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CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 5 - MCQExams.com
CBSE
Class 12 Commerce Maths
Inverse Trigonometric Functions
Quiz 5
If
f
(
x
)
=
√
sec
−
1
(
2
−
|
x
|
4
)
,
then the domain of
f
(
x
)
is ______________.
Report Question
0%
[
−
2
,
2
]
0%
[
−
6
,
6
]
0%
(
−
∞
,
−
6
]
∪
[
6
,
∞
)
0%
None Of These
If
3
tan
−
1
(
1
2
+
√
3
)
−
tan
−
1
1
x
=
tan
−
1
1
3
then
x
=
Report Question
0%
1
0%
2
0%
3
0%
√
3
Explanation
Let
y
=
tan
−
1
(
1
2
+
√
3
)
⟹
tan
y
=
1
2
+
√
3
tan
3
y
=
3
tan
y
−
tan
3
y
1
+
3
tan
2
y
=
1
2
+
√
3
(
3
−
1
(
2
+
√
3
)
2
)
1
+
3
(
2
+
√
3
)
2
=
1
2
+
√
3
(
20
+
12
√
3
)
4
+
4
√
3
=
20
+
12
√
3
20
+
12
√
3
=
1
⟹
3
tan
−
1
(
1
2
+
√
3
)
=
tan
−
1
(
1
)
tan
−
1
(
1
x
)
=
tan
−
1
(
1
)
−
tan
−
1
1
3
=
tan
−
1
1
−
1
3
1
+
1
3
=
1
2
⟹
x
=
2
n
∑
r
=
1
t
a
n
−
1
(
2
r
−
1
1
+
2
2
r
−
1
)
is equal to :
Report Question
0%
t
a
n
−
1
(
2
n
)
0%
t
a
n
−
1
(
2
n
)
−
π
4
0%
t
a
n
−
1
(
2
n
+
1
)
0%
t
a
n
−
1
(
2
n
+
1
)
−
π
4
Explanation
n
∑
r
=
1
tan
−
1
(
2
r
−
1
1
+
2
2
r
−
1
)
=
n
∑
r
=
1
tan
−
1
(
2
r
−
1
(
2
−
1
)
1
+
2
r
.2
r
−
1
)
=
n
∑
r
=
1
tan
−
1
(
2
r
−
2
r
−
1
1
+
2
r
.2
r
−
1
)
=
n
∑
r
=
1
tan
−
1
2
r
−
tan
−
1
(
2
r
−
1
)
=
tan
−
1
2
−
tan
−
1
1
+
tan
−
1
2
2
−
tan
−
1
2
1
+
⋯
+
tan
−
1
2
n
−
tan
−
1
2
n
−
1
=
tan
−
1
(
2
n
)
−
π
4
For
0
<
x
<
1
the value of
cos
−
1
x
+
cos
−
1
(
−
x
)
=
?
Report Question
0%
0
0%
π
0%
−
π
0%
none of these
Explanation
We've for
|
x
|
≤
1
,
cos
−
1
(
−
x
)
=
π
−
cos
−
1
x
.
It also holds in this case.
So for
0
<
x
<
1
we've,
cos
−
1
x
+
cos
−
1
(
−
x
)
=
cos
−
1
x
+
π
−
cos
−
1
x
=
π
.
The value of
sin
h
(
cos
h
−
1
x
)
is
Report Question
0%
√
x
2
+
1
0%
1
/
√
x
2
+
1
0%
√
x
2
−
1
0%
n
o
n
e
o
f
t
h
e
s
e
The value of
∞
∑
r
=
0
t
a
n
1
(
1
1
+
r
+
r
2
)
is equal to
Report Question
0%
π
4
0%
−
π
2
0%
π
0%
π
2
Explanation
given,
∞
∑
r
=
0
tan
−
1
(
1
1
+
r
+
r
2
)
Consider
tan
−
1
(
1
1
+
r
+
r
2
)
=
tan
−
1
(
1
1
+
r
(
r
+
1
)
)
=
tan
−
1
(
(
r
+
1
)
−
r
1
+
(
r
+
1
)
r
)
=
tan
−
1
(
r
+
1
)
−
tan
−
1
r
G.E
=
∞
∑
r
=
0
tan
−
1
(
r
+
1
)
−
tan
−
1
(
r
)
=
tan
−
1
1
−
tan
−
1
0
+
tan
−
1
2
−
tan
−
1
1
+
.
