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Given that sin πœƒ = a/b, then tan πœƒ  is equal to

(a) b/√(a 2 + b 2 )  (b) b/√(b 2 - a 2 )  (c) a/√(a 2 + b 2 )   (d) b/√(b^ 2 + a 2 )

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Question 26 Given that sin πœƒ = π‘Ž/𝑏, then tan πœƒ is equal to (a) 𝑏/√(π‘Ž^2 + 𝑏^2 ) (b) 𝑏/√(𝑏^2 βˆ’ π‘Ž^2 ) (c) π‘Ž/√(π‘Ž^2 + 𝑏^2 ) (d) 𝑏/√(𝑏^2 + π‘Ž^2 ) Let Ξ” ABC be the right angled triangle Given sin πœƒ = 𝒂/𝒃 Since sin πœƒ = π‘Άπ’‘π’‘π’π’”π’Šπ’•π’†/π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† Therefore, BC = a, AC = b Finding side AB By Pythagoras theorem AC2 = AB2 + BC2 b2 = AB2 + a2 b2 βˆ’ a2 = AB2 AB2 = b2 βˆ’ a2 AB = √(𝒃^πŸβˆ’π’‚^𝟐 ) Now, tan πœƒ = 𝐡𝐢/𝐴𝐡 = 𝒂/√(𝒃^𝟐 βˆ’ 𝒂^𝟐 ) So, the correct answer is (d)

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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.