12th Grade > Mathematics
INVERSE TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 60
| Page 1 of 6 pages
Answer: Option A. -> 0
:
A
cos−1√124=cos−1√32=π6,sec−1√2=π4
sin−1√2−√34=sin−1√4−2√38=sin−1√3−12√2=π12
∴ sin−1√2−√34+cos−1√124+sec−1√2=π2
and sin−1(cotπ2)=sin−10=0
:
A
cos−1√124=cos−1√32=π6,sec−1√2=π4
sin−1√2−√34=sin−1√4−2√38=sin−1√3−12√2=π12
∴ sin−1√2−√34+cos−1√124+sec−1√2=π2
and sin−1(cotπ2)=sin−10=0
Answer: Option A. -> 1
:
A
(sin−1y)2+cos−1x=nπ24
(sin−1y)2−cos−1x=π216
⇒(sin−1y)2=(4n+1)π232,cos−1x=π2(4n−1)32
⇒0≤(4n+1)32π2≤π24,0≤(4n−1)32π2≤π
⇒−14≤n≤74,14,≤n≤8π+14
:
A
(sin−1y)2+cos−1x=nπ24
(sin−1y)2−cos−1x=π216
⇒(sin−1y)2=(4n+1)π232,cos−1x=π2(4n−1)32
⇒0≤(4n+1)32π2≤π24,0≤(4n−1)32π2≤π
⇒−14≤n≤74,14,≤n≤8π+14
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 D. -> 17
:
D
tan[tan−112−tan−113]=tan[tan−112−131+16]
=tantan−1(16×67)=17
:
D
tan[tan−112−tan−113]=tan[tan−112−131+16]
=tantan−1(16×67)=17
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. -> 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. -> 17
:
D
tan[tan−112−tan−113]=tan[tan−112−131+16]
=tantan−1(16×67)=17
:
D
tan[tan−112−tan−113]=tan[tan−112−131+16]
=tantan−1(16×67)=17
Answer: Option D. -> 5665
:
D
Givensin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Usingsin(A+B)=sinAcosB+cosAsinB⇒C=35×1213+√1−925√1−144169⇒C=5665.
:
D
Givensin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Usingsin(A+B)=sinAcosB+cosAsinB⇒C=35×1213+√1−925√1−144169⇒C=5665.
Answer: Option D. -> 5665
:
D
Givensin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Usingsin(A+B)=sinAcosB+cosAsinB⇒C=35×1213+√1−925√1−144169⇒C=5665.
:
D
Givensin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Usingsin(A+B)=sinAcosB+cosAsinB⇒C=35×1213+√1−925√1−144169⇒C=5665.
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)
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email