10th Grade > Mathematics
INTRODUCTION TO TRIGONOMETRY MCQs
Total Questions : 52
| Page 1 of 6 pages
Answer: Option B. -> 0
:
B
We know that, cos90∘=0
The given expression
cos1∘ × cos2∘ × cos3∘ ×....× cos90∘ ×……..× cos180∘ reduces to zero as it containscos90∘ which is equal to 0.
:
B
We know that, cos90∘=0
The given expression
cos1∘ × cos2∘ × cos3∘ ×....× cos90∘ ×……..× cos180∘ reduces to zero as it containscos90∘ which is equal to 0.
Answer: Option A. -> 0.5
:
A
We know thattan30∘=1√3.
⇒1−tan230∘1+tan230∘=1−(1√3)21+(1√3)2
=1−(13)1+(13)
=3−13+1
=24 = 12=0.5
:
A
We know thattan30∘=1√3.
⇒1−tan230∘1+tan230∘=1−(1√3)21+(1√3)2
=1−(13)1+(13)
=3−13+1
=24 = 12=0.5
:
3sinθcosθ+cosθsinθ =5sinθ
3sin2θ+cos2θsinθcosθ=5sinθ
3sin2θ+cos2θ=5cosθ
3(1 -cos2θ) +cos2θ = 5cosθ
2cos2θ + 5cosθ - 3 = 0
2cosθ[cosθ + 3] - 1(cosθ + 3) = 0
(cosθ + 3) (2cosθ - 1) = 0
cosθ= -3 or cosθ = 12
Note that cosθ= -3 is not possible as−1≤cosθ≤1
Thus, θ = 60∘
Answer: Option D. -> 1
:
D
x=asec Acos B⇒xa=sec Acos B
y=bsec Asin B⇒yb=sec Asin B
z=ctan A⇒zc=tan A
Squaring each of the equations we get,
x2a2=sec2Acos2B,y2b2=sec2Asin2B,z2c2=tan2A
∴x2a2+y2b2−z2c2=sec2Acos2B+sec2Asin2B−tan2A
=sec2A(cos2B+sin2B)−tan2A
=sec2A−tan2A……(cos2B+sin2B=1)
=1……(sec2A−tan2A=1)
:
D
x=asec Acos B⇒xa=sec Acos B
y=bsec Asin B⇒yb=sec Asin B
z=ctan A⇒zc=tan A
Squaring each of the equations we get,
x2a2=sec2Acos2B,y2b2=sec2Asin2B,z2c2=tan2A
∴x2a2+y2b2−z2c2=sec2Acos2B+sec2Asin2B−tan2A
=sec2A(cos2B+sin2B)−tan2A
=sec2A−tan2A……(cos2B+sin2B=1)
=1……(sec2A−tan2A=1)
Answer: Option D. -> 4
:
D
⇒(sinA+cosec A)2−(sinA−cosec A)2
=4sinAcosec A
as[(a+b)2−(a−b)2=4ab]
now, (sinA×cosec A=sinA×1sinA=1)
∴4.sinAcosec A=4×1=4
:
D
⇒(sinA+cosec A)2−(sinA−cosec A)2
=4sinAcosec A
as[(a+b)2−(a−b)2=4ab]
now, (sinA×cosec A=sinA×1sinA=1)
∴4.sinAcosec A=4×1=4
Answer: Option B. -> 1
:
B
x=asecθ+btanθ,y=atanθ+bsecθ
Squaring both sides we get,
x2=(asecθ+btanθ)2,y2=(atanθ+bsecθ)2
∴x2=a2sec2θ+2absecθtanθ+b2tan2θy2=a2tan2θ+2absecθtanθ+b2sec2θ−−−–––––––––––––––––––––––––––––––––––––––––––––Subtracting,x2−y2=a2(sec2θ−tan2θ)+b2(tan2θ−sec2θ)
