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If 1+sin^2⁡θ=3sin⁡θcos⁡θ, then prove that tan⁡θ=1 or 1/2

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1+sin^2⁡𝜃=3 sin⁡〖𝜃 cos⁡𝜃 〗 Dividing by 〖𝒄𝒐𝒔〗^𝟐 𝜽 on both sides 1/(〖𝑐𝑜𝑠〗^2 𝜃) + sin^2⁡𝜃/(〖𝑐𝑜𝑠〗^2 𝜃) = (3 sin⁡〖𝜃 cos⁡𝜃 〗)/(〖𝑐𝑜𝑠〗^2 𝜃) 〖𝒔𝒆𝒄〗^𝟐 𝜽 + 〖𝑡𝑎𝑛〗^2 𝜃 = 3 tan 𝜃 Putting 〖𝒔𝒆𝒄〗^𝟐 𝜽 = 1 + 〖𝑡𝑎𝑛〗^2 𝜃 1 + 〖𝒕𝒂𝒏〗^𝟐 𝜽+ 〖𝑡𝑎𝑛〗^2 𝜃 = 3 tan 𝜃 1 + 2 〖𝑡𝑎𝑛〗^2 𝜃 = 3 tan 𝜃 Putting tan 𝜽 = x 1 + 2𝑥^2 = 3x 2𝒙^𝟐 – 3x + 1 = 0 Solving by Spitting the middle term 2𝑥^2 – 2x – x + 1 = 0 2𝑥(x – 2) -1(x – 2) = 0 (2x – 1)(x - 2) = 0 Therefore, x = 𝟏/𝟐 , x = 2 Replacing x = tan 𝜃 back we get, tan 𝜽 = 𝟏/𝟐 or 2 Hence proved

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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.