If 1+ sin⁡2α  = 3 sin⁡α  cos⁡α, then values of cot⁡α are

(a) −1, 1   (b) 0, 1   (c)1, 2   (d) −1, −1

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Question 28 If 1+ sin⁡2𝛼 = 3 sin⁡𝛼 cos⁡𝛼, then values of cot⁡𝛼 are (a) −1, 1 (b) 0, 1 (c)1, 2 (d) −1, −1 Now, 1 + sin^2⁡𝛼 = 3 sin⁡𝛼 cos⁡𝛼 Putting 〖𝒔𝒊𝒏〗^𝟐⁡𝜶 + 〖𝒄𝒐𝒔〗^𝟐⁡𝜶 = 1 〖𝒔𝒊𝒏〗^𝟐⁡𝜶 + 〖𝒄𝒐𝒔〗^𝟐⁡𝜶 + sin^2⁡𝛼 = 3 sin⁡𝛼 cos⁡𝛼 2 〖𝑠𝑖𝑛〗^2⁡𝛼 + 〖𝑐𝑜𝑠〗^2⁡𝛼 = 3 𝑠𝑖𝑛⁡𝛼 𝑐𝑜𝑠⁡𝛼 2 〖𝑠𝑖𝑛〗^2⁡𝛼 + 〖𝑐𝑜𝑠〗^2⁡𝛼 − 3 𝑠𝑖𝑛⁡𝛼 𝑐𝑜𝑠⁡𝛼 = 0 𝟐 〖𝒔𝒊𝒏〗^𝟐⁡𝜶 − 3 𝒔𝒊𝒏⁡𝜶 𝒄𝒐𝒔⁡𝜶 + 〖𝒄𝒐𝒔〗^𝟐⁡𝜶 = 0 2 〖𝑠𝑖𝑛〗^2⁡𝛼 − 2 𝑠𝑖𝑛⁡𝛼 𝑐𝑜𝑠⁡𝛼 − 𝑠𝑖𝑛⁡𝛼 𝑐𝑜𝑠⁡𝛼 + 〖𝑐𝑜𝑠〗^2⁡𝛼 = 0 2 𝑠𝑖𝑛⁡𝛼 "(" 𝑠𝑖𝑛⁡𝛼 − 𝑐𝑜𝑠⁡𝛼) − 𝑐𝑜𝑠⁡𝛼 "(" 𝑠𝑖𝑛⁡𝛼 − 𝑐𝑜𝑠⁡𝛼) = 0 "(" 𝟐 𝒔𝒊𝒏⁡𝜶 − 𝒄𝒐𝒔⁡𝜶) "(" 𝒔𝒊𝒏⁡𝜶 − 𝒄𝒐𝒔⁡𝜶) = 0 𝟐 𝒔𝒊𝒏⁡𝜶 − 𝒄𝒐𝒔⁡𝜶 = 0 2 𝑠𝑖𝑛⁡𝛼 = 𝑐𝑜𝑠⁡𝛼 𝑐𝑜𝑠⁡𝛼 "= " 2 𝑠𝑖𝑛⁡𝛼 𝑐𝑜𝑠⁡𝛼/sin⁡𝛼 "= " 2 𝐜𝐨𝐭 𝜶 "= " 𝟐 𝒔𝒊𝒏⁡𝜶 − 𝒄𝒐𝒔⁡𝜶 = 0 𝑠𝑖𝑛⁡𝛼 = 𝑐𝑜𝑠⁡𝛼 𝑐𝑜𝑠⁡𝛼 "= " 2 𝑠𝑖𝑛⁡𝛼 𝑐𝑜𝑠⁡𝛼/sin⁡𝛼 "= " 1 𝐜𝐨𝐭 𝜶 "= " 𝟏 So, the correct answer is (c)

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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 and Computer Science at Teachoo