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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 12
cos
2
θ
(
1
+
tan
2
θ
)
=
1
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Explanation
As we know that
sec
2
A
−
tan
2
A
=
1
cos
2
θ
(
1
+
tan
2
θ
)
=
cos
2
θ
×
sec
2
θ
=
cos
2
θ
×
1
cos
2
θ
=
1
So the relation is
True
csc
6
θ
−
cot
6
θ
=
1
+
3
csc
2
θ
cot
2
θ
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Explanation
csc
6
θ
−
cot
6
θ
=
(
csc
2
θ
)
3
−
(
cot
2
θ
)
3
=
(
csc
2
θ
−
cot
2
θ
)
(
csc
4
θ
+
cot
4
θ
+
csc
2
θ
cot
2
θ
)
=
(
1
)
(
(
csc
2
θ
)
2
+
(
cot
2
θ
)
2
−
2
csc
2
θ
cot
2
θ
+
3
csc
2
θ
cot
2
θ
)
=
(
csc
2
−
cot
2
θ
)
2
+
3
csc
2
θ
cot
2
θ
=
1
+
3
csc
2
θ
cot
2
θ
So the relation is
True
If
x
=
a
cos
θ
+
b
sin
θ
and
y
=
a
sin
θ
−
b
cos
θ
, then
a
2
+
b
2
=
x
2
+
y
2
.
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Explanation
Given,
x
=
a
cos
θ
+
b
sin
θ
......(1) and
y
=
a
sin
θ
−
b
cos
θ
......(2).
Now squaring (1) and (2) and then adding we get,
x
2
+
y
2
=
a
2
(
cos
2
θ
+
sin
2
θ
)
+
2
a
b
sin
θ
.
cos
θ
−
2
a
b
sin
θ
.
cos
θ
+
b
2
(
sin
2
θ
+
cos
2
θ
)
or,
x
2
+
y
2
=
a
2
+
b
2
.
(
cos
A
−
csc
A
)
2
+
(
sin
A
−
sec
A
)
2
=
(
1
−
sec
A
.
csc
A
)
2
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Explanation
Given
(
cos
A
−
csc
A
)
2
+
(
sin
A
−
sec
A
)
2
=
(
1
−
sec
A
csc
A
)
2
Taking LHS
(
cos
A
−
csc
A
)
2
+
(
sin
A
−
sec
A
)
2
⟹
cos
2
A
+
csc
2
A
−
2
cos
A
csc
A
+
sin
2
A
−
2
sin
A
sec
A
+
sec
2
A
⟹
(
sin
2
A
+
cos
2
A
)
+
csc
2
A
+
sec
2
A
−
2
(
cos
A
sin
A
)
−
2
(
sin
A
cos
A
)
⟹
1
+
1
sin
2
A
+
1
cos
2
A
−
2
(
cos
2
A
+
sin
2
sin
A
cos
A
)
⟹
1
+
(
cos
2
A
+
sin
2
sin
2
A
cos
2
A
)
−
2
(
1
sin
A
cos
A
)
⟹
1
+
(
1
sin
2
A
cos
2
A
)
−
2
sec
A
csc
A
⟹
1
+
sec
2
A
csc
2
A
−
2
sec
A
csc
A
⟹
(
1
−
sec
A
csc
A
)
2
∴
LHS
=
RHS
Hence Proved.
If
sin
θ
+
sin
2
θ
+
sin
3
θ
=
1
, then
cos
6
θ
−
4
cos
4
θ
+
8
cos
2
θ
=
4
.
