Opposite angles of a cyclic quadrilateral are supplementry
So, A+C=π and B+D=π
⇒A=π−C and B=π−D
Option (a)
B=π−D⇒secB=sec(π−D)
⇒secB=−secD
Option (b)
A=π−C⇒cosA=cos(π−C)
⇒cosA+cosC=0
Option (c)
A=π−C⇒cosec A=cosec (π−C)
⇒cosec A=cosec C
Option (d)
B=π−D⇒tanB=tan(π−D)
⇒tanB=−tanD