12th Grade > Mathematics
INVERSE TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 60
| Page 4 of 6 pages
Answer: Option C. -> 0
:
C
Putting θ=cos−1x in R.H.S., we have
R.H.S =tan−1√1−x2x=tan−1√1−cos2θcosθ(0≤θ≤π)=tan−1sinθcosθ=tan−1tanθ=θ=cos−1x when −π2<θ<π2
i.e., when 0≤θ<π2
i.e., when 0<cosθ≤1 i.e., 0<x≤1.
:
C
Putting θ=cos−1x in R.H.S., we have
R.H.S =tan−1√1−x2x=tan−1√1−cos2θcosθ(0≤θ≤π)=tan−1sinθcosθ=tan−1tanθ=θ=cos−1x when −π2<θ<π2
i.e., when 0≤θ<π2
i.e., when 0<cosθ≤1 i.e., 0<x≤1.
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 D. -> 12
:
D
Let α=cos−1√p:β=cos−1√1−p
and γ=cos−1√1−q
or
cosα=√p:cosβ=√1−p
and cosγ=√1−q.
Therefore sinα=√1−p,sinβ=√p and sinγ=√q.
The given equation may be written as α+β+γ=3π4
or,α+β=3π4−γ
or, cos(α+β)=cos(3π4−γ)
⇒cosαcosβ−sinαsinβ = cos{π−(π4+γ)}=−cos(π4+γ)
⇒√p√1−p−√1−p√p=−(1√2√1−q−1√2.√q)
⇒0=√1−q−√q⇒1−q=q⇒q=12
:
D
Let α=cos−1√p:β=cos−1√1−p
and γ=cos−1√1−q
or
cosα=√p:cosβ=√1−p
and cosγ=√1−q.
Therefore sinα=√1−p,sinβ=√p and sinγ=√q.
The given equation may be written as α+β+γ=3π4
or,α+β=3π4−γ
or, cos(α+β)=cos(3π4−γ)
⇒cosαcosβ−sinαsinβ = cos{π−(π4+γ)}=−cos(π4+γ)
⇒√p√1−p−√1−p√p=−(1√2√1−q−1√2.√q)
⇒0=√1−q−√q⇒1−q=q⇒q=12
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 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 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 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.
Answer: Option D. -> −717
:
D
tan[2tan−1(15)−π4]=tan[tan−1251−125−tan−1(1)]
=tan[tan−1512−tan−1(1)]=tantan−1(512−11+512)=−717
:
D
tan[2tan−1(15)−π4]=tan[tan−1251−125−tan−1(1)]
=tan[tan−1512−tan−1(1)]=tantan−1(512−11+512)=−717
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.
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email