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Explanation
Given
sin
θ
+
sin
2
θ
+
sin
3
θ
=
1
⟹
sin
θ
(
1
+
sin
2
θ
)
=
1
−
sin
2
θ
⟹
sin
θ
(
2
−
cos
2
θ
)
=
cos
2
θ
Squaring on both sides
⟹
sin
2
θ
(
2
−
cos
2
θ
)
2
=
cos
4
θ
⟹
(
1
−
cos
2
θ
)
(
4
+
cos
4
θ
−
4
cos
2
θ
)
=
cos
4
θ
⟹
4
+
cos
4
θ
−
4
cos
2
θ
−
4
cos
2
θ
−
cos
6
θ
+
4
cos
4
θ
=
cos
4
θ
⟹
cos
6
θ
−
4
cos
4
θ
+
8
cos
2
θ
=
4
So the relation is
True
√
sec
A
−
1
sec
A
+
1
+
√
sec
A
+
1
sec
A
−
1
=
2
csc
A
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Explanation
Given
√
sec
A
−
1
sec
A
+
1
+
√
sec
A
+
1
sec
A
−
1
Mulitiplying and Diving first term by
sec
A
−
1
Mulitiplying and Diving second term by
sec
A
+
1
√
(
sec
A
−
1
)
2
(
sec
A
+
1
)
(
sec
A
−
1
)
+
√
(
sec
A
+
1
)
2
(
sec
A
−
1
)
(
sec
A
+
1
)
√
(
sec
A
−
1
)
2
(
sec
2
A
−
1
)
+
√
(
sec
A
+
1
)
2
(
sec
2
A
−
1
)
(
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
)
√
(
sec
A
−
1
tan
A
)
2
+
√
(
sec
A
+
1
tan
A
)
2
(
sec
2
A
−
1
=
tan
2
A
)
sec
A
−
1
tan
A
+
sec
A
+
1
tan
A
2
sec
A
tan
A
2
(
1
cos
A
)
(
sin
A
cos
A
)
2
sin
A
2
csc
A
State true or false.
tan
2
θ
sec
2
θ
+
cot
2
θ
c
o
s
e
c
2
θ
=
1
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Explanation
Given,
t
a
n
2
θ
s
e
c
2
θ
+
c
o
t
2
θ
c
s
c
2
θ
=
sin
2
θ
cos
2
θ
×
cos
2
θ
+
cos
2
θ
sin
2
θ
×
sin
2
θ
=
sin
2
θ
+
cos
2
θ
=
1
State whether the following statement is true or false.
sin
A
sec
A
+
tan
A
−
1
+
cos
A
cosec
A
+
cot
A
−
1
=
1
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Explanation
⇒
L
.
H
.
S
=
sin
A
sec
A
+
tan
A
−
1
+
c
o
t
A
c
o
s
e
c
A
+
cot
A
−
1
=
sin
A
1
cos
A
+
sin
A
cos
A
−
1
+
cos
A
1
sin
A
+
cos
A
sin
A
−
1
=
sin
A
1
+
sin
A
−
cos
A
cos
A
+
cos
A
1
+
cos
A
−
sin
A
sin
A
=
sin
A
×
cos
A
1
+
sin
A
−
cos
A
+
sin
A
×
cos
A
1
+
cos
A
−
sin
A
=
sin
A
cos
A
[
1
1
+
sin
A
−
cos
A
+
1
1
+
cos
A
−
sin
A
]
=
sin
A
cos
A
[
1
+
cos
A
−
sin
A
+
1
+
sin
A
−
cos
A
(
1
+
sin
A
−
cos
A
)
(
1
+
cos
A
−
sin
A
)
]
=
sin
A
cos
A
[
2
1
+
cos
A
−
sin
A
+
sin
A
+
sin
A
cos
A
−
sin
2
A
−
cos
A
−
cos
2
A
+
cos
A
sin
A
]
=
sin
A
cos
A
[
2
1
−
sin
2
A
−
cos
2
A
+
2
sin
A
cos
A
]
=
sin
A
cos
A
[
2
1
−
(
sin
2
A
+
cos
2
A
)
+
2
sin
A
cos
A
]
=
sin
A
cos
A
[
2
1
−
1
+
2
sin
A
cos
A
]
[ Since,
sin
2
θ
+
cos
2
θ
=
1
]
=
sin
A
cos
A
×
2
2
sin
A
cos
A
=
1
=
R
.
H
.
S
(
csc
A
−
sin
A
)
(
sec
A
−
cos
A
)
=
1
tan
A
+
cot
A
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Check whether following statement is true or false.
cos
4
θ
−
sin
4
θ
=
1
+
2
sin
2
θ
Report Question
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State whether the following statement is true or false.
