[SQP] If cosθ + sinθ = 1 , then prove that cosθ - sinθ = ±1

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

Question 29 - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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