.
.
…
+
tan
−
1
∞
=
tan
−
1
∞
−
tan
−
1
0
=
π
2
−
0
=
π
2
.
The value of
s
i
n
−
1
(
s
i
n
12
)
+
c
o
s
−
1
(
c
o
s
12
)
is equal to :
Report Question
0%
Z
e
r
o
0%
24
−
2
π
0%
4
π
−
24
0%
None of these
Explanation
Given,
sin
−
1
(
sin
12
)
+
cos
−
1
(
cos
12
)
The principle values,
sin
−
1
x
∈
[
−
π
2
,
π
2
]
,
∀
x
∈
[
−
1
,
1
]
cos
−
1
x
∈
[
0
,
π
]
,
∀
x
∈
[
−
1
,
1
]
∴
sin
−
1
(
sin
12
)
≠
12
∉
[
−
π
2
,
π
2
]
∴
cos
−
1
(
cos
12
)
≠
12
∉
[
0
,
π
]
∴
sin
−
1
(
sin
12
)
+
cos
−
1
(
cos
12
)
=
sin
−
1
(
sin
(
12
−
4
π
)
)
+
cos
−
1
(
cos
(
4
π
−
12
)
)
=
(
12
−
4
π
)
+
(
4
π
−
12
)
=
0
∴
sin
−
1
(
sin
12
)
+
cos
−
1
(
cos
12
)
=
0
The simplified form of
c
o
s
−
1
(
3
5
c
o
s
x
+
4
5
s
i
n
x
)
is :
Report Question
0%
t
a
n
−
1
4
3
−
x
0%
t
a
n
−
1
1
3
−
x
0%
t
a
n
−
1
4
3
+
x
0%
None of these
∞
∑
r
=
1
tan
−
1
(
3
r
2
−
r
+
9
)
is -
Report Question
0%
π
3
0%
π
6
0%
π
2
0%
π
12
I
f
cos
(
2
sin
−
1
x
)
=
1
9
,
t
h
e
n
x
i
s
e
q
u
a
l
t
o
Report Question
0%
O
n
l
y
2
3
0%
O
n
l
y
−
2
3
0%
2
3
,
−
2
3
0%
1
3
If
c
o
s
−
1
3
5
−
s
i
n
−
1
4
5
=
c
o
s
−
1
x
, then x is equal to -
Report Question
0%
0
0%
1
0%
-1
0%
None of these
Explanation
Given,
cos
−
1
3
5
+
sin
−
1
4
5
=
cos
−
1
x
apply
cos
on both sides, we get,
cos
(
cos
−
1
3
5
+
sin
−
1
4
5
)
=
cos
(
cos
−
1
x
)
U
s
e
t
h
e
f
o
l
l
o
w
i
n
g
i
d
e
n
t
i
t
y
:
cos
(
s
+
t
)
=
cos
(
s
)
cos
(
t
)
−
sin
(
s
)
sin
(
t
)
cos
(
cos
−
1
(
3
5
)
)
cos
(
sin
−
1
(
4
5
)
)
−
sin
(
cos
−
1
(
3
5
)
)
sin
(
sin
−
1
(
4
5
)
)
=
x
3
5
(
√
1
−
(
4
5
)
2
)
−
4
5
(
√
1
−
(
3
5
)
2
)
=
x
3
5
×
3
5
−
4
5
×
4
5
=
x
9
25
−
16
25
=
x
∴
x
=
−
7
25
Number of solution of the equation
tan
−
1
(
1
x
−
1
+
1
x
−
2
+
1
x
−
3
+
1
x
−
4
)
+
cos
−
1
(
x
)
=
3
π
4
−
sin
−
1
(
x
)
are
Report Question
0%
0
0%
1
0%
2
0%
3
The range of the function
f
(
x
)
=
ℓ
n
(
sin
−
1
(
x
2
+
x
)
)
is
Report Question
0%
[
−
ℓ
n
(
sin
−
1
1
4
)
,
ℓ
n
π
4
]
0%
[
−
ℓ
n
π
4
,
ℓ
n
π
2
]
0%
(
0
,
ℓ
n
π
