Last updated at Dec. 13, 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
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo