Question

sin A + cos A =p; sec A + cosec A = q; Prove that q(p^2-1)=2p

Answer 1

q(p^2-1) = 2p

=} LHS = q(p^2-1) =sec +cosec [(sin + cos)^2-1

= (sec+cosec) [sin^2+cos^2+2sin.cos -1]

= (1/cos +1/sin) [1+2sin.cos-1] {{identity: sin2=cos2=1}}

taking LCM of 1/sin + 1/cos

= (sin +cos /sin.cos)[2sin.cos]

= sin + cos / sin.cos * 2sin.cos

= 2(sin + cos)

= 2(p) {{sin + cos = p :(given in question)}}

Therefore 2p i.e. = RHS

hence, 2p = 2p