2
]
0%
(
−
∞
,
ℓ
n
π
2
]
The trigonometric equation
sin
−
1
x
=
2
sin
−
1
a
, has a solution for-
Report Question
0%
|
a
|
≤
1
√
2
0%
1
2
<
|
a
|
<
1
√
2
0%
a
l
l
r
e
a
l
v
a
l
u
e
s
o
f
a
0%
|
a
|
<
1
2
Explanation
The trigonometric equation
sin
−
1
x
=
2
sin
−
1
a
will have a solution if
2
sin
−
1
a
∈
[
−
π
2
,
π
2
]
(
∵
sin
−
1
x
∈
[
−
π
2
,
π
2
]
)
⟹
sin
−
1
a
∈
[
−
π
4
,
π
4
]
⟹
a
∈
[
−
sin
π
4
,
sin
π
4
]
⟹
a
∈
[
−
1
√
2
,
1
√
2
]
⟹
|
a
|
≤
1
√
2
If
f
(
x
)
=
cos
−
1
(
√
2
x
2
+
1
x
2
+
1
)
, then range of
f
(
x
)
is
Report Question
0%
[
0
,
π
]
0%
(
0
,
π
4
]
0%
(
0
,
π
3
]
0%
[
0
,
π
2
)
Explanation
Given,
f
(
x
)
=
cos
−
1
(
√
2
x
2
+
1
x
2
+
1
)
Range of
√
2
x
2
+
1
x
2
+
1
0
<
f
(
x
)
≤
1
cos
−
1
(
1
)
≤
cos
−
1
(
√
2
x
2
+
1
x
2
+
1
)
<
cos
−
1
(
0
)
0
≤
cos
−
1
(
√
2
x
2
+
1
x
2
+
1
)
<
π
2
Therefore the range is,
0
≤
f
(
x
)
<
π
2
Number of solution(s) to the equation
cos
−
1
x
+
sin
−
1
(
x
2
)
=
π
6
is/are
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
Hence,
option
(
B
)
is correct answer.
c
o
s
−
1
(
π
3
+
s
e
c
−
1
(
−
2
)
)
=
Report Question
0%
-1
0%
1
0%
0
0%
None of these
Explanation
As we know that
sec
−
1
(
−
2
)
=
2
π
3
∴
cos
(
π
3
+
sec
−
1
(
−
2
)
)
=
cos
(
π
3
+
2
π
3
)
=
cos
π
=
−
1
The value of
cot
(
c
o
s
e
c
−
1
5
3
+
tan
−
1
2
3
)
is equal to-
Report Question
0%
6
17
0%
3
17
0%
4
17
0%
5
17
∫
π
0
[
c
o
t
x
]
d
x
,
w
h
e
r
e
[
⋅
]
denotes the greatest integer function, is equal to:
Report Question
0%
1
0%
-1
0%
−
π
2
0%
π
2
s
e
c
h
−
1
(
s
i
n
θ
)
=
Report Question
0%
l
o
g
(
t
a
n
θ
2
)
0%
l
o
g
(
s
i
n
θ
2
)
0%
l
o
g
(
c
o
s
θ
2
)
0%
l
o
g
(
c
o
t
θ
2
)
cos
−
1
[
cos
(
−
17
15
π
)
]
is equal to
Report Question
0%
−
17
π
15
0%
17
π
15
0%
13
π
15
0%
−
2
π
15
Find
∫
x
.
sin
x
d
x
Report Question
0%
sin
x
+
x
cos
x
=
C
0%
sin
x
−
x
cos
x
+
C
0%
sin
x
+
x
sin
x
+
C
0%
sin
x
−
x
sin
x
+
C
If
A
=
tan
1
−
(
x
√
3
2
k
−
x
)
and
B
=
tan
−
1
(
2
x
−
k
k
√
3
)
then
A
.
