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Question 6 If tanβ‘πœƒ=5/2 then (4 sinβ‘πœƒ+ cosβ‘πœƒ)/(4 sinβ‘πœƒβˆ’ cosβ‘πœƒ) is equal to (A) 11/9 (B) 3/2 (C) 9/11 (D) 4Given tan ΞΈ = πŸ“/𝟐 Now, (4 sinβ‘γ€–πœƒ + cosβ‘πœƒ γ€—)/(4 sinβ‘γ€–πœƒ βˆ’ cosβ‘πœƒ γ€— ) Dividing numerator and denominator by cos ΞΈ = (4 sinβ‘πœƒ/cosβ‘πœƒ + coπ‘ β‘πœƒ/cosβ‘πœƒ )/(4 sinβ‘πœƒ/cosβ‘πœƒ βˆ’ coπ‘ β‘πœƒ/cosβ‘πœƒ ) = (πŸ’ π­πšπ§β‘γ€–πœ½ + πŸγ€—)/(πŸ’ π­πšπ§β‘γ€–πœ½ βˆ’ πŸγ€— ) Putting tan ΞΈ = πŸ“/𝟐 = (4 Γ— 5/2 + 1)/(4 Γ— 5/2 βˆ’ 1) = (10 + 1)/(10 βˆ’ 1) = 𝟏𝟏/πŸ— So, the correct answer is (A)

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