12th Grade > Mathematics
TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 30
| Page 3 of 3 pages
Answer: Option D. -> None of these
:
D
14√3cos23∘−sin23∘
=12cos30∘cos23∘−sin30∘sin23∘
=12 cos(30∘+23∘) =12 cos53∘.
:
D
14√3cos23∘−sin23∘
=12cos30∘cos23∘−sin30∘sin23∘
=12 cos(30∘+23∘) =12 cos53∘.
Answer: Option C. -> 316
:
C
sin20osin40∘sin60∘sin80∘
=12sin20∘sin60∘(2sin40∘sin80∘)
=12sin20∘sin60∘(cos40∘−cos120∘)
=12.√32sin20∘(1−2sin220∘+12)
=√34sin20∘(32−2sin220∘)
=√38(3sin20∘−4sin320∘)
=√38sin60∘=√38.√32=316
:
C
sin20osin40∘sin60∘sin80∘
=12sin20∘sin60∘(2sin40∘sin80∘)
=12sin20∘sin60∘(cos40∘−cos120∘)
=12.√32sin20∘(1−2sin220∘+12)
=√34sin20∘(32−2sin220∘)
=√38(3sin20∘−4sin320∘)
=√38sin60∘=√38.√32=316
Answer: Option C. -> 4cos28∘ cos29∘ sin33∘
:
C
1+cos56∘+cos58∘−cos66∘=
=2cos228∘+2sin62∘.sin4∘
=2cos228∘+2cos28∘.sin4∘
=2cos28∘(cos28∘+cos86∘)
=2cos28∘.2cos57∘.cos29∘
=4cos28∘cos29∘sin33∘
Aliter: Apply the condition identify
cosA+cosB−cosC=−1+4cosA2cosB2sinc2
[∴56∘+58∘+66∘=180∘]
We get the value of required expression equal to 4cos28∘cos29∘sin33∘.
:
C
1+cos56∘+cos58∘−cos66∘=
=2cos228∘+2sin62∘.sin4∘
=2cos228∘+2cos28∘.sin4∘
=2cos28∘(cos28∘+cos86∘)
=2cos28∘.2cos57∘.cos29∘
=4cos28∘cos29∘sin33∘
Aliter: Apply the condition identify
cosA+cosB−cosC=−1+4cosA2cosB2sinc2
[∴56∘+58∘+66∘=180∘]
We get the value of required expression equal to 4cos28∘cos29∘sin33∘.
Answer: Option B. -> G.P.
:
B
cos 2B =
cos(A+C)cos(A−C) =
cosAcosC−sinAsinCcosAcosC+sinAsinC
⇒1−tan2B1+tan2B =
1−tanAtanC1+tanAtanC
⇒1+tan2B−tanAtanC−tanAtanCtan2B
= 1−tan2B+tanAtanC−tanAtanCtan2B
⇒ 2tan2B=2tanAtanC⇒ tan2B = tan A tan C
Hence, tan A, tan B and tan C will be in G.P.
:
B
cos 2B =
cos(A+C)cos(A−C) =
cosAcosC−sinAsinCcosAcosC+sinAsinC
⇒1−tan2B1+tan2B =
1−tanAtanC1+tanAtanC
⇒1+tan2B−tanAtanC−tanAtanCtan2B
= 1−tan2B+tanAtanC−tanAtanCtan2B
⇒ 2tan2B=2tanAtanC⇒ tan2B = tan A tan C
Hence, tan A, tan B and tan C will be in G.P.
Answer: Option C. -> 4
:
C
√3cosec20∘−sec20∘=√3sin20∘−1cos20∘
=√3cos20∘−sin20∘sin20∘cos20∘=2[√32cos20∘−12sin20∘]22sin20∘cos20∘
=4cos(20∘+30∘)sin40∘=4cos50∘sin40∘=4sin40∘sin40∘=4.
:
C
√3cosec20∘−sec20∘=√3sin20∘−1cos20∘
=√3cos20∘−sin20∘sin20∘cos20∘=2[√32cos20∘−12sin20∘]22sin20∘cos20∘
=4cos(20∘+30∘)sin40∘=4cos50∘sin40∘=4sin40∘sin40∘=4.
Answer: Option A. -> a
:
A
Given that tan θ = ba.
Now, a cos 2θ + b sin 2 θ = a(1−tan2θ1+tan2θ) + b(2tanθ1+tan2θ)
Putting tanθ = ba, we get
= a⎛⎝1−b2a21+b2a2⎞⎠ + b⎛⎝2ba1+b2a2⎞⎠ = a(a2−b2a2+b2) + b(2baa2+b2)
= 1(a2+b2) a3−ab2+2ab2 = a(a2+b2)a2+b2 = a.
:
A
Given that tan θ = ba.
Now, a cos 2θ + b sin 2 θ = a(1−tan2θ1+tan2θ) + b(2tanθ1+tan2θ)
Putting tanθ = ba, we get
= a⎛⎝1−b2a21+b2a2⎞⎠ + b⎛⎝2ba1+b2a2⎞⎠ = a(a2−b2a2+b2) + b(2baa2+b2)
= 1(a2+b2) a3−ab2+2ab2 = a(a2+b2)a2+b2 = a.
Answer: Option D. -> ∞
:
D
We have cos A =
√32 ⇒ A = 30∘
⇒ tan 3A = tan90∘ = ∞.
:
D
We have cos A =
√32 ⇒ A = 30∘
⇒ tan 3A = tan90∘ = ∞.
Answer: Option A. -> tan
A2
:
A
(sin2A1+cos2A)(cosA1+cosA)
=
2sinAcosA2cos2A
cosA1+cosA =
sinA1+cosA = tan
A2
:
A
(sin2A1+cos2A)(cosA1+cosA)
=
2sinAcosA2cos2A
cosA1+cosA =
sinA1+cosA = tan
A2
Answer: Option A. -> 2
:
A
sinx+cosecx=2⇒(sinx−1)2=0⇒sinx=1
∴sinnx+cosecnx=1+1=2
:
A
sinx+cosecx=2⇒(sinx−1)2=0⇒sinx=1
∴sinnx+cosecnx=1+1=2
Answer: Option C. -> π5 radian
:
C
Given, diameter of circular wire = 10cm, therefore length of wire = 10π.
Hence required angle = arcradius=10π50=π5 radian.
:
C
Given, diameter of circular wire = 10cm, therefore length of wire = 10π.
Hence required angle = arcradius=10π50=π5 radian.
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