B
=
Report Question
0%
0
o
0%
π
/
6
0%
π
/
4
0%
π
/
3
if x
>
0
t
h
e
n
t
a
n
h
−
1
(
x
2
−
1
x
2
+
1
)
Report Question
0%
l
o
g
e
(
2
x
)
0%
l
o
g
e
x
0%
l
o
g
e
(
3
x
)
0%
l
o
g
e
(
5
x
)
i
f
x
>
0
t
h
e
n
t
a
n
h
−
1
(
x
2
−
1
x
2
+
1
)
Report Question
0%
l
o
g
e
(
2
x
)
0%
l
o
g
e
x
0%
l
o
g
e
(
3
x
)
0%
l
o
g
e
(
5
x
)
If
cos
−
1
x
−
cos
−
1
(
y
2
)
=
α
a
x
2
−
4
x
y
cos
α
+
y
2
=
Report Question
0%
−
4
sin
2
α
0%
4
sin
2
α
0%
4
0%
2
sin
2
α
If
cot
−
1
x
+
tan
−
1
(
1
3
)
=
π
2
, then
x
will be
Report Question
0%
1
0%
3
0%
1
3
0%
None of these
Explanation
Given
cot
−
1
x
+
tan
−
1
(
1
3
)
=
π
2
⟹
cot
−
1
x
=
π
2
−
tan
−
1
(
1
3
)
⟹
cot
−
1
x
=
cot
−
1
(
1
3
)
⟹
x
=
1
3
If
x
=
sin
−
1
(
sin
10
)
and
y
=
cos
−
1
(
cos
10
)
, then find
y
−
x
.
Report Question
0%
π
0%
7
π
0%
0
0%
10
Explanation
Given,
x
=
s
i
n
−
1
(
s
i
n
10
)
x
=
3
π
−
10
Also,
y
=
c
o
s
−
1
(
c
o
s
10
)
y
=
4
π
−
10
Therefore,
y
−
x
=
4
π
−
10
−
3
π
+
10
=
π
State true or false.
sin
−
1
x
+
cos
−
1
x
=
π
2
Report Question
0%
True
0%
False
Explanation
As we know that
For
y
∈
[
−
1
,
1
]
,
sin
−
1
y
+
cos
−
1
y
=
π
2
So
sin
−
1
x
+
cos
−
1
x
=
π
2
So the relation is
True
sin
−
1
(
3
5
)
+
cos
−
1
(
12
13
)
=
sin
−
1
(
56
65
)
Report Question
0%
True
0%
False
cos
−
1
{
1
2
x
2
+
√
1
−
x
2
√
1
−
x
2
4
}
=
cos
−
1
x
2
−
cos
−
1
x
holds for:
Report Question
0%
|
x
|
≤
1
0%
x
∈
R
0%
0
≤
x
≤
1
0%
−
1
≤
x
≤
0
Explanation
c
l
e
a
r
l
y
,
x
/
2
ϵ
[
−
1
,
1
]
&
x
ϵ
[
−
1
,
1
]
x
ϵ
[
−
2
,
2
]
&
x
ϵ
[
−
1
,
1
]
Intersection of above sells is [-1,1]
Thus
,
[
|
x
|
≤
1
]
cos
−
1
(
41
/
49
)
sin
−
1
(
2
/
7
)
=
Report Question
0%
4
0%
3
0%
2
0%
1
Explanation
Step-1: Rewrite the Numerator observing Denominator
& apply multiple - angles of trigonometric function.
