If A,B and C are the angles of a triangle ,such that sec (A-B), sec A and sec (A+B) are in arithmatic progression ,then

Options

A. `cosec^(2)A=2cosec^(2)(B)/(2)`
B. `2sec^(2)A="sec"^(2)(B)/(2)`
C. `cosec^(2)A=cosec^(2)(B)/(2)`
D. `sec^(2)B="sec"^(2)(A)/(2)`

Answer

B

Explanation

`becausesecA=(sec(A-B)+sec(A+B))/(2)`
`=(cos(A+B)+cos(A-B))/(2cos(A+B)cos(A-B))`
`=(cosAcosB)/([cos^(2)Acos^(2)B-sin^(2)Asin^(2)B])`
`rArrsecA=(cosAcosB)/([cos^(2)Acos^(2)B-(1-cos^(2)AcosB)])`
`rArrcos^(2)A+cos^(2)B-1=cos^(2)AcosB`
`rArrcos^(2)A=1+cosBrArrcos^(2)A=2"cos"^(2)(B)/(2)`
`rArrsec^(2)A=(1)/(2)"sec"^(2)(B)/(2)rArr2sec^(2)A="sec"^(2)(B)/(2)`