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
CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
Question 2
Question 3
Question 4 Important
Question 5 Important
Question 6
Question 7
Question 8 Important
Question 9 Important
Question 10
Question 11 Important
Question 12
Question 13 Important
Question 14
Question 15 Important
Question 16
Question 17 Important
Question 18
Question 19 Important
Question 20
Question 21 Important
Question 22 Important
Question 23 Important
Question 24
Question 25 Important
Question 26
Question 27
Question 28 Important You are here
Question 29 Important
Question 30
Question 31 Important
Question 32 Important
Question 33
Question 34 Important
Question 35
Question 36 Important
Question 37 Important
Question 38
Question 39 Important
Question 40
Question 41 (Case Based Question) Important
Question 42 (Case Based Question)
Question 43 (Case Based Question) Important
Question 44 (Case Based Question)
Question 45 (Case Based Question)
Question 46 (Case Based Question)
Question 47 (Case Based Question) Important
Question 48 (Case Based Question)
Question 49 (Case Based Question) Important
Question 50 (Case Based Question) Important
CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
Last updated at Dec. 16, 2024 by Teachoo
Question 28 If 1+ sin2𝛼 = 3 sin𝛼 cos𝛼, then values of cot𝛼 are (a) −1, 1 (b) 0, 1 (c)1, 2 (d) −1, −1 Now, 1 + sin^2𝛼 = 3 sin𝛼 cos𝛼 Putting 〖𝒔𝒊𝒏〗^𝟐𝜶 + 〖𝒄𝒐𝒔〗^𝟐𝜶 = 1 〖𝒔𝒊𝒏〗^𝟐𝜶 + 〖𝒄𝒐𝒔〗^𝟐𝜶 + sin^2𝛼 = 3 sin𝛼 cos𝛼 2 〖𝑠𝑖𝑛〗^2𝛼 + 〖𝑐𝑜𝑠〗^2𝛼 = 3 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 2 〖𝑠𝑖𝑛〗^2𝛼 + 〖𝑐𝑜𝑠〗^2𝛼 − 3 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 = 0 𝟐 〖𝒔𝒊𝒏〗^𝟐𝜶 − 3 𝒔𝒊𝒏𝜶 𝒄𝒐𝒔𝜶 + 〖𝒄𝒐𝒔〗^𝟐𝜶 = 0 2 〖𝑠𝑖𝑛〗^2𝛼 − 2 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 − 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 + 〖𝑐𝑜𝑠〗^2𝛼 = 0 2 𝑠𝑖𝑛𝛼 "(" 𝑠𝑖𝑛𝛼 − 𝑐𝑜𝑠𝛼) − 𝑐𝑜𝑠𝛼 "(" 𝑠𝑖𝑛𝛼 − 𝑐𝑜𝑠𝛼) = 0 "(" 𝟐 𝒔𝒊𝒏𝜶 − 𝒄𝒐𝒔𝜶) "(" 𝒔𝒊𝒏𝜶 − 𝒄𝒐𝒔𝜶) = 0 𝟐 𝒔𝒊𝒏𝜶 − 𝒄𝒐𝒔𝜶 = 0 2 𝑠𝑖𝑛𝛼 = 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛼 "= " 2 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼/sin𝛼 "= " 2 𝐜𝐨𝐭 𝜶 "= " 𝟐 𝒔𝒊𝒏𝜶 − 𝒄𝒐𝒔𝜶 = 0 𝑠𝑖𝑛𝛼 = 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛼 "= " 2 𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼/sin𝛼 "= " 1 𝐜𝐨𝐭 𝜶 "= " 𝟏 So, the correct answer is (c)