We have,
cos
−
1
(
41
/
49
)
sin
−
1
(
2
/
7
)
.....(i)
Now,
cos
−
1
(
1
−
2
(
2
7
)
2
)
Put
sin
θ
=
2
7
∴
cos
−
1
(
1
−
2
sin
2
θ
)
cos
−
1
(
cos
2
θ
)
\boldsymbol{[\cos 2 \theta = 1 - 2 \sin^{2} \theta]}
= 2\theta
\textbf{Step-2: Use the above results & get the required unknown.}
\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\
= \dfrac{2 \theta}{\theta} = 2
[\textbf{using (i)}]
\textbf{Hence, option - C is the answer}
The value of
tan(\frac { 1 }{ 2 } { cos }^{ -1 }(\frac { \sqrt { 5 } }{ 3 } ))
is
Report Question
0%
\frac { 3+\sqrt { 5 } }{ 2 }
0%
3-\sqrt { 5 }
0%
\frac { 1 }{ 2 } (3-\sqrt { 5 } )
0%
none of these
Explanation
Given,
y=\tan \left [ \dfrac{1}{2}\cos ^{-1}\left ( \dfrac{\sqrt{5}}{3} \right ) \right ]
Let,
x=\cos ^{-1}\left ( \dfrac{\sqrt{5}}{3} \right )
\Rightarrow \cos x=\dfrac{\sqrt{5}}{3}
\therefore y=\tan \dfrac{1}{2}x
y=\tan \dfrac{x}{2}
y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}
=\sqrt{\dfrac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}
=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}
rationalizing the factor, we get,
y=\dfrac{3-\sqrt{5}}{2}
\sin^{-1}\left(\dfrac{4}{5}\right)+\sin^{-1}\left(\dfrac{7}{25}\right)=\sin^{-1}\left(\dfrac{117}{125}\right)
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0%
True
0%
False
If tan (x + y) = 33 and x =
{ tan }^{ -1 }3
, then y will be
Report Question
0%
0.3
0%
{ tan }^{ -1 }(1.3)
0%
{ tan }^{ -1 }(0.3)
0%
{ tan }^{ -1 }(\frac { 1 }{ 18 } )
Explanation
Given,
\tan (x+y)=33
x=\tan ^{-1}3
\Rightarrow \tan x=3
Formula,
\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}
33=\dfrac{3+\tan y}{1-(3)\tan y}
33(1-3\tan y)=3+\tan y
33-99\tan y=3+\tan y
30=100\tan y
\tan y=\dfrac{30}{100}=0.3
\therefore y=\tan ^{-1}0.3
If
sin^{-1}(\dfrac{1}{3}) + sin^{-1}(\dfrac{2}{3}) = sin^{-1} x
, then x is equal to
Report Question
0%
0
0%
\dfrac{\sqrt{5}+4\sqrt{2}}{9}
0%
\dfrac{5\sqrt{2}-4\sqrt{5}}{9}
0%
\dfrac{\pi}{2}
Explanation
\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{2}{3}\\\sin ^{-1}\left\{\dfrac{1}{3}\sqrt{1-\left(\dfrac{2}{3}\right)^2}+\dfrac{2}{3}\sqrt{1-\left(\dfrac{1}{3}\right)^2}\right\}\\\sin^{-1}\left\{\dfrac{1}{3}\left(\dfrac{\sqrt5}{3}\right)+\dfrac{2}{3}\left(\dfrac{\sqrt8}{3}\right)\right\}\\\sin^{-1}\left\{\dfrac{\sqrt5+4\sqrt2}{9}\right\}\\x=\dfrac{\sqrt5+4\sqrt2}{9}
The value of
\sin ^{-1}(\sin 5\frac {\pi}{3})=
Report Question
0%
-\frac {\pi}3
0%
\frac {\pi}3
0%
\frac {4\pi}3
0%
\frac {3\pi}3
Explanation
Given,
\sin ^{-1}\left (\sin 5\dfrac{\pi }{3} \right )
Let
\theta =\sin ^{-1}\left (\sin 5\dfrac{\pi }{3} \right )
