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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 4 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 4
Given
sin
A
=
3
5
and
tan
A
is
3
m
, then
m
is:
Report Question
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1
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2
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4
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3
Explanation
Given
sin
A
=
3
5
sin
A
=
P
B
=
3
5
Thus,
P
=
3
,
H
=
5
By Pythagoras Theorem,
H
2
=
P
2
+
B
2
5
2
=
3
2
+
B
2
B
=
4
cm
Now,
tan
A
=
P
B
=
3
4
Find the smallest positive number p for which the equation
c
o
s
(
p
s
i
n
x
)
=
s
i
n
(
p
c
o
s
x
)
has a solution
x
ε
[
0
,
2
π
]
Report Question
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√
2
π
/
4
0%
√
2
π
/
2
0%
√
2
π
0%
π
/
6
Explanation
cos
(
p
sin
x
)
=
sin
(
p
cos
x
)
⇒
cos
(
p
sin
x
)
=
cos
(
π
2
−
p
cos
x
)
⇒
p
sin
x
=
2
n
π
±
(
π
2
−
p
cos
x
)
⇒
p
sin
x
+
p
cos
x
=
2
n
π
+
π
2
or
p
sin
x
−
p
cos
x
=
2
n
π
−
π
2
⇒
p
√
2
(
sin
(
x
+
π
4
)
)
=
(
4
n
+
1
)
π
2
Or
p
√
2
(
sin
(
x
−
π
4
)
)
=
(
4
n
−
1
)
π
2
As
−
1
≤
sin
(
x
+
π
4
)
≤
1
⇒
−
p
√
2
≤
p
√
2
sin
(
x
+
π
4
)
≤
p
√
2
⇒
−
p
√
2
≤
(
4
n
+
1
)
π
2
≤
p
√
2
...(1)
And as
−
1
≤
sin
(
x
−
π
4
)
≤
1
⇒
−
p
√
2
≤
p
√
2
sin
(
x
−
π
4
)
≤
p
√
2
⇒
−
p
√
2
≤
(
4
n
−
1
)
π
2
≤
p
√
2
...(2)
(2) is always a subset of of first, therefore we have to consider only first
It is sufficient to consider
n
≥
0
, because for
n
>
0
, the solution will be same for
n
≥
0
If
n
≥
0
−
√
2
p
≤
(
4
n
+
1
)
π
2
⇒
(
4
n
+
1
)
π
2
≤
√
2
p
√
2
p
≥
π
2
⇒
p
≥
π
2
√
2
=
π
√
2
4
Number of solutions of the equation
t
a
n
x
+
s
e
c
x
=
2
c
o
s
x
lying in the interval
[
0
,
2
π
]
is
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0
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1
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2
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3
Explanation
t
a
n
x
+
s
e
c
x
=
2
c
o
s
x
⇒
s
i
n
x
+
1
=
2
c
o
s
2
x
⇒
s
i
n
x
+
1
=
2
−
2
s
i
n
2
x
⇒
2
s
i
n
2
x
+
s
i
n
x
−
1
=
0
⇒
(
2
s
i
n
x
−
1
)
(
s
i
n
x
+
1
)
=
0
but
s
i
n
x
=
−
1
⇒
x
=
3
π
2
...........(1)
s
i
n
x
=
1
2
=
s
i
n
π
6
therefore general solution is,
x
=
n
π
+
(
−
1
)
n
.
π
6
x
=
.
.
.
.
.
.
,
π
6
,
5
π
6
,
.
.
.
.
.
.
..........(2)
Therefore, number of solutions in the given interval are
3
.
Hence, option 'D' is correct.
The value of
sin
2
15
∘
+
sin
2
30
∘
+
sin
2
45
∘
+
sin
2
60
∘
+
sin
2
75
∘
is
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1
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3
2
0%
5
2
0%
3
Explanation
(
sin
2
75
0
+
sin
2
(
15
0
)
)
+
(
sin
2
30
0
+
sin
2
60
0
)
+
sin
2
45
0
=
(
cos
2
15
0
+
sin
2
15
0
)
+
(
cos
2
60
0
+
sin
2
60
0
)
+
1
2
=
1
+
1
+
1
2
=
2
+
1
2
=
5
2
Given that
tan
(
A
+
B
)
=
tan
A
+
tan
B
1
−
tan
A
tan
B
where
A
and
B
are acute angle.
