If A,B and C are the angles of a triangle, such that sec(A-B) , secA and sec(A+B) are in arithmetic progression, then
cosec2A=2cosec2B2
2sec2A=sec2B2
2cosec2A=cosec2B2
2sec2A=2sec2A2
Explanation for the correct option:
Step 1. Given that, A,B and C are the angles of a triangle
sec(A-B) , secA and sec(A+B) are in A.P
⇒2secA=sec(A–B)+sec(A+B)
⇒2secA=1cos(A–B)+1cos(A+B)
⇒ secA=cos(A+B)+cos(A–B)2cos(A+B)cos(A–B)
Step 2. Using identities cos(θ+ϕ)+cos(θ-ϕ)=2cosθcosϕ ,
cos(θ+ϕ)=cos2θcos2ϕ, cos(θ-ϕ)=sin2θsin2ϕ
⇒ 1cosA=(2cosAcosB)2(cos2Acos2B–sin2Asin2B)
⇒ (cos2AcosB)cos2Acos2B–(1–cos2A)(1–cos2B)=1 ∵sin2θ+cos2θ=1
⇒cos2Acos2B–1+cos2A+cos2B-cos2Acos2B=cos2AcosB
⇒ cos2A–cos2AcosB=1–cos2B
⇒ cos2A(1–cosB)=(1+cosB)(1–cosB)
⇒ cos2A=1+cosB
⇒ cos2A=2cos2B2
⇒ sec2A=12sec2B2
∴2sec2A=sec2(B2)
Hence, option ‘B’ is correct.