12th Grade > Mathematics
TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option D. -> 12
:
D
A.M. ≥ G.M.
⇒ 9tan2θ+4cot2θ2 ≥ √4cot2θ.9tan2θ
⇒ 9tan2θ+4cot2θ ≥ 12
Therefore, the minimum value is 12.
:
D
A.M. ≥ G.M.
⇒ 9tan2θ+4cot2θ2 ≥ √4cot2θ.9tan2θ
⇒ 9tan2θ+4cot2θ ≥ 12
Therefore, the minimum value is 12.
Answer: Option B. -> 3π4
:
B
We have tan A = -
12 and tan B = -
13
Now, tan(A + B) =
tanA+tanB1−tanAtanB =
−12−131−12.13 = -1
⇒ tan(A + B) = tan
3π4. Hence, A + B =
3π4.
:
B
We have tan A = -
12 and tan B = -
13
Now, tan(A + B) =
tanA+tanB1−tanAtanB =
−12−131−12.13 = -1
⇒ tan(A + B) = tan
3π4. Hence, A + B =
3π4.
Answer: Option B. -> √5+18
:
B
cos2A−sin2B= cos(A + B). Cos(A - B)
∴cos248∘−sin212∘=cos60∘.cos36∘
=12(√5+14)=√5+18
:
B
cos2A−sin2B= cos(A + B). Cos(A - B)
∴cos248∘−sin212∘=cos60∘.cos36∘
=12(√5+14)=√5+18
Answer: Option B. -> 2cos3θ
:
B
We have x + 1x = 2cosθ,
Now x3 + 1x3 = (x+1x)3 - 3x1x(x+1x)
= (2cosθ)3−3(2cosθ)=8cos3θ−6cosθ
=2(4cos3θ−3cosθ)=2cos3θ.
Trick:Put x = 1 ⇒ θ=0∘.
Then x3 + 1x3 = 2 = 2cos3θ.
:
B
We have x + 1x = 2cosθ,
Now x3 + 1x3 = (x+1x)3 - 3x1x(x+1x)
= (2cosθ)3−3(2cosθ)=8cos3θ−6cosθ
=2(4cos3θ−3cosθ)=2cos3θ.
Trick:Put x = 1 ⇒ θ=0∘.
Then x3 + 1x3 = 2 = 2cos3θ.
Answer: Option A. -> √2cosθ,√2sinθ
:
A
We have tanθ = sinα−cosαsinα+cosα
⇒ tanθ = sin(α−π4)cos(α−π4) ⇒ tanθ = tan(α−π4)
⇒ θ = α - π4⇒ α = θ + π4
Hence, sinα+cosα = sin(θ+π4)+cos(θ+π4)
= √2cosθ
And sinα−cosα = sin(θ+π4) - cos(θ+π4)
= 1√2 sinθ + 1√2 cosθ - 1√2 cosθ + 1√2 sinθ
= 2√2 sinθ = √2sinθ=√2sinθ.
:
A
We have tanθ = sinα−cosαsinα+cosα
⇒ tanθ = sin(α−π4)cos(α−π4) ⇒ tanθ = tan(α−π4)
⇒ θ = α - π4⇒ α = θ + π4
Hence, sinα+cosα = sin(θ+π4)+cos(θ+π4)
= √2cosθ
And sinα−cosα = sin(θ+π4) - cos(θ+π4)
= 1√2 sinθ + 1√2 cosθ - 1√2 cosθ + 1√2 sinθ
= 2√2 sinθ = √2sinθ=√2sinθ.
Answer: Option C. -> √32
:
C
cos2(π3−x) -cos2(π3+x)
= {cos(π3−x)+cos(π3+x)} {cos(π3−x)−cos(π3+x)}
= {2cosπ3cosx}{2sinπ3sinx}
= sin 2π3 sin 2x =√32 sin 2x
Its maximum value is√32, {- 1 ≤ sin 2x ≤ 1}.
:
C
cos2(π3−x) -cos2(π3+x)
= {cos(π3−x)+cos(π3+x)} {cos(π3−x)−cos(π3+x)}
= {2cosπ3cosx}{2sinπ3sinx}
= sin 2π3 sin 2x =√32 sin 2x
Its maximum value is√32, {- 1 ≤ sin 2x ≤ 1}.
Answer: Option D. -> 49
:
D
Given that tan
A2 =
32.
1+cosA1−cosA =
2cos2A22sin2A2 = cot2A2 = (23)2 =
49
:
D
Given that tan
A2 =
32.
1+cosA1−cosA =
2cos2A22sin2A2 = cot2A2 = (23)2 =
49
Answer: Option D. -> 192
:
D
Given expression is
sin25∘+sin210∘+sin215∘+..............+sin285∘+sin290∘.
We know that sin90∘=1orsin290∘=1
Similarly, sin45∘=1√2 or sin245∘=12 and the angles are in A.P. of 18 terms. We also know that
sin285∘=[sin(90∘−5∘)]2=cos25∘.
Therefore from the complementary rule, we find sin25∘+sin285∘=sin25∘+cos25∘=1
Therefore,
sin25∘+sin210∘+sin215∘+.............+sin285∘+sin290∘
=(1+1+1+1+1+1+1+1)+1+12=912.
:
D
Given expression is
sin25∘+sin210∘+sin215∘+..............+sin285∘+sin290∘.
We know that sin90∘=1orsin290∘=1
Similarly, sin45∘=1√2 or sin245∘=12 and the angles are in A.P. of 18 terms. We also know that
sin285∘=[sin(90∘−5∘)]2=cos25∘.
Therefore from the complementary rule, we find sin25∘+sin285∘=sin25∘+cos25∘=1
Therefore,
sin25∘+sin210∘+sin215∘+.............+sin285∘+sin290∘
=(1+1+1+1+1+1+1+1)+1+12=912.
Answer: Option A. -> 13
:
A
Given that secθ = 54
secθ = 1+tan2(θ2)1−tan2(θ2) ⇒ 54 = 1+tan2(θ2)1−tan2(θ2)
⇒ 5−5tan2(θ2)=4+4tan2(θ2)
⇒ 9tan2(θ2) = 1 ⇒ tan(θ2) = 13.
:
A
Given that secθ = 54
secθ = 1+tan2(θ2)1−tan2(θ2) ⇒ 54 = 1+tan2(θ2)1−tan2(θ2)
⇒ 5−5tan2(θ2)=4+4tan2(θ2)
⇒ 9tan2(θ2) = 1 ⇒ tan(θ2) = 13.
Answer: Option D. -> 12
:
D
cotA1+cotA. cotB1+cotB = 1(1+tanA)(1+tanB)
= 1tanA+tanB+1+tanAtanB
[ ∵ tan(A + B) = tan225∘]
⇒ tanA + tan B = 1 - tan A tan B
= 11−tanAtanB+1+tanAtanB = 12.
:
D
cotA1+cotA. cotB1+cotB = 1(1+tanA)(1+tanB)
= 1tanA+tanB+1+tanAtanB
[ ∵ tan(A + B) = tan225∘]
⇒ tanA + tan B = 1 - tan A tan B
= 11−tanAtanB+1+tanAtanB = 12.
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