\sin \theta =\left (\sin 5\dfrac{\pi }{3} \right )=\sin \left ( 2\pi - \dfrac{\pi }{3}\right )
\Rightarrow \sin \theta =-\sin \dfrac{\pi }{3}
\therefore \theta =-\dfrac{\pi }{3}
The value of
\displaystyle sec\left [ sin^{-1}\left (sin \dfrac{50\pi }{9} \right ) + cos^{-1}cos\left ( \dfrac{31\pi }{9} \right ) \right ]
is equal to
Report Question
0%
sec\dfrac{10\pi}{9}
0%
sec \ 9{\pi}
0%
-1
0%
1
Explanation
Given,
\sec \left [ \sin ^{-1}\left ( \sin \dfrac {50\pi }{9} \right ) +\cos ^{-1}\left ( \cos \dfrac {31\pi }{9} \right ) \right ]
=\sec \left [ \dfrac {50\pi }{9} + \dfrac {31\pi }{9}\right ]
=\sec \dfrac {81\pi }{9}
=\sec 9\pi
\tan { ^{ -1 }\left( 3/5 \right) } +\tan { ^{ -1 }\left( 1/4 \right) } =
Report Question
0%
0
0%
\pi /4
0%
3n/4
0%
None of these
Explanation
given,
{\tan}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}
={\tan}^{-1}{\left(\dfrac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{5}\times\frac{1}{4}}\right)}
={\tan}^{-1}{\left(\dfrac{\frac{12+5}{20}}{1-\frac{3}{20}}\right)}
={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{20-3}{20}}\right)}
={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{17}{20}}\right)}
={\tan}^{-1}{\left(1\right)}
=\dfrac{\pi}{4}
\tan { ^{ -1 }\left( \tan { 3\pi /4 } \right) = }
Report Question
0%
5\pi /4
0%
\pi /4
0%
-\pi /4
0%
None of these
Explanation
Let
y={\tan}^{-1}{\left(\tan{\dfrac{3\pi}{4}}\right)}
\Rightarrow \tan{y}=\tan{\left(\pi-\dfrac{\pi}{4}\right)}
\Rightarrow \tan{y}=\tan{\dfrac{\pi}{4}}
\therefore y=\dfrac{\pi}{4},\pi+\dfrac{\pi}{4}
\Rightarrow y=\dfrac{\pi}{4},\dfrac{4\pi+\pi}{4}
\therefore y=\dfrac{\pi}{4},\dfrac{5\pi}{4}
The value of
\sin^{-1}(\cos (\log_{2}(4\alpha -44)))
is
Report Question
0%
\dfrac {\pi}{3}
0%
\dfrac {\pi}{2}
0%
0
0%
1
\tan (2\cos ^{-1}\frac 35)=
_____
Report Question
0%
\frac 83
0%
\frac {24}{25}
0%
\frac 7{25}
0%
\frac {-24}7
Explanation
Given,
\tan \left(2\cos ^{-1}\left(\frac{3}{5}\right)\right)
as
\tan \left(2x\right)=\dfrac{2\tan \left(x\right)}{1-\tan ^2\left(x\right)}
=\dfrac{2\tan \left(\cos ^{-1} \left(\frac{3}{5}\right)\right)}{1-\tan ^2\left(\cos ^{-1} \left(\frac{3}{5}\right)\right)}
as
\tan \left(\cos ^{-1}\left(x\right)\right)=\frac{\sqrt{1-x^2}}{x}
=\dfrac{2\left(\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\left(\frac{3}{5}\right)}\right)}{1-\left(\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\left(\frac{3}{5}\right)}\right)^2}
=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\left(\frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}\right)^2}
=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\frac{4^2}{3^2}}
=\dfrac{\frac{8}{3}}{1-\frac{4^2}{3^2}}
