Processing math: 14%
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
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
]
∴
\therefore \cos ^{-1}(\cos 12)\neq 12\notin \left [ 0,\pi \right ]
\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)
=\sin ^{-1}(\sin (12-4\pi ))+\cos ^{-1}(\cos (4\pi -12))
=(12-4\pi ) + (4\pi -12)
=0
\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)=0
The simplified form of
{ cos }^{ -1 }\left( \frac { 3 }{ 5 } { cos }x+\frac { 4 }{ 5 } { sin }x \right)
is :
Report Question
0%
{ tan }^{ -1 }\frac { 4 }{ 3 } -x
0%
{ tan }^{ -1 }\frac { 1 }{ 3 } -x
0%
{ tan }^{ -1 }\frac { 4 }{ 3 } +x
0%
None of these
\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {3}{r^{2}-r+9}\right)
is -
Report Question
0%
\dfrac {\pi}{3}
0%
\dfrac {\pi}{6}
0%
\dfrac {\pi}{2}
0%
\dfrac {\pi}{12}
If\,\cos \left( {2{{\sin }^{ - 1}}x} \right) = \frac{1}{9},\,then\,\,x\,\,is\,\,equal\,\,to
Report Question
0%
Only\,\frac{2}{3}\,
0%
Only - \frac{2}{{3\,}}
0%
\,\frac{2}{3}, - \frac{2}{3}
0%
\frac{1}{3}
If
{ cos }^{ -1 }\dfrac { 3 }{ 5 } -{ sin }^{ -1 }\dfrac { 4 }{ 5 } ={ cos }^{ -1 }x
, then x is equal to -
Report Question
0%
0
0%
1
0%
-1
0%
None of these
Explanation
Given,
\cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5}=\cos ^{-1}x
apply
\cos
on both sides, we get,
\cos \left ( \cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5} \right )=\cos (\cos ^{-1}x)
\mathrm{Use\:the\:following\:identity}:\quad \cos \left(s+t\right)=\cos \left(s\right)\cos \left(t\right)-\sin \left(s\right)\sin \left(t\right)
\cos \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\cos \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)-\sin \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\sin \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)=x
\dfrac{3}{5}\left ( \sqrt{1-\left (\dfrac{4}{5} \right )^2} \right )-\dfrac{4}{5}\left ( \sqrt{1-\left (\dfrac{3}{5} \right )^2} \right )=x
\dfrac{3}{5}\times \dfrac{3}{5}-\dfrac{4}{5}\times \dfrac{4}{5}=x
\dfrac{9}{25}-\dfrac{16}{25}=x
\therefore x=-\dfrac{7}{25}
Number of solution of the equation
\tan^{-1}\left(\dfrac {1}{x-1}+\dfrac {1}{x-2}+\dfrac {1}{x-3}+\dfrac {1}{x-4} \right)+\cos^{-1}(x)=\dfrac {3\pi}{4}-\sin^{-1}(x)
are
Report Question
0%
0
0%
1
0%
2
0%
3
The range of the function
f(x)=\ell n\ (\sin^{-1}(x^{2}+x))
is
Report Question
0%
\left[-\ell n \left(\sin^{-1}\dfrac {1}{4}\right),\ell n \dfrac {\pi}{4}\right]
0%
\left[-\ell n \dfrac {\pi}{4},\ell n \dfrac {\pi}{2}\right]
0%
\left(0,\ell n \dfrac {\pi}{2}\right]
0%
\left(-\infty ,\ell n \dfrac {\pi}{2}\right]
The trigonometric equation
\sin^{-1}x=2\sin^{-1}a
, has a solution for-
Report Question
0%
\left|a\right|\le\dfrac{1}{\sqrt{2}}
0%
\dfrac{1}{2}<\left|a\right|<\dfrac{1}{\sqrt{2}}
0%
all\ real\ values\ of\ a
0%
\left|a\right|<\dfrac{1}{2}
Explanation
The trigonometric equation
\text{sin}^{-1} x=2\text{sin}^{-1} a
will have a solution if
2\text{sin}^{-1} a\in \bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigg]
(\because\text{sin}^{-1} x\in [-\frac{\pi}{2},\frac{\pi}{2}])
\implies \text{sin}^{-1} a\in \bigg[-\dfrac{\pi}{4},\dfrac{\pi}{4}\bigg]
\implies a\in \bigg[-\sin \dfrac{\pi}{4},\sin \dfrac{\pi}{4}\bigg]
\implies a\in \bigg[-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\bigg]
\implies |a|\le \dfrac{1}{\sqrt{2}}
If
f(x)=\cos^{-1}\left(\dfrac {\sqrt {2x^{2}+1}}{x^{2}+1}\right)
, then range of
f(x)
is
Report Question
0%
[0,\pi]
0%
\left(0,\dfrac {\pi}{4}\right]
0%
\left(0,\dfrac {\pi}{3}\right]
0%
\left[0,\dfrac {\pi}{2}\right)
Explanation
Given,
f\left(x\right)=\cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)
Range of
\dfrac{\sqrt{2x^2+1}}{x^2+1}
0< f(x)\leq1
\cos^{-1}(1)\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)<\cos^{-1}(0)
0\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)< \dfrac{\pi}{2}
Therefore the range is,
0\leq f(x)< \dfrac{\pi}{2}
Number of solution(s) to the equation
{\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{\pi }{6}\,
is/are
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
Hence,
option
(B)
is correct answer.