Calculate
A
+
B
when
tan
A
=
1
2
,
tan
B
=
1
3
.
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A
+
B
=
30
∘
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A
+
B
=
45
∘
0%
A
+
B
=
60
∘
0%
A
+
B
=
75
∘
Explanation
Given that
tan
A
=
1
/
2
,
tan
B
=
1
/
3
tan
(
A
+
B
)
=
tan
A
+
tan
B
1
−
tan
A
tan
B
=
1
2
+
1
3
1
−
(
1
2
)
(
1
3
)
=
5
6
×
6
5
=
1
∴
If
A = 60^{\small\circ}\ and\ B = 30^{\small\circ}
, then verify each of the following:
(i)\ \cos(A-B) = \cos A \cos B + \sin A \sin B
(ii)\cot(A+B) = \displaystyle\frac{\cot A\cot B - 1}{\cot A + \cot B}
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(i)\ True
(ii)\ False
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(i)\ False
(ii)\ False
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(i)\ True
(ii)\ True
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(i)\ False
(ii)\ True
Explanation
=cos(60^{0}-30^{0})
=cos(30^{0})
=\frac{\sqrt{3}}{2}
=LHS
Now
RHS
=cos60^{0}cos30^{0}+sin30^{0}sin60^{0}
=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}
=\frac{\sqrt{3}}{2}
=RHS
Hence verified.
=cot(60^{0}+30^{0})
=cot(90^{0})
=0
=LHS
RHS
=\frac{cot60^{0}.cot30^{0}-1}{cot60^{0}+cot30^{0}}
=\dfrac{1-1}{\frac{1}{\sqrt{3}}+\sqrt{3}}
=0
Hence\ verified.
Is LHS=RHS?
\quad \sqrt{\displaystyle\frac{cosec\theta - 1}{cosec\theta + 1}}+\sqrt{\displaystyle\frac{cosec\theta + 1}{cosec\theta - 1}}=2\cos\theta
Say true or false?
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True
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False
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Ambiguous
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Data insufficient
Explanation
Simplifying the given expression, we get
\dfrac{cosec\theta-1+cosec\theta+1}{\sqrt{cosec^{2}\theta-1}}
=\dfrac{2cosec\theta}{cot\theta}
=\dfrac{2cosec\theta}{cos\theta.cosec\theta}
=\dfrac{2}{cos\theta}
=2sec\theta
Evaluate :
\displaystyle\frac{2\tan30^{\small\circ}}{1-\tan^230^{\small\circ}}
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0
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1
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\sqrt2
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\sqrt3
Explanation
We know that
\tan 2A
=\dfrac{2\tan A}{1-\tan^{2}A}
Here
A=30^{0}
Substituting, we get
\dfrac{2\tan30^{0}}{1-\tan^{2}30^{0}}
=\tan(2(30^{0}))
=\tan(60^{0})
=\sqrt{3}
Is LHS=RHS?
\quad \displaystyle\frac{\cos\theta}{1+\sin\theta}+\displaystyle\frac{\cos\theta}{1-\sin\theta}=2cosec\theta
Say true or false.
