10th Grade > Mathematics
INTRODUCTION TO TRIGONOMETRY MCQs
Total Questions : 52
| Page 2 of 6 pages
Answer: Option C. -> 2 sec2 A
:
C
⇒(1+tanA)2+(1−tanA)2
=(1+tan2A+2tanA)+(1+tan2A−2tanA)
=(2+2tan2A+2tanA−2tanA)
=(2+2tan2A)
=2(1+tan2A)
=2sec2A
:
C
⇒(1+tanA)2+(1−tanA)2
=(1+tan2A+2tanA)+(1+tan2A−2tanA)
=(2+2tan2A+2tanA−2tanA)
=(2+2tan2A)
=2(1+tan2A)
=2sec2A
Answer: Option D. -> 1
:
D
sec17∘cosec73∘+tan68∘cot22∘−[cos244∘+cos246∘]
=sec17∘cosec(90∘−17∘)+tan68∘cot(90∘−68∘)−[cos244∘+cos2(90∘−44∘)]
=sec17∘sec17∘+tan68∘tan68∘−[cos244∘+sin244∘]....(complementary angles)
=1+1−1
=1
:
D
sec17∘cosec73∘+tan68∘cot22∘−[cos244∘+cos246∘]
=sec17∘cosec(90∘−17∘)+tan68∘cot(90∘−68∘)−[cos244∘+cos2(90∘−44∘)]
=sec17∘sec17∘+tan68∘tan68∘−[cos244∘+sin244∘]....(complementary angles)
=1+1−1
=1
Answer: Option C. -> tan2θ
:
C
The given expression is
(1+tanθ1+cotθ)2=(1+tanθ1+1tanθ)2
=(1+tanθtanθ+1tanθ)2
=((tanθ)(1+tanθ)tanθ+1)2
=tan2θ
:
C
The given expression is
(1+tanθ1+cotθ)2=(1+tanθ1+1tanθ)2
=(1+tanθtanθ+1tanθ)2
=((tanθ)(1+tanθ)tanθ+1)2
=tan2θ
Answer: Option C. -> tanA
:
C
sinA−2sin3A2cos3A−cosA=sinA(1−2sin2A)cosA(2cos2A−1)
=sinA(sin2A+cos2A−2sin2A)cosA(2cos2A−(sin2A+cos2A))
=sinA(cos2A−sin2A)cosA(cos2A−sin2A)
=tanA
:
C
sinA−2sin3A2cos3A−cosA=sinA(1−2sin2A)cosA(2cos2A−1)
=sinA(sin2A+cos2A−2sin2A)cosA(2cos2A−(sin2A+cos2A))
=sinA(cos2A−sin2A)cosA(cos2A−sin2A)
=tanA
Answer: Option A. -> cot θ
:
A
1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ)[1+sin(90∘−θ)]
=1+cosθ−sin2θsinθ(1+cosθ)
=(1−sin2θ)+cosθsinθ(1+cosθ)
=cos2θ+cosθsinθ(1+cosθ)
=cosθ(cosθ+1)sinθ(1+cosθ)=cotθ
:
A
1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ)[1+sin(90∘−θ)]
=1+cosθ−sin2θsinθ(1+cosθ)
=(1−sin2θ)+cosθsinθ(1+cosθ)
=cos2θ+cosθsinθ(1+cosθ)
=cosθ(cosθ+1)sinθ(1+cosθ)=cotθ
Answer: Option B. -> cosA2
:
B
A+B+C=180∘
(sum of angles of a triangle)
⟹B+C=180∘−A
∴B+C2=180∘−A2
B+C2=90∘−A2
Taking sineon both sides we get,
sin(B+C2)=sin(90∘−A2)
∴sin(B+C2)=cosA2
:
B
A+B+C=180∘
(sum of angles of a triangle)
⟹B+C=180∘−A
∴B+C2=180∘−A2
B+C2=90∘−A2
Taking sineon both sides we get,
sin(B+C2)=sin(90∘−A2)
∴sin(B+C2)=cosA2
Answer: Option B. -> x2a2−y2b2=1
:
B
x=acosecθ⇒xa=cosecθ.....(1)
y=bcotθ⇒yb=cotθ.....(2)
Squaring the equations and subtracting (2) from (1) we get,
x2a2−y2b2=cosec2θ−cot2θ
∴x2a2−y2b2=1....(1+cot2θ=cosec2θ)
:
B
x=acosecθ⇒xa=cosecθ.....(1)
y=bcotθ⇒yb=cotθ.....(2)
Squaring the equations and subtracting (2) from (1) we get,
x2a2−y2b2=cosec2θ−cot2θ
∴x2a2−y2b2=1....(1+cot2θ=cosec2θ)
Answer: Option A. -> 1
:
A
(1+tan A tan B)2+(tan A-tan B)2sec2Asec2B
=1+2tan A tan B+tan2Atan2B+tan2A−2tan A tan B+tan2Bsec2Asec2B
=1+tan2Atan2B+tan2A+tan2Bsec2Asec2B
=1+tan2A+tan2B+tan2Atan2Bsec2Asec2B
=1(1+tan2A)+tan2B(1+tan2A)sec2Asec2B
=(1+tan2A)(1+tan2B)sec2Asec2B
=(sec2A)(sec2B)sec2Asec2B.…(1+tan2θ=sec2θ)
=1
:
A
(1+tan A tan B)2+(tan A-tan B)2sec2Asec2B
=1+2tan A tan B+tan2Atan2B+tan2A−2tan A tan B+tan2Bsec2Asec2B
=1+tan2Atan2B+tan2A+tan2Bsec2Asec2B
=1+tan2A+tan2B+tan2Atan2Bsec2Asec2B
=1(1+tan2A)+tan2B(1+tan2A)sec2Asec2B
=(1+tan2A)(1+tan2B)sec2Asec2B
=(sec2A)(sec2B)sec2Asec2B.…(1+tan2θ=sec2θ)
=1
Answer: Option B. -> 10°
:
B
Given: cos3θ=√32
We know that cos30∘=√32
Comparing the two we get,
3θ=30∘.... (given0∘ < 3θ < 90∘)
⇒θ=10∘
:
B
Given: cos3θ=√32
We know that cos30∘=√32
Comparing the two we get,
3θ=30∘.... (given0∘ < 3θ < 90∘)
⇒θ=10∘
Answer: Option A. -> tan6 θ+3 tan2 θ sec2 θ+1
:
A
sec6θ=(sec2θ)3
=(tan2θ+1)3……(1+tan2θ=sec2θ)
=tan6θ+3tan4θ+3tan2θ+1
=tan6θ+3tan2θ(tan2θ)+3tan2θ+1
=tan6θ+3tan2θ(sec2θ−1)+3tan2θ+1
=tan6θ+3tan2θsec2θ−3tan2θ+3tan2θ+1
=tan6θ+3tan2θsec2θ+1
:
A
sec6θ=(sec2θ)3
=(tan2θ+1)3……(1+tan2θ=sec2θ)
=tan6θ+3tan4θ+3tan2θ+1
=tan6θ+3tan2θ(tan2θ)+3tan2θ+1
=tan6θ+3tan2θ(sec2θ−1)+3tan2θ+1
=tan6θ+3tan2θsec2θ−3tan2θ+3tan2θ+1
=tan6θ+3tan2θsec2θ+1
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