=\dfrac{8}{3\left(1-\frac{16}{9}\right)}
=-\dfrac{8}{\frac{7}{3}}
=-\dfrac{24}{7}
{ \tan }^{ -1 }\left( \dfrac { 1 }{ 7 } \right) +{ \tan }^{ -1 }\left( \dfrac { 1 }{ 13 } \right) =\cos ^{ -1 }{ \left( \dfrac { 9 }{ 2 } \right) }
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0%
True
0%
False
The solution set of the equation
2 cos^{ -1 } x = cot^{ -1 } \left(\dfrac { 2x^{ 2 } - 1 }{ 2x \sqrt { 1 x^{ 2 } } }\right)
is
Report Question
0%
\left(0, 1\right)
0%
\left(-1, 1\right)-{ 0 }
0%
\left(-1, 0\right)
0%
\left[-1, 1\right]
{ \cos }^{ -1 }\left[ \cos \left( 2{ \cot }^{ -1 }\left( \sqrt { 2 } -1 \right) \right) \right]
is equal to
Report Question
0%
\sqrt{2}-1
0%
1-\sqrt{2}
0%
\pi/4
0%
3\pi/4
The number of solutions of the equation
3\cos^{-1}x-\pi x-\dfrac {\pi}{2}=0
Report Question
0%
0
0%
1
0%
2
0%
infinite
If
\alpha =\cos^{-1}\left(\dfrac{3}{5}\right),\beta=\tan^{-1}\left(\dfrac{1}{3}\right)
where
0<\alpha,beta <\dfrac{\pi}{2},
then
\alpha -\beta
is equal to :
Report Question
0%
\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)
0%
\tan^{-1}\left(\dfrac{9}{14}\right)
0%
\cos^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)
0%
\tan^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)
Explanation
\cos \alpha=\dfrac{3}{5},\tan \beta =\dfrac{1}{5}
\Rightarrow \tan \alpha =\dfrac{4}{3}
\Rightarrow tan (\alpha-\beta)=\dfrac{\dfrac{4}{3}-\dfrac{1}{3}}{1+\dfrac{4}{3}.\dfrac{1}{3}}=\dfrac{9}{3}
\Rightarrow \sin(\alpha-\beta)=\dfrac{9}{5\sqrt{10}}
\Rightarrow \alpha-\beta =\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)
If
\;\sin {\;^{ - 1}}\dfrac{1}{x} = 2\;{\tan ^{ - 1}}\dfrac{1}{7} + {\cos ^{ - 1}}\dfrac{3}{5}
, then
x =
___
Report Question
0%
\dfrac {24}{117}
0%
\dfrac {7}{3}
0%
\dfrac {125}{117}
0%
None of these
Explanation
2\tan ^{-1}\left(\dfrac{1}{7}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)
=\tan ^{-1}\left(\dfrac{1}{7}\right)+\tan ^{-1}\left(\dfrac{1}{7}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)
=\tan ^{-1}\left(\dfrac{\dfrac{1}{7}+\dfrac{1}{7}}{1-\dfrac{1}{7}\times \dfrac{1}{7}}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)
\left[\because \tan ^{-1}+\tan ^{-1}y \tan ^{-1x}=\tan ^{-1}\left(\dfrac{x+y}{1-xy}\right)\right]
=\tan ^{-1}\left(\dfrac{\dfrac{2}{7}}{1-\dfrac{1}{49}}\right)+\theta
=\tan ^{-1}\left(\dfrac{7}{24}\right)+\tan ^{-1}\left(\dfrac{4}{3}\right)
=\tan ^{-1}\left(\dfrac{\dfrac{7}{24}+\dfrac{4}{3}}{1-\dfrac{7}{24}\times \dfrac{4}{3}}\right)
=\tan ^{-1}\left(\dfrac{117}{44}\right)
\because \sin ^{-1}\left(\dfrac{1}{x}\right)=\tan ^{-1}\left(\dfrac{117}{44}\right)
Let
\theta =\tan ^{-1}\left(\dfrac{117}{44}\right)
