Trigonometric Functions(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
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.
Explanation:-
Answer: Option C. ->
a2+b2
:
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).
Explanation:-
Answer: Option C. ->
2a23
:
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.
Explanation:-
Answer: Option D. ->
12
:
D
A.M. ≥ G.M.
⇒ 9tan2θ+4cot2θ2 ≥ √4cot2θ.9tan2θ
⇒ 9tan2θ+4cot2θ ≥ 12
Therefore, the minimum value is 12.
Explanation:-
Answer: Option D. ->
-5
:
D
Minimum value of (3sinθ+4cosθ) is -√32+42 i.e., -5.
Explanation:-
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.
Explanation:-
Answer: Option B. ->
√32
:
B
cot215∘−1cot215∘+1 = cos215∘sin215∘−1cos215∘sin215∘+1
= cos215∘−sin215∘cos215∘+sin215∘ = cos(30∘) = √32
Explanation:-
Answer: Option D. ->
- 1665
:
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).
Explanation:-
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.
Explanation:-
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.