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Ex 8.1, 7 - If cot = 7/8, evaluate (i) (1 + sin) (1 - sin) - Finding values of expressions

Ex 8.1, 7 - Chapter 8 Class 10 Introduction to Trignometry - Part 2
Ex 8.1, 7 - Chapter 8 Class 10 Introduction to Trignometry - Part 3 Ex 8.1, 7 - Chapter 8 Class 10 Introduction to Trignometry - Part 4

Ex 8.1, 7 - Chapter 8 Class 10 Introduction to Trignometry - Part 5

 

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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 = 1/(7/8) 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 (Hypotenuse)2 = (Height)2 + (Base)2 AC2 = AB2 + BC2 AC2 = (7x)2 + (8x)2 AC2 = 49x2 + 64x2 AC2 = 113x2 AC = √113𝑥2 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 ) Here, We have to evaluate ((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 − 8^2/(√113)2))/((1 − 72/(√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 )

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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.