12th Grade > Mathematics
INVERSE TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 60
| Page 2 of 6 pages
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
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≤tanθ≤1
i.e.,−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≤tanθ≤1
i.e.,−1≤x≤1
Answer: Option C. -> ϕ
:
C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀xϵR
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of pϵR.
∴ Set of values of p=ϕ
:
C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀xϵR
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of pϵR.
∴ Set of values of p=ϕ
Answer: Option A. -> 0
:
A
Let x=tanθ,−π2<θ<π22θ+cos−1cos2θ
(i)0≤2θ<π2cos−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<π2cos−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 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 A. -> 0
:
A
cos−1(cos5π3)+sin−1(cos5π3)=cos−1[cos(2π−π3)]+sin−1[sin(2π−π3)]=π3−π3=0.
:
A
cos−1(cos5π3)+sin−1(cos5π3)=cos−1[cos(2π−π3)]+sin−1[sin(2π−π3)]=π3−π3=0.
Answer: Option B. -> 1√3
:
B
3sin−12x1+x2−4cos−11−x21+x2+2tan−12x1+x2=π3
Putting x=tanθ
3sin−1(2tanθ1+tan2θ)−4cos−1(1−tan2θ1+tan2θ)
+2tan−1(2tanθ1−tan2θ)=π3
⇒3sin−1(sin2θ)−4cos−1(cos2θ)
+2tan−1(tan2θ)=π3
⇒3(2θ)−4(2θ)+2(2θ)=π3⇒6θ−8θ+4θ=π3
⇒θ=π6⇒tan−1x=π6⇒x=tanπ6=1√3
:
B
3sin−12x1+x2−4cos−11−x21+x2+2tan−12x1+x2=π3
Putting x=tanθ
3sin−1(2tanθ1+tan2θ)−4cos−1(1−tan2θ1+tan2θ)
+2tan−1(2tanθ1−tan2θ)=π3
⇒3sin−1(sin2θ)−4cos−1(cos2θ)
+2tan−1(tan2θ)=π3
⇒3(2θ)−4(2θ)+2(2θ)=π3⇒6θ−8θ+4θ=π3
⇒θ=π6⇒tan−1x=π6⇒x=tanπ6=1√3
Answer: Option B. -> 1415
:
B
sin[2tan−1(13)]+cos[tan−1(2√2)]
=sin[tan−1231−19]+cos[tan−1(2√2)]
=sin[tan−134]+cos[tan−12√2]
=tan−12√2=cos−113
Alsotan−134=sin−135
=35+13=1415.
:
B
sin[2tan−1(13)]+cos[tan−1(2√2)]
=sin[tan−1231−19]+cos[tan−1(2√2)]
=sin[tan−134]+cos[tan−12√2]
=tan−12√2=cos−113
Alsotan−134=sin−135
=35+13=1415.
Answer: Option D. -> None of these
:
D
cos−1(cos12)−sin−1(sin14)⇒4π−12+5π−14=9π−26
:
D
cos−1(cos12)−sin−1(sin14)⇒4π−12+5π−14=9π−26
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