Question

1] prove that:( 1 -sin² Θ) sec² Θ=1.
2] If sinA + cos A= p and sec A+ consec A=q show that q (p²-1)=2p.

Answer 1

1)
 LHS:
        (1-sin²Ф)sec²Ф= cos²Ф×sec²Ф=1=RHS
2)
p²-1=sin²a+cos²+2sinAcosA-1
     1+2sinAcosA-1=2sinAcosA
q(p²-1)=(secA+cosecA)(2sinAcosA)
       = 2(sinA+cosA)


2p=2(sinA+cosA)

q(p²-1)=2p
hence proved
LHS=RHS