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Question 29 If cosθ + sinθ = 1 , then prove that cosθ - sinθ = ±1Given cos⁡θ+sin⁡θ=1 Squaring both sides (cos⁡θ+sin⁡θ )^2=1^2 〖𝒄𝒐𝒔〗^𝟐⁡𝜽+〖𝒔𝒊𝒏〗^𝟐⁡𝜽+2 𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃=1 Putting 〖𝑐𝑜𝑠〗^2⁡𝜃+〖𝑠𝑖𝑛〗^2⁡𝜃 = 1 𝟏+2 𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃=1 2 𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃=1−1 2 𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃=0 𝒄𝒐𝒔⁡𝜽 𝒔𝒊𝒏⁡𝜽=𝟎 Since we need to find cosθ - sinθ , lets square it also Let cos θ – sin θ = x Squaring both sides (cos⁡θ−sin⁡θ )^2=𝑥^2 〖𝒄𝒐𝒔〗^𝟐⁡𝜽+〖𝒔𝒊𝒏〗^𝟐⁡𝜽−𝟐 𝒄𝒐𝒔⁡𝜽 𝒔𝒊𝒏⁡𝜽=𝒙^𝟐 Putting 〖𝑐𝑜𝑠〗^2⁡𝜃+〖𝑠𝑖𝑛〗^2⁡𝜃 = 1 𝟏−2 𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃=𝑥^2 From (1), putting 𝒄𝒐𝒔⁡𝜽 𝒔𝒊𝒏⁡𝜽=𝟎 1−2 × 𝟎=𝑥^2 1=x^2 𝐱^𝟐=𝟏 x=±1 Thus, 𝒄𝒐𝒔⁡𝜽−𝒔𝒊𝒏⁡𝜽=±𝟏 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.