x2−y2=a2(sec2θ−tan2θ)−b2(sec2θ−tan2θ)
x2−y2=a2−b2……(sec2θ−tan2θ=1)
∴x2−y2a2−b2=1
:
B
x=asecθ+btanθ,y=atanθ+bsecθ
Squaring both sides we get,
x2=(asecθ+btanθ)2,y2=(atanθ+bsecθ)2
∴x2=a2sec2θ+2absecθtanθ+b2tan2θy2=a2tan2θ+2absecθtanθ+b2sec2θ−−−–––––––––––––––––––––––––––––––––––––––––––––Subtracting,x2−y2=a2(sec2θ−tan2θ)+b2(tan2θ−sec2θ)
x2−y2=a2(sec2θ−tan2θ)−b2(sec2θ−tan2θ)
x2−y2=a2−b2……(sec2θ−tan2θ=1)
∴x2−y2a2−b2=1
Answer: Option D. -> 1+cos θ1−cos θ
:
D
tan2θ(secθ−1)2=sec2θ−1(secθ−1)2
=(secθ−1)(secθ+1)(secθ−1)(secθ−1)
=1cosθ+11cosθ−1
=1+cosθcosθ1−cosθcosθ
=1+cosθ1−cosθ
:
D
tan2θ(secθ−1)2=sec2θ−1(secθ−1)2
=(secθ−1)(secθ+1)(secθ−1)(secθ−1)
=1cosθ+11cosθ−1
=1+cosθcosθ1−cosθcosθ
=1+cosθ1−cosθ
Answer: Option B. -> 0
:
B
sinA−sinBcosA+cosB+cosA−cosBsinA+sinB
= (sinA−sinB)(sinA+sinB)+(cosA−cosB)(cosA+cosB)(cosA+cosB)(sinA+sinB)
=sin2A−sin2B+cos2A−cos2B(cosA+cosB)(sinA+sinB)
= sin2A+cos2A−(sin2B‘+cos2B)(cosA+cosB)(sinA+sinB)
= 1−1(cosA+cosB)(sinA+sinB)
= 0
:
B
sinA−sinBcosA+cosB+cosA−cosBsinA+sinB
= (sinA−sinB)(sinA+sinB)+(cosA−cosB)(cosA+cosB)(cosA+cosB)(sinA+sinB)
=sin2A−sin2B+cos2A−cos2B(cosA+cosB)(sinA+sinB)
= sin2A+cos2A−(sin2B‘+cos2B)(cosA+cosB)(sinA+sinB)
= 1−1(cosA+cosB)(sinA+sinB)
= 0
Answer: Option C. -> 1
:
C
We know that,1+tan2θ=sec2θand1+cot2θ=cosec2θ
11+tan2θ+11+cot2θ=1sec2θ+1cosec2θ
=cos2θ+sin2θ
=1
:
C
We know that,1+tan2θ=sec2θand1+cot2θ=cosec2θ
11+tan2θ+11+cot2θ=1sec2θ+1cosec2θ
=cos2θ+sin2θ
=1
Answer: Option D. -> 2
:
D
(sec A+tan A−1)(sec A−tan A+1)tan A
=[sec A+(tan A−1)][sec A−(tan A−1)]tan A
=sec2A−(tan A−1)2tan A
=sec2A−(tan2A−2tan A+1)tan A
=sec2A−tan2A+2tan A−1tan A
=1+2tan A−1tan A……(∵1+tan2A=sec2A)
=2tan Atan A
=2
:
D
(sec A+tan A−1)(sec A−tan A+1)tan A
=[sec A+(tan A−1)][sec A−(tan A−1)]tan A
=sec2A−(tan A−1)2tan A
=sec2A−(tan2A−2tan A+1)tan A
=sec2A−tan2A+2tan A−1tan A
=1+2tan A−1tan A……(∵1+tan2A=sec2A)
=2tan Atan A
=2
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email