1
sec
A
+
tan
A
−
1
cos
A
=
1
cos
A
−
1
sec
A
−
tan
A
Report Question
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Explanation
Consider
R
H
S
=
1
s
e
c
A
−
t
a
n
A
−
1
c
o
s
A
Multiplying by
s
e
c
A
+
t
a
n
A
in the numerator and denominator of
first term,
we get
s
e
c
A
+
t
a
n
A
(
s
e
c
A
+
t
a
n
A
)
(
s
e
c
A
−
t
a
n
A
)
−
1
c
o
s
A
= secA + tanA - secA........................ (Since,\ sec²A - tan²A = 1)
= tanA
Adding and subtracting
secA
, we get
=secA + tanA - secA
= \dfrac{1}{cosA} - (secA - tanA)
Now multiplying and dividing
(secA - tanA)
by
(secA + tanA)
,
we
get
=\dfrac{1}{cosA} - \dfrac{(sec²A - tan²A)}{(secA + tanA)}
=\dfrac{ 1}{ cosA} - \dfrac{1}{secA + tanA}
= L.H.S
\dfrac{\cos\theta-\sin \theta+1}{\cos\theta+\sin \theta-1}=\text{cosec} \theta+\cot \theta
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Explanation
Given
LHS
\dfrac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}
Dividing Numerator and Denominator by
\sin \theta
\implies \dfrac{\cot \theta +\text{cosec} \theta-1}{\cot \theta-\text{cosec} \theta+1}
\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} ^2\theta-\cot^2\theta)}{\cot \theta-\text{cosec} \theta+1}
\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} \theta-\cot\theta)(\text{cosec} \theta+\cot \theta)}{\cot \theta-\text{cosec} \theta+1}
\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} \theta-\cot\theta)(\text{cosec} \theta+\cot \theta)}{\cot \theta-\text{cosec} \theta+1}
\implies \dfrac{(\cot \theta +\text{cosec} \theta)(\cot \theta-\text{cosec} \theta+1)}{\cot \theta-\text{cosec} \theta+1}
\implies \text{cosec} \theta+\cot \theta
State whether the following statement is true or false.
\dfrac { \cos A-\sin A+1 }{ \cos A+\sin A-1 }= \text{cosec} A+\cot A
. (by using the identity
\text{cosec}^{ 2 }A=1+{ \cot }^{ 2 }A.)\quad
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Explanation
LHS=\dfrac{(cosA-sinA)+1}{(cosA+sinA)-1}
taking conjugate
=\dfrac{(cosA-sinA)+1}{(cosA+sinA)-1 } \times \dfrac{(cosA+sinA)+1}{(cosA+sinA)+1}
multiplying we get
=\dfrac{cos^2A-sin^2A+2cosA+1}{2sinAcosA}
=\dfrac{cos^2A-sin^2A+2cosA+sin^2A+cos^2A}{2sinAcosA}
on solving we get
=\dfrac{2cos^2A+2cosA}{2sinAcosA}
taking
2
common and divided,
we get
=\dfrac{cosA+cos^2A}{sinAcosA}
we can also write in the form of
=\dfrac{cosA}{sinAcosA} +\dfrac{cos^2A}{sinAcosA}
on solving we get
cosecA+cotA =R.H.S
The value of
\dfrac{\cos^{2}\theta+\tan^{2}\theta\ -1}{\sin^{2}\theta}
is
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\cos^{2}\theta
0%
\tan^{2}\theta
0%
\dfrac{1}{\sin^{2}\theta}
Explanation
\dfrac{\cos^2 \theta+\tan^2 \theta-1}{\sin^2 \theta}=\dfrac{\cos^2 \theta}{\sin^2 \theta}+\dfrac{\sin^2 \theta}{\sin^2 \theta\cos^2 \theta}-\dfrac{1}{\sin^2 \theta}
=\cot^2 \theta+\text{sec}^2 \theta-\text{cosec}^{2}\theta
=\text{sec}^2 \theta-(\text{cosec}^2 \theta-\cot^2 \theta)
=\text{sec}^2 \theta-1
=\tan^2 \theta
State whether the following statement is true or false.
