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

  1. Chapter 8 Class 10 Introduction to Trignometry
  2. Serial order wise
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Ex 8.1, 7 If cot ΞΈ = 7/8 , evaluate : (i) ((1 + π‘ π‘–π‘›β‘πœƒ)(1 βˆ’ π‘ π‘–π‘›β‘πœƒ))/((1 + π‘π‘œπ‘ β‘πœƒ)(1 βˆ’ π‘π‘œπ‘ β‘πœƒ)) We will first calculate the value of sin ΞΈ & cos ΞΈ Now, tan ΞΈ = 1/cotβ‘πœƒ tan ΞΈ = 1/(7/8) tan ΞΈ = 8/7 (𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ βˆ πœƒ)/(𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ βˆ πœƒ)=8/7 𝐡𝐢/𝐴𝐡=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 = √113π‘₯2 AC = √113 x Now, we need to find sin ΞΈ and cos ΞΈ sin πœƒ = (𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ βˆ πœƒ)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = 𝐡𝐢/𝐴𝐢 = 8π‘₯/(√113 π‘₯) = 8/(√113 ) Here, We have to evaluate ((1 + sinβ‘πœƒ ) (1 βˆ’γ€– sinγ€—β‘πœƒ ))/((1 + cosβ‘πœƒ ) (1 βˆ’γ€– cosγ€—β‘πœƒ ) ) Using (a + b) (a – b) = a2 – b2 = ( (12 βˆ’ 𝑠𝑖𝑛2 πœƒ))/( (12 βˆ’ π‘π‘œπ‘ 2πœƒ)) = ( (1 βˆ’ 𝑠𝑖𝑛2 πœƒ))/( (1 βˆ’ π‘π‘œπ‘ 2πœƒ)) Putting sin πœƒ = 8/(√113 ) & cos ΞΈ = 7/(√113 ) = ((1 βˆ’ (8/(√113 ))^2 ))/((1 βˆ’ (7/(√113 ))^2 ) ) = ((1 βˆ’ 8^2/(√113)2))/((1 βˆ’ 72/(√113)2) ) = ((1 βˆ’ 64/113))/((1 βˆ’ 49/113) ) = (((113 βˆ’ 64)/113))/(((113 βˆ’ 49)/113) ) = (113 βˆ’ 64)/(113 βˆ’ 49 ) = 49/(64 ) Hence, ((1 + π‘ π‘–π‘›β‘πœƒ)(1 βˆ’ π‘ π‘–π‘›β‘πœƒ))/((1 + π‘π‘œπ‘ β‘πœƒ)(1 βˆ’ π‘π‘œπ‘ β‘πœƒ)) = 49/(64 )

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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