Problem 1 :
If (cos4α/cos2β) + (sin4α/sin2β) = 1, prove that
(i) sin4α + sin4β = 2sin2αsin2β
(ii) (cos4β/cos2α) + (sin4β/sin2α) = 1
Solution :
Given that :
(cos4α/cos2β) + (sin4α/sin2β) = 1
((cos4α sin2β + sin4α cos2β) / cos2β sin2β) = 1
((cos4α sin2β + sin4α cos2β) / cos2β sin2β) = 1
cos4α sin2β + sin4α cos2β = cos2β sin2β
cos4α (1- cos2β) + cos2β (1 - cos2α)2 = cos2β sin2β
cos4α (1- cos2β) + cos2β(1 + cos4α - 2cos2α) = cos2β sin2β
cos4α - cos4α cos2β + cos2β + cos2β cos4α - 2cos2β cos2α
= cos2β(1 - cos2β)
cos4α - cos4α cos2β + cos2β + cos2β cos4α - 2cos2β cos2α - cos2β + cos4β = 0
cos4α - 2cos2β cos2α + cos4β = 0
(cos2α - cos2β)2 = 0
cos2α = cos2β
1 - sin2α = 1 - sin2β
sin2α = sin2β
(i) sin4α + sin4β = 2sin2α sin2β
L.H.S :
= sin4α + sin4β
= (sin2α - sin2β)2 + 2sin2α sin2β
= (sin2 α - sin2 α)2 + 2sin2 α sin2 β
= 2sin2α sin2β ----> R.H.S
(ii) (cos4β/cos2α) + (sin4β/sin2α) = 1.
= (cos2 β cos2 β/cos2α) + (sin2β sin2β/sin2α)
= (cos2β cos2α/cos2α) + (sin2β sin2α/sin2α)
= cos2β + sin2β
= 1
= R.H.S
Problem 2 :
If y = 2 sinα/(1 + cosα + sinα), then prove that
(1 − cosα + sinα)/(1 + sinα) = y
Solution :
(1 − cos α + sin α)/(1 + sin α)
Multiply both numerator and denominator by (1 + cosα + sinα).
= 2sinα/(1 + cosα + sinα)
Hence proved.
Problem 3 :
Solution :
x = 1 + cos2θ + cos4θ + cos6θ + .............
x = 1 / (1 - cos2θ) = 1/sin2θ
y = 1 + sin2θ + sin4θ + sin6θ + .............
y = 1 / (1 - sin2θ) = 1/cos2θ
y = 1 + cos2θ sin2θ + cos4θ sin4θ + .............
z = 1 / (1 - sin2θ cos2θ) = 1/(1 - sin2θcos2θ)
L.H.S
xyz = (1/sin2θ) (1/cos2θ) (1/(1 - sin2θ cos2θ))
= 1/[sin2θ cos2θ(1 - sin2θ cos2θ)]
R.H.S
x + y + z = (1/sin2θ) + (1/cos2θ) + (1/(1 - sin2θ cos2θ))
Hence proved.
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