Ex 8.1

Chapter 8 Class 10 Introduction to Trignometry
Serial order wise

### Transcript

Ex 8.1, 8 If 3 cot A = 4, check whether ((1 β π‘ππ2π΄))/((1 + π‘ππ2π΄))= cos2 A β sin2A or not. Given 3 cot A = 4 cot A = π/π So, tan A = 1/cotβ‘π΄ tan A = 1/((4/3) ) tan A = π/π Now, tan A = 3/4 (πΊπππ ππππππππ β π¨)/(πΊπππ ππππππππ β π¨) = π/π π©πͺ/π¨π© = π/π Let BC = 3x & AB = 4x We find AC using Pythagoras theorem In right triangle ABC (Hypotenuse)2 = (Height)2 + (Base)2 (AC)2 = (AB)2 + (BC)2 (AC)2 = (4x)2 + (3x)2 (AC)2 = 16x2 + 9x2 (AC)2 = 25x2 AC = β(25"x2" ) AC = 5x Now, sin π¨ = (π πππ πππππ ππ‘π π‘π β π΄)/π»π¦πππ‘πππ’π π = π΅πΆ/π΄πΆ = 3π₯/5π₯ = π/π Similarly, cos A = (π πππ ππππππππ‘ π‘π π΄)/π»π¦πππ‘πππ’π π = π΄π΅/π΄πΆ = 4π₯/5π₯ = π/π We have to check whether , (1 β π‘ππ2 π΄)/(1 + π‘ππ2 π΄ ) = cos2 A β sin2 A (π β ππππ π¨)/(π + ππππ π¨ ) Putting tan A = 3/4 = (π β (π/π)^π)/(π + (π/π)^π ) = ((1 β 9/16))/((1 + 9/16) ) = (((16 β 9)/16))/(((16 + 9)/16) ) = (16 β 9)/(16 + 9) = π/ππ cos2 A β sin2 A Putting cos A = 4/5 & sin A = 3/5 = (π/π)^πβ(π/π)^π = 16/25 β 9/25 = (16 β 9)/25 = π/ππ Since L.H.S = R.H.S Hence proved