\therefore \tan \theta=\dfrac{117}{44}
\therefore \sin \theta=\dfrac{117}{\sqrt{117^2+44^2}}=\dfrac{117}{125}
\therefore \theta =\sin ^{-1}\left(\dfrac{117}{125}\right)
\therefore \sin ^{-1}\left(\dfrac{1}{x}\right)=\sin ^{-1}\left(\dfrac{117}{125}\right)
On comparing
\dfrac{1}{x}=\dfrac{117}{125}
\boxed{\therefore x=\dfrac{125}{117}}
If
\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha
where
-1-1\le x\le 1,-2\le y\le 2,x\le \cfrac { y }{ 2 }
then for all
4{ x }^{ 2 }-4xy\cos { \alpha } +{ y }^{ 2 }
is equal to
Report Question
0%
4\sin ^{ 2 }{ \alpha } -2{ x }^{ 2 }{ y }^{ 2 }
0%
4\cos ^{ 2 }{ \alpha } +2{ x }^{ 2 }{ y }^{ 2 }
0%
4\sin ^{ 2 }{ \alpha }
0%
2\sin ^{ 2 }{ \alpha }
Explanation
\cos { \left( \cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } \right) } =\cos { \alpha }
\cos { \alpha } \Rightarrow x\times \cfrac { y }{ 2 } +\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { y }^{ 2 } }{ 4 } } \Rightarrow { \left( \cos { \alpha } -\cfrac { xy }{ 2 } \right) }^{ 2 }=\left( 1-{ x }^{ 2 } \right) \left( 1-\cfrac { { y }^{ 2 } }{ 4 } \right)
{ x }^{ 2 }+\cfrac { { y }^{ 2 } }{ 4 } -xy\cos { \alpha } =1-\cos ^{ 2 }{ \alpha } =\sin ^{ 2 }{ \alpha } \quad
The value of
tan \left (\cos^{-1} \dfrac {3}{5} + \tan^{-1} \dfrac {1}{4}\right )
is
Report Question
0%
\dfrac {19}{8}
0%
\dfrac {8}{19}
0%
\dfrac {19}{12}
0%
\dfrac {3}{4}
Explanation
Given,
\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}
Let,
{\cos}^{-1}{\dfrac{3}{5}}=a
\Rightarrow\,\cos{a}=\dfrac{3}{5}
\Rightarrow\,{\sin}^{2}{a}=1-{\cos}^{2}{a}=1-\dfrac{9}{25}=\dfrac{25-9}{25}=\dfrac{16}{25}
\therefore\,\sin{a}=\dfrac{4}{5}
\Rightarrow\,\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}
\Rightarrow\,a={\tan}^{-1}{\dfrac{4}{3}}
\therefore\,\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}
=\tan{\left(a+{\tan}^{-1}{\dfrac{1}{4}}\right)}
where
a={\cos}^{-1}{\dfrac{3}{5}}
=\tan{\left({\tan}^{-1}{\dfrac{4}{3}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}
where
a={\cos}^{-1}{\dfrac{3}{5}}={\tan}^{-1}{\dfrac{4}{3}}
We know that
{\tan}^{-1}{A}+{\tan}^{-1}{B}={\tan}^{-1}{\left(\dfrac{A+B}{1-AB}\right)}
=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{4}{3}+\dfrac{1}{4}}{1-\dfrac{4}{3}\times\dfrac{1}{4}}\right)}\right)}
=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{16+3}{12}}{1-\dfrac{4}{12}}\right)}\right)}
=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{19}{12}}{\dfrac{12-4}{12}}\right)}\right)}
=\tan{\left({\tan}^{-1}{\dfrac{19}{8}}\right)}
=\dfrac{19}{8}
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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