12th Grade > Mathematics
INVERSE TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 60
| Page 3 of 6 pages
Answer: Option B. -> tan−1√x
:
B
Let x=tan2θ⇒θ=tan−1√x
Now, 12cos−1(1−x1+x)
=12cos−1(1−tan2θ1+tan2θ)
=12cos−1cos2θ=2θ2=θ=tan−1√x.
:
B
Let x=tan2θ⇒θ=tan−1√x
Now, 12cos−1(1−x1+x)
=12cos−1(1−tan2θ1+tan2θ)
=12cos−1cos2θ=2θ2=θ=tan−1√x.
Answer: Option D. -> π4
:
D
Since 2tan−1x=tan−12x1−x2
∴4tan−115=2[2tan−115]=2tan−1251−125
=2tan−11024=tan−120241−100576=tan−1120119
So, 4tan−115−tan−11239=tan−1120119−tan−11239
=tan−1120119−12391+120119.1239=tan−1(120×239)−119(119×239)+120
⇒tan−11=π4.
:
D
Since 2tan−1x=tan−12x1−x2
∴4tan−115=2[2tan−115]=2tan−1251−125
=2tan−11024=tan−120241−100576=tan−1120119
So, 4tan−115−tan−11239=tan−1120119−tan−11239
=tan−1120119−12391+120119.1239=tan−1(120×239)−119(119×239)+120
⇒tan−11=π4.
Answer: Option C. -> 3
:
C
Given cos−1x+cos−1y+cos−1z=3π
∵0≤cos−1x≤π
∴0≤cos−1y≤π and 0≤cos−1z≤π
Here cos−1x=cos−1y=cos−1z=π
⇒x=y=z=cosπ=−1
∴xy+yz+zx=(−1)(−1)+(−1)(−1)+(−1)(−1)
=1+1+1=3.
:
C
Given cos−1x+cos−1y+cos−1z=3π
∵0≤cos−1x≤π
∴0≤cos−1y≤π and 0≤cos−1z≤π
Here cos−1x=cos−1y=cos−1z=π
⇒x=y=z=cosπ=−1
∴xy+yz+zx=(−1)(−1)+(−1)(−1)+(−1)(−1)
=1+1+1=3.
Answer: Option A. -> 1√10
:
A
Let 0≤x≤π
So,cos−145=x⇒cosx=45 ....(i)
Now sin(12cos−145)=sin(x2) ....(ii)
From (i), cosx=45⇒1−2sin2x2=45
⇒2sin2x2=1−45=15⇒sinx2=√110.
:
A
Let 0≤x≤π
So,cos−145=x⇒cosx=45 ....(i)
Now sin(12cos−145)=sin(x2) ....(ii)
From (i), cosx=45⇒1−2sin2x2=45
⇒2sin2x2=1−45=15⇒sinx2=√110.
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. -> 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 D. -> y2=1−x
:
D
Let tan−12=α⇒x=sin2α=45y=sin(β2)=√1−cosβ2=√1−352=1√5
:
D
Let tan−12=α⇒x=sin2α=45y=sin(β2)=√1−cosβ2=√1−352=1√5
Answer: Option B. -> 3π−10
:
B
3π<10<3π+π2∴10rad∈Q3∴−π2<3π−10<0∴3π−10∈Q4Alsosin(3π−10)=sin10Hencesin−1(sin10)=sin−1sin(3π−10)=3π−10
:
B
3π<10<3π+π2∴10rad∈Q3∴−π2<3π−10<0∴3π−10∈Q4Alsosin(3π−10)=sin10Hencesin−1(sin10)=sin−1sin(3π−10)=3π−10
Answer: Option A. -> 0
:
A
sin−1|cosx|−cos−1|sinx|=π2−cos−1|cosx|−π2+sin−1|sinx|=a⇒sin−1|sinx|−cos−1|cosx|=a⇒a=0∀x
:
A
sin−1|cosx|−cos−1|sinx|=π2−cos−1|cosx|−π2+sin−1|sinx|=a⇒sin−1|sinx|−cos−1|cosx|=a⇒a=0∀x
Answer: Option D. -> −12
:
D
sin(cot−1(1+x))=cos(tan−1x)⇒cot−1(1+x)=[π2±tan−1x]⇒π2−tan−1(1+x)=π2±tan−1x⇒1+x=±x
x=−12Which, on verification, satisfies the equation.
:
D
sin(cot−1(1+x))=cos(tan−1x)⇒cot−1(1+x)=[π2±tan−1x]⇒π2−tan−1(1+x)=π2±tan−1x⇒1+x=±x
x=−12Which, on verification, satisfies the equation.
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