Grade 11 Trigonometry Questions and Answers PDF download

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Grade 11 Trigonometry Questions and Answers PDF download, grade 11 trigonometry past papers, trigonometry grade 11 study notes pdf download, grade 11 trigonometry notes, grade 11 trigonometry graphs questions and answers pdf, advanced trigonometry questions and answers pdf, trigonometric functions pdf, grade 12 trigonometry questions and answers pdf download, grade 10 trigonometry questions and answers pdf

Trigonometry is a branch of mathematics that focuses on relationships between the sides and angles of triangles. The word trigonometry comes from the Latin derivative of Greek words for triangle (trigonon) and measure (metron). Trigonometry (Trig) is an intricate piece of other branches of mathematics such as, Geometry, Algebra, and Calculus.

Grade 11 Trigonometry Questions and Answers PDF download

Question 1 :

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

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 = 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)

(a2+ b2) sec2θ + (c2+ d2) tan2θ

+ c2tan2θ – 2ac secθ tan θ

sec2θ + d2tan2θ – 2bd secθ tan θ  = b+ c2

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