\dfrac { \cot A-\cos A }{ \cot A+\cos A } =\dfrac { \text{cosec}A-1 }{ \text{cosec}A+1 }
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Explanation
LHS = \dfrac{(cotA - cosA)}{(cotA+cosA)}
=\dfrac{(\dfrac{cosA}{sinA} - cosA)}{(\dfrac{cosA}{sinA} + cosA)}
=\dfrac{cosA(\dfrac{1}{sinA} - 1)}{cosA(1/sinA + 1)}
=\dfrac{ (\dfrac{1}{sinA} -1)}{(\dfrac{1}{sinA}+1)}
=\dfrac{(cosecA-1)}{(cosecA+1)}
..............[ since
1/sinA= cosecA
]
=RHS
State whether the following statement is true or false.
\dfrac{1}{secA-1}+\dfrac{1}{secA+1}=2cosecAcotA
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Explanation
LHS=\dfrac{1}{secA-1}+\dfrac{1}{secA+1}
=\dfrac{secA+1+secA-1}{sec^2A-1}
=\dfrac{2secA}{tan^2A}=\dfrac{2cos^2A}{cosAsin^2A}=2\dfrac{1}{sinA}\dfrac{cosA}{sinA}=2cosecAcotA=RHS
State whether the following statement is true or false.
\sec { ^{ 4 } } A\left( 1-\sec { ^{ 4 } } A \right) -2\tan { ^{ 2 } } A=1
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Explanation
sec^4A(1-sin^4A)-2tan^2A
=sec^4A-sec^4Asin^4A-2tan^2A
=sec^4A-sin^4A(1/cos^4A)-2tan^2A
=sec^4A-tan^4A-2tan^2A
={(sec^2A)^2-(tan^2A)^2}-2tan^2A
=(sec^2A+tan^2A)(sec^2A-tan^2A)-2tan^2A
=sec^2A+tan^2A-2tan^2A \quad [∵ sec^2A-tan^2A=1]
=sec^2A-tan^2A
=1
State whether the following statement is true or false.
If
\cos A+{ \cos }^{ 2 }A=1,
then
{ \sin }^{ 2 }A+{ \sin }^{ 4 }A=1.
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Explanation
Given:
cosA+cos^2A=1
\implies cosA=1-cos^2A=sin^2A
..........[1]
Now,
sin^2A+sin^4A=cosA+cos^2A=1
.....................[from
1
]
If
{ cosec }\theta -cot\theta =p
, then
cosec\theta +cot\theta =
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1/p
0%
-1/p
0%
-p
0%
{ p }^{ 2 }
Explanation
\csc{\theta}-\cot{\theta}=p
\Rightarrow \left(\csc{\theta}-\cot{\theta}\right)\left(\csc{\theta}+\cot{\theta}\right)=p\left(\csc{\theta}+\cot{\theta}\right)
\Rightarrow {\csc}^{2}{\theta}-{\cot}^{2}{\theta}=p\left(\csc{\theta}+\cot{\theta}\right)
\Rightarrow 1=p\left(\csc{\theta}+\cot{\theta}\right)
\therefore \csc{\theta}+\cot{\theta}=\dfrac{1}{p}
State whether the following statement is true or false.
