Search for: secθ=a2+b2a2−b2, where a,b,∈R, gives real values of θ if and only ifsecθ=a2+b2a2−b2, where a,b,∈R, gives real values of θ if and only ifAa=b≠0B|a|≠|b|≠0Ca+b=0,a≠0Dnone of these Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:Clearly,a2+b2≥a2−b2 for all ||a|≠|b∣≠0,⇒ a2+b2a2−b2≥1 or. a2+b2a2−b2≤−1∴ secθ=a2+b2a2−b2 is meaningful.Thus, secθ=a2+b2a2−b2 gives real values of θ if and only |a|≠|b|≠0Post navigationPrevious: If in a triangle ABC, cosAa=cosBb=cosCc,then the triangle isNext: In any △ABC, if cotA2,cotB2,cotC2 are in A.P., then a, b, c are in Related content NEET Rank Assurance Program | NEET Crash Course 2023 JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 JEE Crash Course – JEE Crash Course 2023 NEET Crash Course – NEET Crash Course 2023