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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 8.1, 7 If cot θ = 7/8 , evaluate : (i) ((1 + 𝑠𝑖𝑛⁡𝜃)(1 − 𝑠𝑖𝑛⁡𝜃))/((1 + 𝑐𝑜𝑠⁡𝜃)(1 − 𝑐𝑜𝑠⁡𝜃)) We will first calculate the value of sin θ & cos θ Now, tan θ = 1/cot⁡𝜃 tan 𝛉 = 𝟖/𝟕 We can write tan 𝜃 = 8/7 (𝒔𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 ∠𝜽)/(𝒔𝒊𝒅𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 ∠𝜽)=𝟖/𝟕 𝐵𝐶/𝐴𝐵=8/7 Let BC = 8x & AB = 7x We find AC using Pythagoras Theorem In right triangle ABC Using Pythagoras theorem Hypotenuse2 = Height2 + Base2 AC2 = AB2 + BC2 AC2 = (7x)2 + (8x)2 AC2 = 49x2 + 64x2 AC2 = 113x2 AC = √113𝑥2 AC = √𝟏𝟏𝟑 x AC = √𝟏𝟏𝟑 x Now, we need to find sin θ and cos θ sin 𝜽 = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝜃)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝐵𝐶/𝐴𝐶 = 8𝑥/(√113 𝑥) = 𝟖/(√𝟏𝟏𝟑 ) cos 𝜽 = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠𝜃)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝐴𝐵/𝐴𝐶 = 7𝑥/(√113 𝑥) = 𝟕/(√𝟏𝟏𝟑 ) We have to find ((𝟏 + 𝐬𝐢𝐧⁡𝜽 ) (𝟏 −〖 𝐬𝐢𝐧〗⁡𝜽 ))/((𝟏 + 𝐜𝐨𝐬⁡𝜽 ) (𝟏 −〖 𝐜𝐨𝐬〗⁡𝜽 ) ) Using (a + b) (a – b) = a2 – b2 = ( (12 − 𝑠𝑖𝑛2 𝜃))/( (12 − 𝑐𝑜𝑠2𝜃)) = ( (1 − 𝑠𝑖𝑛2 𝜃))/( (1 − 𝑐𝑜𝑠2𝜃)) Putting sin 𝜽 = 𝟖/(√𝟏𝟏𝟑 ) & cos θ = 𝟕/(√𝟏𝟏𝟑 ) = ((1 − (8/(√113 ))^2 ))/((1 − (7/(√113 ))^2 ) ) = ((𝟏 − 𝟔𝟒/𝟏𝟏𝟑))/((𝟏 − 𝟒𝟗/𝟏𝟏𝟑) ) = (((113 − 64)/113))/(((113 − 49)/113) ) = (113 − 64)/(113 − 49 ) = 𝟒𝟗/(𝟔𝟒 ) Hence, ((𝟏 + 𝒔𝒊𝒏⁡𝜽)(𝟏 − 𝒔𝒊𝒏⁡𝜽))/((𝟏 + 𝒄𝒐𝒔⁡𝜽)(𝟏 − 𝒄𝒐𝒔⁡𝜽)) = 𝟒𝟗/(𝟔𝟒 ) Ex 8.1, 7 If cot θ = 7/8 , evaluate : (ii) cot2 θ Given cot θ = 7/8 So, cot2 θ = (7/8)^2 = 72/82 = 𝟒𝟗/𝟔𝟒

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.