GRADE 11 TRIGONOMETRY QUESTIONS AND ANSWERS

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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