12th Grade > Mathematics
INVERSE TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 60
| Page 6 of 6 pages
Answer: Option A. -> (−1−1√2)∪(1√2,1)
:
A
tan2(sin−1x)>1
⇒tan(sin−1x)<−1 or tan(sin−1x)>1
⇒−π2<sin−1x<−π4 or π4<sin−1x<π2
⇒−1<x<−1√2 or 1√2<x<1
⇒xϵ(−1−−1√2)∪(1√2,1)
:
A
tan2(sin−1x)>1
⇒tan(sin−1x)<−1 or tan(sin−1x)>1
⇒−π2<sin−1x<−π4 or π4<sin−1x<π2
⇒−1<x<−1√2 or 1√2<x<1
⇒xϵ(−1−−1√2)∪(1√2,1)
Answer: Option A. -> 0
:
A
Let x=tanθ,−π2<θ<π22θ+cos−1cos2θ
(i)0≤2θ<π2.2cos−1cos2θ=2θ
So given expression =4θ=4tan−1x
(ii)−π2<θ≤0⇒−π<2θ≤0cos−1cos2θ=−2θ
So, given expression becomes independent of θ
For −π2<θ≤0⇒−∞<x≤0
:
A
Let x=tanθ,−π2<θ<π22θ+cos−1cos2θ
(i)0≤2θ<π2.2cos−1cos2θ=2θ
So given expression =4θ=4tan−1x
(ii)−π2<θ≤0⇒−π<2θ≤0cos−1cos2θ=−2θ
So, given expression becomes independent of θ
For −π2<θ≤0⇒−∞<x≤0
Answer: Option A. -> (n−1)d1+a1an
:
A
θ=tan−1a2−a11+a1a2+tan−1a3−a21+a2a3+⋯+tan−1an−an−11+an−1an
=(tan−1a2−tan−1a1)+(tan−1a3−tan−1a2)+⋯+tan−1an−an−11+an−1an+⋯+(tan−1an−tan−1an−1)
=tan−1an−tan−1a1
=tan−1an−a11+a1an=tan−1(n−1)d1+a1an
∴tanθ=(n−1)d1+a1an
:
A
θ=tan−1a2−a11+a1a2+tan−1a3−a21+a2a3+⋯+tan−1an−an−11+an−1an
=(tan−1a2−tan−1a1)+(tan−1a3−tan−1a2)+⋯+tan−1an−an−11+an−1an+⋯+(tan−1an−tan−1an−1)
=tan−1an−tan−1a1
=tan−1an−a11+a1an=tan−1(n−1)d1+a1an
∴tanθ=(n−1)d1+a1an
Answer: Option A. -> 0
:
A
cos−1(cos5π3)+sin−1(sin5π3)=cos−1[cos(2π−π3)]+sin−1[sin(2π−π3)]=π3−π3=0.
:
A
cos−1(cos5π3)+sin−1(sin5π3)=cos−1[cos(2π−π3)]+sin−1[sin(2π−π3)]=π3−π3=0.
Answer: Option A. -> tan2(α2)
:
A
tan−1[1√cosα]−tan−1[√cosα]=x
⇒tan−1⎡⎢⎣1√cosα−√cosα1+√cosα√cosα⎤⎥⎦=x⇒tanx=1−cosα2√cosα
∴sinx=1−cosα1+cosα=2sin2α22cos2α2=tan2(α2).
:
A
tan−1[1√cosα]−tan−1[√cosα]=x
⇒tan−1⎡⎢⎣1√cosα−√cosα1+√cosα√cosα⎤⎥⎦=x⇒tanx=1−cosα2√cosα
∴sinx=1−cosα1+cosα=2sin2α22cos2α2=tan2(α2).
Answer: Option B. -> 0
:
B
cos{cos−1(−17)+sin−1(−17)}=cosπ2=0
:
B
cos{cos−1(−17)+sin−1(−17)}=cosπ2=0
Answer: Option C. -> 23
:
C
Let Sec−13=α,cosec−14=β⇒tan2α+cot2β=9−1+16−1=23
:
C
Let Sec−13=α,cosec−14=β⇒tan2α+cot2β=9−1+16−1=23
Answer: Option D. -> −1≤x≤1
:
D
Putting θ=tan−1x,−π2<θ<π2, we have
R.H.S. sin−12tanθ1+tan2θ=sin−1sin2θ=2θ=2tan−1x
When −π2≤2θ≤π2 i.e., −π4≤θ≤π4
i.e., −1≤x=tanθ≤1.
:
D
Putting θ=tan−1x,−π2<θ<π2, we have
R.H.S. sin−12tanθ1+tan2θ=sin−1sin2θ=2θ=2tan−1x
When −π2≤2θ≤π2 i.e., −π4≤θ≤π4
i.e., −1≤x=tanθ≤1.
Answer: Option B. -> 1
:
B
tan−1(x)+tan−1(y)+tan−1(z)=π
⇒tan−1x+tan−1y=π−tan−1z
⇒x+y1−xy=−z⇒x+y=−z+xyz
⇒x+y+z=xyz
Dividing by xyz, we get
1yz+1xz+1xy=1.
Note: Students should remember this question as a formula.
:
B
tan−1(x)+tan−1(y)+tan−1(z)=π
⇒tan−1x+tan−1y=π−tan−1z
⇒x+y1−xy=−z⇒x+y=−z+xyz
⇒x+y+z=xyz
Dividing by xyz, we get
1yz+1xz+1xy=1.
Note: Students should remember this question as a formula.
Answer: Option B. -> 4π−10
:
B
cos−1(cos10)=cos−1(cos(4π−10))=4π−10
:
B
cos−1(cos10)=cos−1(cos(4π−10))=4π−10
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