11th Grade > Mathematics
TRIGONOMETRIC FUNCTIONS MCQs
:
C
We have sin(α−β) = sin(θ−β−¯¯¯¯¯¯¯¯¯¯¯¯¯θ−α)
= sin(θ−β)cos(θ−α)−cos(θ−β)sin(θ−α)
= ba - √1−b2√1−a2
And cos(α−β)=cos(θ−β−¯¯¯¯¯¯¯¯¯¯¯¯¯θ−α)
= cos(θ−β)cos(θ−α)+sin(θ−β)sin(θ−α)
= a√1−b2+b√1−a2
∴ Given expression is cos2(α−β)+2absin(α−β)
= (a√1−b2+b√1−a2)2 + 2ab{ab−√1−a2√1−b2}
= a2+b2.
Trick: Put α=30∘, β=60∘ and θ=90∘,
then a = 12, b = 12
∴ cos2(α−β)+2absin(α−β) = 34 + 12 × (- 12) = 12
which is given by option (c).
:
C
Adding and subtracting the given relation,
we get (m+ n) = acos3α+3acosαsin2α
+ 3acos2α.sinα+asin3α
= a(cosα+sinα)3
and similarly (m - n) = a(cosα−sinα)3
Thus, (m+n)23+(m−n)23
= a23 {(cosα+sinα)2 + (cosα−sinα)2}
= a23 {2(cos2α+sin2α)} = 2a23.
:
D
Minimum value of (3sinθ+4cosθ) is -√32+42 i.e., -5.
:
D
A.M. ≥ G.M.
⇒ 9tan2θ+4cot2θ2 ≥ √4cot2θ.9tan2θ
⇒ 9tan2θ+4cot2θ ≥ 12
Therefore, the minimum value is 12.
:
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.
:
B
cot215∘−1cot215∘+1 = cos215∘sin215∘−1cos215∘sin215∘+1
= cos215∘−sin215∘cos215∘+sin215∘ = cos(30∘) = √32
:
D
We have sin A = 45 and cos B = - 1213
Now, cos(A + B) = cos A cos B - sin A sin B
= √1−1625(−1213) - 45(−√1−144169)
= - 35 × 1213 - 45(−513) = - 1665
(Since A lies in first quadrant and B lies in third quadrant).
:
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.
:
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.
:
D
14√3cos23∘−sin23∘
= 12cos30∘cos23∘−sin30∘sin23∘
= 12 cos(30∘+23∘) = 12 cos53∘.