Ex 8.1, 8 - Chapter 8 Class 10 Introduction to Trignometry

Transcript

Ex 8.1, 8 If 3 cot A = 4, check whether ((1 − 𝑡𝑎𝑛2𝐴))/((1 + 𝑡𝑎𝑛2𝐴))= cos2 A – sin2A or not. Given 3 cot A = 4 cot A = 𝟒/𝟑 So, tan A = 1/cot⁡𝐴 tan A = 1/((4/3) ) tan A = 𝟑/𝟒 Now, tan A = 3/4 (𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 ∠𝑨)/(𝑺𝒊𝒅𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 ∠𝑨) = 𝟑/𝟒 𝑩𝑪/𝑨𝑩 = 𝟑/𝟒 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 = √(25"x2" ) AC = 5x Now, sin 𝑨 = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝐴)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝐵𝐶/𝐴𝐶 = 3𝑥/5𝑥 = 𝟑/𝟓 Similarly, cos A = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝐴)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝐴𝐵/𝐴𝐶 = 4𝑥/5𝑥 = 𝟒/𝟓 We have to check whether , (1 − 𝑡𝑎𝑛2 𝐴)/(1 + 𝑡𝑎𝑛2 𝐴 ) = cos2 A – sin2 A (𝟏 − 𝒕𝒂𝒏𝟐 𝑨)/(𝟏 + 𝒕𝒂𝒏𝟐 𝑨 ) Putting tan A = 3/4 = (𝟏 − (𝟑/𝟒)^𝟐)/(𝟏 + (𝟑/𝟒)^𝟐 ) = ((1 − 9/16))/((1 + 9/16) ) = (((16 − 9)/16))/(((16 + 9)/16) ) = (16 − 9)/(16 + 9) = 𝟕/𝟐𝟓 cos2 A – sin2 A Putting cos A = 4/5 & sin A = 3/5 = (𝟒/𝟓)^𝟐−(𝟑/𝟓)^𝟐 = 16/25 − 9/25 = (16 − 9)/25 = 𝟕/𝟐𝟓 Since L.H.S = R.H.S Hence proved