CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard

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Question 25 (Choice 2) - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at August 10, 2026 by

Find an acute angle θ when cos⁡θ - sin⁡θ/cos⁡θ  + sin⁡θ  = 1+ √3/1 + √3

[Class 10] Find an acute angle θ when cosθ − sin θ cosθ + sin θ = 1−√3 - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard

part 2 - Question 25 (Choice 2) - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

 

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Question 25 (Choice 2) Find an acute angle θ when (cos⁡θ − sin⁡θ)/(cos⁡θ + sin⁡θ ) = (1 − √3)/(1 + √3)Given (cos⁡θ − sin⁡θ)/(cos⁡θ + sin⁡θ ) = (1 − √3)/(1 + √3) Cross multiplying (1 + √3) (cos θ − sin θ) = (1 − √3) (cos θ + sin θ) 1 (cos θ − sin θ) + √3(cos θ − sin θ) = 1 (cos θ + sin θ) − √3 (cos θ + sin θ) cos θ − sin θ + √3cos θ − √𝟑 sin θ = cos θ + sin θ − √3cos θ − √𝟑sin θ − sin θ + √3cos θ = sin θ − √3cos θ √3cos θ + √3cos θ = sin θ + sin θ 2√𝟑cos θ = 2 sin θ √3cos θ = sin θ √3 = sin⁡〖θ 〗/cos⁡〖θ 〗 tan θ = √3 Since tan 60° = (cos⁡θ − sin⁡θ)/(cos⁡θ + sin⁡θ ) = (1 − √3)/ So, the correct answer is (c) √3 = sin⁡〖θ 〗/cos⁡〖θ 〗 tan θ = √𝟑 Since tan 60° = √3 Therefore, θ = 60°

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