\dfrac { 1+\cos\theta -\sin^{ 2 }\theta }{ \sin { \theta \left( 1+\cos { \theta } \right) } } =\cot { \theta }
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Explanation
\dfrac { 1+\cos\theta -\sin^{ 2 }\theta }{ \sin { \theta \left( 1+\cos { \theta } \right) } } =\cot { \theta }
\Rightarrow
L.H.S=\dfrac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}
=\dfrac{1+\cos\theta-(1-\cos^2\theta)}{\sin\theta(1+\cos\theta)}
=\dfrac{1+\cos\theta-1+\cos^2\theta}{\sin\theta(1+\cos\theta)}
=\dfrac{\cos\theta+\cos^2\theta}{\sin\theta(1+\cos\theta}
=\dfrac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)}
=\dfrac{\cos\theta}{\sin\theta}
=\cot\theta
=R.H.S
\dfrac{{\tan\theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta .\text{cosec}\theta
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Explanation
\dfrac{\tan \theta}{1-\cot \theta}+\dfrac{\cot \theta}{1-\tan \theta}=\dfrac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\dfrac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}
=\dfrac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)}-\dfrac{\cos^2 \theta}{\sin \theta(\sin \theta-\cos \theta)}
=\dfrac{1}{\sin \theta-\cos \theta}\bigg(\dfrac{\sin^2 \theta}{\cos \theta}-\dfrac{\cos^2 \theta}{\sin \theta}\bigg)
=\dfrac{\sin^3 \theta-\cos^3 \theta}{(\sin \theta-\cos \theta)\sin \theta\cos \theta}
=\dfrac{(\sin \theta-\cos \theta)(\sin^2 \theta+\cos^2 \theta+\sin \theta\cos \theta)}{(\sin \theta-\cos \theta)\sin \theta\cos \theta}
(\because a^3-b^3=(a-b)(a^2+b^2+a b))
=\dfrac{1+\sin \theta\cos \theta}{\sin \theta\cos \theta}=1+\text{sec}\theta\text{cosec}\theta
So the relation is
\text{True}
The value of
\sin \theta
is not equal to
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\pm \frac{1}{2}
0%
\frac{1}{\sqrt 2}
0%
\pm \frac{1}{3}
0%
2
The maximum value of
2 \sin x-4 \cos x
is
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-\sqrt { 12 }
0%
2
0%
\sqrt { 20 }
0%
None of these
Explanation
\textbf{Step – 1: Convert }\mathbf{2\sin x - 4\cos x}\textbf{ into single sine function.}
\text{Let }f(x) = 2\sin x - 4\cos x
\text{Here, }a = 2\ \text{and }b = -\ 4
\therefore
\sqrt {{a^2} + {b^2}} = \sqrt {{2^2} + {{( - 4)}^2}} \\
= \sqrt {4 + 16} \\
= \sqrt {20} \\
{\text{Divide }}f(x){\text{ by }}\sqrt {{a^2} + {b^2}}
\dfrac{{f(x)}}{{\sqrt {20} }} = \dfrac{2}{{\sqrt {20} }}\sin x - \dfrac{4}{{\sqrt {20} }}\cos x
{\text{By using triangular method,}}
\text{Let,}
\sin \theta = \dfrac{4}{{\sqrt {20} }}{\text{ and }}\cos \theta = \dfrac{2}{{\sqrt {20} }}
\therefore \dfrac{{f(x)}}{{\sqrt {20} }} = \cos \theta \sin x - \sin \theta \cos x
\Rightarrow \dfrac{{f(x)}}{{\sqrt {20} }} = \sin (x - \theta )
\mathbf{[\because \sin (a - b) = \sin a\cos b - \cos a\sin b]}
\Rightarrow f(x) = \sqrt {20} \sin (x - \theta )
\textbf{Step – 2: Sine function lies in the range}
\mathbf{[- 1,1]}
-1\leq \sin (x - \theta ) \leq 1
\Rightarrow - \sqrt {20} \leq \sqrt{20}\sin (x - \theta ) \leq \sqrt {20}
\textbf{ Hence, the maximum value of} \ \mathbf{2 \sin x - 4\cos x}{\textbf{ is equal to }}\mathbf{\sqrt {20}.}
The value of
\sin\theta
is not equal to
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0%
\pm\dfrac {1}{2}
0%
\dfrac {1}{\sqrt {2}}
0%
\pm\dfrac {1}{3}
0%
2
If
x_{1}
and
x_{2}
are two distinct roots of the equation
a\cos x+b\sin x=c
, then
\tan\dfrac {x_{1}+x_{2}}{2}
is equal to
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0%
a/b
0%
b/a
0%
c/a
0%
a/c
Explanation
a \cos x + b \sin x =c
\dfrac{1-\tan^2\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}} +b \dfrac{2\tan \dfrac{x}{2}}{1+\tan^2 \dfrac{x}{2}}=c
(c+a) \tan^2 \dfrac{x}{2} -2b \tan \dfrac{x}{2} +(c-a) =0
Since,
x_1
and
x_2
are roots,
\tan \dfrac{x_1}{2}+\tan \dfrac{x_2}{2} =\dfrac{2b}{c+a}
and
\tan \dfrac{x_1}{2}.\tan \dfrac{x_2}{2} =\dfrac{c-a}{c+a}
Hence,
\tan (\dfrac{x_1+x_2}{2}) = \dfrac{\tan \dfrac{x_1}{2} +\tan \dfrac{x_2}{2}}{1-\tan \dfrac{x_1}{2} \tan \dfrac{x_2}{2}}
=\dfrac{\dfrac{2b}{c+a}}{1-\dfrac{c-a}{c+a}}=\dfrac{b}{a}
Hence, option (b) is correct.
