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Ex 8.1, 8 If 3 cot A = 4, check whether ((1 βˆ’ π‘‘π‘Žπ‘›2𝐴))/((1 + π‘‘π‘Žπ‘›2𝐴))= cos2 A – sin2A or not. Given 3 cot A = 4 cot A = πŸ’/πŸ‘ So, tan A = 1/cot⁑𝐴 tan A = 1/((4/3) ) tan A = πŸ‘/πŸ’ Now, tan A = 3/4 (π‘Ίπ’Šπ’…π’† π’π’‘π’‘π’π’”π’Šπ’•π’† βˆ π‘¨)/(π‘Ίπ’Šπ’…π’† 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 βˆ π‘¨) = πŸ‘/πŸ’ 𝑩π‘ͺ/𝑨𝑩 = πŸ‘/πŸ’ 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 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ ∠𝐴)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = 𝐡𝐢/𝐴𝐢 = 3π‘₯/5π‘₯ = πŸ‘/πŸ“ Similarly, cos A = (𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ 𝐴)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = 𝐴𝐡/𝐴𝐢 = 4π‘₯/5π‘₯ = πŸ’/πŸ“ We have to check whether , (1 βˆ’ π‘‘π‘Žπ‘›2 𝐴)/(1 + π‘‘π‘Žπ‘›2 𝐴 ) = cos2 A – sin2 A (𝟏 βˆ’ π’•π’‚π’πŸ 𝑨)/(𝟏 + π’•π’‚π’πŸ 𝑨 ) Putting tan A = 3/4 = (𝟏 βˆ’ (πŸ‘/πŸ’)^𝟐)/(𝟏 + (πŸ‘/πŸ’)^𝟐 ) = ((1 βˆ’ 9/16))/((1 + 9/16) ) = (((16 βˆ’ 9)/16))/(((16 + 9)/16) ) = (16 βˆ’ 9)/(16 + 9) = πŸ•/πŸπŸ“ cos2 A – sin2 A Putting cos A = 4/5 & sin A = 3/5 = (πŸ’/πŸ“)^πŸβˆ’(πŸ‘/πŸ“)^𝟐 = 16/25 βˆ’ 9/25 = (16 βˆ’ 9)/25 = πŸ•/πŸπŸ“ Since L.H.S = R.H.S Hence proved

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Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.