If secα and cosec α are the roots of the equation x2−ax+b=0, then
a2=b(b−2)
a2=b(b+2)
a2+b2=2b
none of these
It is given that sec αand cosecα are the roots of the equation x2−ax+b=0
∴ secα+cosecα=a and secαcosecα=b ⇒ sinα+cosα=asinαcosα and sinαcosα=1b ⇒ sinα+cosα=ab and sinαcosα=1b (sinα+cosα)2=1+2sinαcosα ⇒ a2b2=1+23⇒a2=b(b+2)