cos^{-1}\left(\dfrac{\pi}{3}+sec^{-1}(-2)\right)
=
Report Question
0%
-1
0%
1
0%
0
0%
None of these
Explanation
As we know that
\text{sec}^{-1}(-2)=\dfrac{2\pi}{3}
\therefore \cos \bigg(\dfrac{\pi}{3}+\text{sec}^{-1}(-2)\bigg)=\cos \bigg(\dfrac{\pi}{3}+\dfrac{2\pi}{3}\bigg)=\cos \pi=-1
The value of
\cot\left(cosec^{-1}\dfrac{5}{3}+\tan^{-1}\dfrac{2}{3}\right)
is equal to-
Report Question
0%
\dfrac{6}{17}
0%
\dfrac{3}{17}
0%
\dfrac{4}{17}
0%
\dfrac{5}{17}
\int _{ 0 }^{ \pi }{ \left[ cotx \right] dx,where\left[ \cdot \right] }
denotes the greatest integer function, is equal to:
Report Question
0%
1
0%
-1
0%
-\dfrac { \pi }{ 2 }
0%
\dfrac { \pi }{ 2 }
{ sec\quad h }^{ -1 }\left( sin\quad \theta \right) =
Report Question
0%
log\left( tan\frac { \theta }{ 2 } \right)
0%
log\left( sin\frac { \theta }{ 2 } \right)
0%
log\left( cos\frac { \theta }{ 2 } \right)
0%
log\left( cot\frac { \theta }{ 2 } \right)
\cos^{ -1 }\left[\cos\left( -\frac { 17 }{ 15 } \pi \right) \right]
is equal to
Report Question
0%
-\frac { 17\pi }{ 15 }
0%
\frac { 17\pi }{ 15 }
0%
\frac { 13\pi }{ 15 }
0%
\frac { -2\pi }{ 15 }
Find
\displaystyle \int x.\sin xdx
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-}\left(\dfrac {x\sqrt {3}}{2k-x}\right)
and
B=\tan^{-1}\left(\dfrac {2x-k}{k\sqrt {3}}\right)
then
A.B=
Report Question
0%
0^{o}
0%
\pi /6
0%
\pi /4
0%
\pi /3
if x
>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right)
Report Question
0%
{ log }_{ e }\left( 2x \right)
0%
{ log }_{ e }x
0%
{ log }_{ e }\left( 3x \right)
0%
{ log }_{ e }\left( 5x \right)
if\quad x>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right)
Report Question
0%
{ log }_{ e }\left( 2x \right)
0%
{ log }_{ e }x
0%
{ log }_{ e }\left( 3x \right)
0%
{ log }_{ e }\left( 5x \right)
If
\cos^{-1}x-\cos^{-1}(\dfrac {y}{2})=\alpha
ax^{2}-4xy\cos \alpha +y^{2}=
Report Question
0%
-4\sin^{2}\alpha
0%
4\sin^{2}\alpha
0%
4
0%
2\sin 2\alpha
If
\cot^{-1}{x}+\tan^{-1}\left (\dfrac{1}{3}\right)=\dfrac{\pi}{2}
, then
x
will be
Report Question
0%
1
0%
3
0%
\dfrac {1}{3}
0%
None of these
Explanation
Given
\text{cot}^{-1} x+\text{tan}^{-1}\bigg(\dfrac{1}{3}\bigg)=\dfrac{\pi}{2}
\implies \text{cot}^{-1} x=\dfrac{\pi}{2}-\text{tan}^{-1} \bigg(\dfrac{1}{3}\bigg)
\implies \text{cot}^{-1} x=\text{cot}^{-1}\bigg(\dfrac{1}{3}\bigg)
\implies x=\dfrac{1}{3}
If
x={ \sin }^{ -1 }(\sin10)
and
y={ \cos }^{ -1 }(\cos10)
, then find
y - x
.
Report Question
0%
\pi
0%
7\pi
0%
0
0%
10
Explanation
Given,
x=sin^{-1}(sin 10)
x=3\pi -10
Also,
y=cos^{-1}(cos 10)
y=4\pi - 10
Therefore,
y-x=4\pi-10-3\pi+10 = \pi
State true or false.
\sin^{-1}x+\cos^{-1}x=\dfrac {\pi}{2}
Report Question
0%
True
0%
False
Explanation
As we know that
For
y\in [-1,1],\text{sin}^{-1} y+\text{cos}^{-1} y=\dfrac{\pi}{2}
So
\text{sin}^{-1} x+\text{cos}^{-1} x=\dfrac{\pi}{2}
So the relation is
\text{True}
{ \sin }^{ -1 }\left (\dfrac { 3 }{ 5 }\right )+{ \cos }^{ -1 }\left (\dfrac { 12 }{ 13 }\right )={ \sin }^{ -1 }\left (\dfrac { 56 }{ 65 } \right)
Report Question
0%
True
0%
False
\cos ^{ -1 }{ \left\{ \dfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 } } \right\} =\cos ^{ -1 }{ \dfrac { x }{ 2 } -\cos ^{ -1 }{ x } } }
holds for:
Report Question
0%
|x|\le 1
0%
x\in R
0%
0\le x\le 1
0%
-1\le x\le 0
Explanation
clearly, x/2\epsilon[-1,1]
&
x\epsilon[-1,1]
x\epsilon [-2,2]
&
x\epsilon[-1,1]
Intersection of above sells is [-1,1]
Thus
,[\left | x \right |\leq 1]
\frac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}=
Report Question
0%
4
0%
3
0%
2
0%
1
Explanation
\textbf{Step-1: Rewrite the Numerator observing Denominator}
\textbf{& apply multiple - angles of trigonometric function.}
\text{We have,}
\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\
\text{.....(i)}
\text{Now,}\\
\cos^{-1} (1- 2 (\dfrac{2}{7})^2)
\text{Put}
\sin \theta = \dfrac{2}{7}
\therefore
\cos^{-1}(1- 2 \sin^2 \theta)
\cos^{-1} (\cos 2\theta)
\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)
Report Question
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) }
Report Question
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}
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page