Learn all Concepts of Chapter 8 Class 10 (with VIDEOS). Check - Trigonometry - Class 10

Last updated at July 13, 2020 by Teachoo

Transcript

Ex 8.1, 7 If cot ΞΈ = 7/8 , evaluate : (i) ((1 + π ππβ‘π)(1 β π ππβ‘π))/((1 + πππ β‘π)(1 β πππ β‘π)) We will first calculate the value of sin ΞΈ & cos ΞΈ Now, tan ΞΈ = 1/cotβ‘π tan ΞΈ = 1/(7/8) tan ΞΈ = 8/7 (π πππ πππππ ππ‘π β π)/(π πππ ππππππππ‘ β π)=8/7 π΅πΆ/π΄π΅=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 = β113π₯2 AC = β113 x Now, we need to find sin ΞΈ and cos ΞΈ sin π = (π πππ πππππ ππ‘π π‘π β π)/π»π¦πππ‘πππ’π π = π΅πΆ/π΄πΆ = 8π₯/(β113 π₯) = 8/(β113 ) Here, We have to evaluate ((1 + sinβ‘π ) (1 βγ sinγβ‘π ))/((1 + cosβ‘π ) (1 βγ cosγβ‘π ) ) Using (a + b) (a β b) = a2 β b2 = ( (12 β π ππ2 π))/( (12 β πππ 2π)) = ( (1 β π ππ2 π))/( (1 β πππ 2π)) Putting sin π = 8/(β113 ) & cos ΞΈ = 7/(β113 ) = ((1 β (8/(β113 ))^2 ))/((1 β (7/(β113 ))^2 ) ) = ((1 β 8^2/(β113)2))/((1 β 72/(β113)2) ) = ((1 β 64/113))/((1 β 49/113) ) = (((113 β 64)/113))/(((113 β 49)/113) ) = (113 β 64)/(113 β 49 ) = 49/(64 ) Hence, ((1 + π ππβ‘π)(1 β π ππβ‘π))/((1 + πππ β‘π)(1 β πππ β‘π)) = 49/(64 )

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.