\dfrac{1-\sin A}{1+\sin A}=(\sec A -\tan A)^2
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True
0%
False
Explanation
L.H.S. :
\dfrac{1-\sin A}{1+\sin A}
=\dfrac{(1-\sin A)^2}{(1-\sin A)(1+\sin A)}
=\dfrac{(1-\sin A)^2}{1-\sin^2A}
=\dfrac{(1-\sin A)^2}{\cos ^2A}
=[\dfrac{1-\sin A}{\cos A}]^2
=(\sec A -\tan A)^2
= R.H.S.
If
sin{(2cos^{-1}(\dfrac{1}{\sqrt{5}})} + cos{(2tan^{-1}(\dfrac{1}{3}))}=\dfrac{p}{q}
, where p & q are relatively prime then digit at units place of
{(p-q)}^{2k+1}, k\epsilon{N}
, can be ________.
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0%
1
0%
3
0%
7
0%
9
Given
A={ sin }^{ 2 }\theta +{ cos }^{ 4 }\theta
, then for all real
\theta
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0%
1\le A\le 2
0%
\cfrac { 3 }{ 4 } \le A\le 1
0%
\cfrac { 13 }{ 16 } \le A\le 1
0%
\cfrac { 3 }{ 4 } \le A\le \cfrac { 13 }{ 16 }
The values of x in
\left(0, \dfrac{\pi}{2}\right)
satisfying the equation
\sin x\cos x=\dfrac{1}{4}
are ________.
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\dfrac{\pi}{6}, \dfrac{\pi}{12}
0%
\dfrac{\pi}{12}, \dfrac{5\pi}{12}
0%
\dfrac{\pi}{8}, \dfrac{3\pi}{8}
0%
\dfrac{\pi}{8}, \dfrac{\pi}{4}
Explanation
We know that,
\displaystyle \text{sin}(2x) = 2\text{ sin }x\text{ cos }x
.
So, from
\displaystyle \text{sin }x \text{ cos }x = \frac{1}{4}
, we get that
\text{ sin }(2x) = \frac{1}{2}
.
So, we have to find the solutions of the equation
\displaystyle\text{sin} (2x) = \frac{1}{2}
, where
\displaystyle x \in (0, \frac{\pi}{2})
.
From this data, we get that
\displaystyle 2x = \frac{\pi}{6}
or
\dfrac{5\pi}{6}
.
That is,
x = \dfrac{\pi}{12}
or
\dfrac{5\pi}{12}
.
What is
\cot(\dfrac{A}{2})-\tan(\dfrac{A}{2})
equal to ?
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\tan A
0%
\cot A
0%
2\tan A
0%
2\cot A
Explanation
\cot \left(\dfrac{A}{2}\right)-\tan\left(\dfrac{A}{2}\right)
\Rightarrow
\dfrac{\cos \dfrac{A}{2}}{\sin\dfrac{A}{2}}-\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}
\Rightarrow
\dfrac{\cos^2\dfrac{A}{2}-\sin^2\dfrac{A}{2}}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}
\Rightarrow
\dfrac{\cos 2\left(\dfrac{A}{2}\right)}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}
\Rightarrow
\dfrac{\cos A}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}
\Rightarrow
\dfrac{2\cos A}{2\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}
\Rightarrow
\dfrac{2\cos A}{\sin 2\left(\dfrac{A}{2}\right)}
\Rightarrow
\dfrac{2\cos A}{\sin A}
\Rightarrow
2\cot A
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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