If sin πœƒ + cos πœƒ = √2 cos πœƒ, (πœƒ ≠ 90°) then the value of tan πœƒ  is

(a) √2 − 1  (b) √2 + 1  (c) √2  (d) − √2

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Question 6 If sin πœƒ + cos πœƒ = √2 cos πœƒ, (πœƒ β‰  90Β°) then the value of tan πœƒ is (a) √2 βˆ’ 1 (b) √2 + 1 (c) √2 (d) βˆ’ √2 Given sin πœƒ + cos πœƒ = √2 cos πœƒ Dividing both sides by cos ΞΈ (sinβ‘πœƒ + cosβ‘πœƒ)/cosβ‘πœƒ = (√2 cosβ‘πœƒ)/cosβ‘πœƒ sinβ‘πœƒ/cosβ‘πœƒ +cosβ‘πœƒ/cosβ‘πœƒ =√2 tan πœƒ+1=√2 tan πœƒ=√2βˆ’1 ∴ Option (a) is correct

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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.