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Last updated at April 16, 2024 by Teachoo
Ex 3.3, 8 Prove that cosβ‘γ (Ο + π₯) cosβ‘γ(β π₯)γ γ/(sinβ‘γ (Ο β π₯)γ cosβ‘γ"(" Ο/2 " + " π₯")" γ ) = cot2 π₯ Solving L.H.S. πππβ‘γ(π + π) γπππ γβ‘γ(βπ)γ γ/(πππβ‘(π β π) πππβ‘γ"(" π /π " + " π")" γ ) Putting Ο = 180Β° = πππ β‘γ(180Β° + π₯) γπππ γβ‘γ(βπ₯)γ γ/(π ππβ‘(180Β° β π₯) πππ β‘γ"(" 90Β° "+ " π₯")" γ ) Using cos (180Β° + x) = βcos x cos (βx) = cos x sin (180Β° β x) = sin x & cos(90Β° + x) = βsin x = (βπππβ‘γπ Γ πππβ‘π γ)/((πππβ‘γπ) Γγ(βπππγβ‘π) γ ) = (βπππ 2π₯)/(βπ ππ2π₯) = cot2x = R.H.S. Hence proved