Question 1 :

If cot θ (1 + sin θ) = 4m and cot θ (1 − sin θ) = 4n, then prove that (m2 − n2)= mn

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)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 sinθ)/16

=  [(cosθ/sinθ)  sinθ]/16

=  (1/16) (cosθ/sinθ)  ---(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]

=  cotθ(1 - sinθ)/16

=  cotθ(cosθ)/16

=  (cosθ/sinθ)(cosθ) (1/16)

=  (1/16) (cosθ/sinθ)  ---(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.

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/ a= (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.

a sec θ − c tan θ = b

Take squares on both sides

(a sec θ − c tan θ)2 = b

a2sec2θ + c2tan2θ - 2ac secθ tan θ  =  b ----(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)

(a2b2sec2θ + (c2+ d2) tan2θ

+ c2tan2θ - 2ac secθ tan θ

sec2θ + d2tan2θ - 2bd secθ tan θ  = b2 c2

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