If SinA + CosA = 1 and (SinA,CosA>0)

Question

Show that (1 + SecA) (1 + CosecA) ≥ 9

Answer 1

sinA+ cosA = 1, sinA>0, cosA>0, A is in quadrant I

0<A<90 degrees

-sinA -cosA =-1

(1-secA)(1-cscA)

= 1-secA-cscA +secAcscA

= 1-1/cosA -1/sinA + 1/(cosA)(1/sinA)

= (cosAsinA - sinA -cosA+ 1)/cosAsinA

replace -sinA-cosA with -1

= (cosAsinA -1+1)/cosAsinA

= cosAsinA/cosAsinA

=1 which is Not > 9,

maybe the problem was to prove (1-secA)(1-cscA)<9?

and the inequality sign got mixed up

if cosA+sinA = 1

above is the proof that (1-secA)(1-cscA) =1 < 9