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Last updated at June 5, 2023 by Teachoo

Misc 16 If cosβ‘π¦=π₯ cosβ‘(π + π¦), with cosβ‘π β Β± 1, prove that ππ¦/ππ₯ = (γπππ γ^2 (π + π¦))/sinβ‘γπ γ Given cosβ‘π¦ = π₯ cosβ‘(π + π¦) cosβ‘π¦/(cosβ‘(π + π¦)) = π₯ π = πππβ‘π/(πππβ‘(π + π)) Differentiating π€.π.π‘.π₯. π(π₯)/ππ₯ = π/ππ₯ (cosβ‘π¦/cosβ‘(π + π¦) ) 1 = π/ππ₯ (cosβ‘π¦/cosβ‘(π + π¦) ) . ππ¦/ππ¦ 1 = π/ππ¦ (cosβ‘π¦/cosβ‘(π + π¦) ) . ππ¦/ππ₯ 1 = ((π(cosβ‘π¦ )/ππ¦ . γ cosγβ‘(π + π¦) β π(γ cosγβ‘(π + π¦) )/ππ¦ . cosβ‘π¦)/(cosβ‘(π + π¦) )^2 ) . ππ¦/ππ₯ 1 = ((βsinβ‘π¦ . γ cosγβ‘(π + π¦) β(γβsin γβ‘(π + π¦) ) π(π + π¦)/ππ¦ . cosβ‘π¦)/γcos^2 γβ‘(π + π¦) ) . ππ¦/ππ₯ Using quotient rule As (π’/π£)β² = (π’^β² π£ β π£^β² π’)/π£^2 where u = cos y & v = cos (π + y) 1 = ((βsinβ‘π¦ . γ cosγβ‘(π + π¦) + γsin γβ‘(π + π¦) (0 + 1) . cosβ‘π¦)/γcos^2 γβ‘(π + π¦) ) . ππ¦/ππ₯ 1 = ((sinβ‘(π + π¦) . γ cosγβ‘π¦ β γcos γβ‘(π + π¦) . sinβ‘π¦)/γcos^2 γβ‘(π + π¦) ) . ππ¦/ππ₯ 1 = πππβ‘((π + π) β π)/γcos^2 γβ‘(π + π¦) . ππ¦/ππ₯ 1 = sinβ‘(π)/γcos^2 γβ‘(π + π¦) . ππ¦/ππ₯ γcos^2 γβ‘(π + π¦)/sinβ‘(π) = ππ¦/ππ₯ π π/π π = γγπππγ^π γβ‘(π + π)/πππβ‘(π) We know that πππβ‘(π βπ)=π ππβ‘π₯ γ πππ γβ‘π¦βγπππ γβ‘π₯ . π ππβ‘π¦