos 𝛿 common from C3 = cos⁡〖δ 〗 |■8(sin⁡α&cos⁡α&cos 𝛼@sin⁡β&cos⁡β&cos⁡𝛽@sin⁡γ&cos⁡γ&cos γ )| + (−sin⁡𝛿) |■8(sin⁡α&cos⁡α&sin⁡𝛼@sin⁡β&cos⁡β&sin⁡𝛽@sin⁡γ&cos⁡γ&sin⁡γ )| = cos⁡〖δ 〗(0) + (−sin⁡δ) (0) = 0 = R.H.S Hence proved Using Property: If any two row or column are identical, then value of determinant is zero

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.