if sin+cos=p
sec+cosec=q then prove dat q(p2-1)=2p
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q(p2-1) = 2p
=
LHS = q(p2-1) =sec +cosec [(sin + cos)2-1
= (sec+cosec) [sin2+cos2+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 = RHS
hence prove= 2p = 2p
=
LHS = q(p2-1) =sec +cosec [(sin + cos)2-1
= (sec+cosec) [sin2+cos2+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 = RHS
hence prove= 2p = 2p
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