Evaluating using Trignometric Identities
Last updated at April 16, 2024 by Teachoo
Example 11 Prove that cotβ‘γπ΄ β cosβ‘π΄ γ/cotβ‘γπ΄ + cosβ‘π΄ γ =(πππ ππ π΄ β 1)/(πππ ππ π΄ + 1) Taking L.H.S cotβ‘γπ΄ βγ cosγβ‘π΄ γ/cotβ‘γπ΄ +γ cosγβ‘π΄ γ Writing everything in terms of sin A and cos A = (cosβ‘γ π΄γ/sinβ‘γ π΄γ βγ cosγβ‘γ π΄γ)/(cosβ‘γ π΄γ/sinβ‘γ π΄γ +γ cosγβ‘γ π΄γ ) = cosβ‘γ π΄ βγ cosγβ‘γ π΄ sinβ‘γ π΄γ γ γ/(sinβ‘γ π΄γ/(cosβ‘γ π΄ + cosβ‘γ π΄ sinβ‘γ π΄γ γ γ/sinβ‘γ π΄γ )) = ( (πππβ‘γ π¨γ β πππβ‘γ π¨γ sinβ‘γ π΄ γ))/((πππβ‘γ π¨γ + πππβ‘γ π¨γ sinβ‘γ π΄ γ)) = (πππβ‘γ π¨γ (1 β sinβ‘γ π΄ γ))/(πππβ‘γ π¨γ (1 + sinβ‘γ π΄ γ)) = ( (1 β sinβ‘γ π΄ γ))/( (1 + sinβ‘γ π΄ γ)) Dividing sin A on numerator and denominator = ( ((1 β sinβ‘γ π΄ γ))/(πππ π¨))/( ((1 + sinβ‘γ π΄ γ))/(πππ π¨ )) = ( 1/(π ππ π΄) β (π ππ π΄)/(π ππ π΄))/(1/(π ππ π΄) + (π ππ π΄)/(π ππ π΄)) = ( 1/(π ππ π΄) β 1)/(1/(π ππ π΄) + 1) = (πππππ π¨ β π)/(πππππ π¨ + π) = R.H.S. So, L.H.S = R.H.S Hence proved