Question 1 :
If cot θ (1 + sin θ) = 4m and cot θ (1 − sin θ) = 4n, then prove that (m2 − n2)2 = mn
Answer :
cot θ (1 + sin θ) = 4m and cot θ (1 − sin θ) = 4n
m = [cot θ (1 + sin θ)] / 4
n = [cot θ (1 - sin θ)] / 4
m2 = [cot θ (1+sin θ)/4]2 = [cot2 θ (1+sin θ)2/16] ---(1)
n2 = [cot θ (1-sin θ)/4]2 = [cot2 θ (1-sin θ)2/16] ----(2)
(1) - (2)
L.H.S
m2 - n2 = [cot2 θ (1+sin θ)2 - cot2 θ (1+sin θ)2]/16
= [cot2 θ (1+sin2θ+2sinθ) - cot2 θ (1+sin2θ-2sinθ)]/16
= (4 cot2 θ sinθ/16)
= (cot2 θ sin θ/4)
(m2 - n2)2 = (cot4 θ sin2 θ)/16
= [(cos4 θ/sin4 θ) ⋅ sin2 θ]/16
= (1/16) (cos4 θ/sin2 θ) ---(1) L.H.S
m = [cot θ (1 + sin θ)] / 4
n = [cot θ (1 - sin θ)] / 4
mn = [cot θ (1 + sin θ)/4] [cot θ (1 - sin θ)/4]
= cot2 θ(1 - sin2 θ)/16
= cot2 θ(cos2 θ)/16
= (cos2 θ/sin2 θ)⋅(cos2 θ) (1/16)
= (1/16) (cos4 θ/sin2 θ) ---(2) R.H.S
Question 2 :
If cosec θ − sin θ = a3 and sec θ − cos θ = b3, then prove that a2b2 (a2 + b2) = 1.
Answer :
Given that :
cosec θ − sin θ = a3
(1/sin θ) - sin θ = a3
(1 - sin2θ) / sin θ = a3
cos2θ / sin θ = a3 -----(1)
sec θ − cos θ = b3
(1/cos θ) - cos θ = b3
(1 - cos2θ) / cos θ = b3
sin2θ / cos θ = b3 -----(2)
(2) / (1)
b3/ a3 = (sin2θ / cos θ) / (cos2θ / sin θ)
= (sin2θ / cos θ) ⋅ (sin θ/cos2θ)
= (sin3θ /cos3θ)
b3/ a3 = tan3θ
tanθ = b/a
sin θ = b/√(b2 +a2)
cos θ = a/√(b2 +a2)
By applying the values of sin θ and cos θ in (1)
cos2θ / sin θ = a3 -----(1)
[a/√(b2 +a2)]2/[b/√(b2 +a2)] = a3
ab√(b2 +a2) = 1
Taking squares on both sides, we get
a2b2(b2 +a2) = 1
Hence proved.
Question 3 :
Eliminate θ from the equations a sec θ − c tan θ = b and bsec θ + d tan θ = c.
Answer :
a sec θ − c tan θ = b
Take squares on both sides
(a sec θ − c tan θ)2 = b2
a2sec2θ + c2tan2θ - 2ac secθ tan θ = b2 ----(1)
bsec θ + d tan θ = c
Take squares on both sides
(b sec θ + d tan θ)2 = c2
b2sec2θ + d2tan2θ - 2bd secθ tan θ = c2 ----(2)
(1) + (2)
(a2+ b2) sec2θ + (c2+ d2) tan2θ
+ c2tan2θ - 2ac secθ tan θ
sec2θ + d2tan2θ - 2bd secθ tan θ = b2 + c2
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