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Yes
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No
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Ambiguous
0%
Data insufficient
Explanation
\dfrac{cos(\theta)(1-sin\theta+1+sin\theta)}{1-sin^{2}\theta}
=\dfrac{cos(\theta)(2)}{cos^{2}\theta}
=\dfrac{2}{cos\theta}
=2sec\theta
If
\alpha + \beta = 90^{\small\circ}
and
\alpha = 2\beta
, then
\cos^2\alpha + \sin^2\beta
equal
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1
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0
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\displaystyle\frac{1}{2}
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2
Explanation
It is given that
\alpha=2\beta
Hence
\alpha+\beta=90^{0}
Substituting, we get
3\beta=90^{0}
Hence
\beta=30^{0}
Therefore
\alpha=60^{0}
Hence
\sin^{2}\beta+\cos^{2}\alpha
=\dfrac{1}{2}^{2}+\dfrac{1}{2}^{2}
=\dfrac{1}{4}+\dfrac{1}{4}
=\dfrac{1}{2}
If
p\sin x = q
. If
x
is acute, then
\displaystyle \sqrt{p^{2}-q^{2}}tan x
is equal to
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p
0%
q
0%
p q
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p + q
Explanation
Given:
p \sin x = q
Now,
\displaystyle \sqrt{p^{2}-q^{2}}\tan x
=\sqrt{p^{2}-p^{2}\sin ^{2}x.}\tan \:x
=\sqrt{p^{2}\left ( 1-\sin ^{2}x \right )}.\tan \:x
=p\sqrt{\left ( 1-\sin ^{2}x \right )}.\tan \:x\left .....( \because \cos ^{2}\theta +\sin ^{2}\theta =1 \right )
=P\sqrt{\cos ^{2}x}.\tan \:x
=p\cos \:x\tan \:x
=p\sin \:x
=q
\displaystyle \left | \tan x \right |=\tan x+\frac{1}{\cos x}(0\leq \times \leq 2\pi)
has
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no solution
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one solution
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two solutions
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three solutions
Explanation
When,
x \in [0, \frac{\pi}{2}]
and
x \in [\pi, \frac{3\pi}{2}]
Thus,
\tan x > 0
hence,
|\tan x | = \tan x
hence,
\displaystyle \left | \tan x \right |=\tan x+\frac{1}{\cos x}(0\leq \times \leq 2\pi)
=
\displaystyle \tan x =\tan x+\frac{1}{\cos x}
=
\cos x = 1
=
\cos x = \cos 90
=
x = 90^{\circ}
But it is not possible since,
\cos x \ne 0
(
\cos x
being the denominator in the equation)
When,
x \in [\frac{\pi}{2}, \pi ]
and
x \in [\frac{3\pi}{2}, 2\pi]
Thus,
\tan x < 0
hence,
|\tan x | = - \tan x
hence,
\displaystyle \left | \tan x \right |=\tan x+\frac{1}{\cos x}
=
2 \tan x = \frac{1}{\cos x}
=
\sin x = \frac{1}{2}
This is not possible, as
\sin x
is not positive in these intervals of value of
x
Hence, the equation has no solution for any value of x.
If
8 \tan A = 15
, then the value of
\displaystyle \frac{\sin A -\cos A}{\sin A+\cos A}
is:
Report Question
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\displaystyle \frac{7}{23}
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\displaystyle \frac{11}{23}
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\displaystyle \frac{13}{23}
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\displaystyle \frac{17}{23}
Explanation
\displaystyle 8\tan A=15 \Rightarrow \tan A=\frac{15}{8}
Now,
\dfrac{\sin A-\cos A}{\sin A+\cos A}
Multiply and divide by
\cos A
=\dfrac{{\dfrac{\sin A}{\cos A}}-\dfrac{\cos A}{\cos A}}{{\dfrac{\sin A}{\cos A}}+\dfrac{\cos A}{\cos A}}
=\dfrac{\tan A-1}{\tan A+1}
Now, substitute the value of
\displaystyle \tan A
= \dfrac{\dfrac{15}{8} -1}{\dfrac{15}{8}+1}
= \dfrac{7}{23}
The number of solutions of the equation
\displaystyle\frac { \sec { x } }{ 1 - \cos { x } } = \displaystyle\dfrac { 1 }{ 1 - \cos { x } }
in
\left[ 0, 2\pi \right]
is equal to
Report Question
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3
0%
2
0%
1
0%
0
Explanation
\displaystyle\frac { \sec { x } }{ 1-\cos { x } } =\frac { 1 }{ 1-\cos { x } }
For
1-\cos { x } \neq 0\Rightarrow \cos { x } \neq 1
..(1)
Gives
\sec { x } =1\Rightarrow \cos { x } =1
...(2)
From (1) and (2) we get no solution.
If
x=2\sin ^{ 2 }{ \theta }
,
y=2\cos ^{ 2 }{ \theta } +1
, then the value of
x+y
is:
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2
0%
3
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\cfrac{1}{2}
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1
Explanation
Given,
x=2{ \sin }^{ 2 }\theta ,y=2{ \cos }^{ 2 }\theta +1
To find:
(x+y)
x+y=2{ \sin }^{ 2 }\theta +2{ \cos }^{ 2 }\theta +1
=2({ \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta )+1
.....(As
\sin^2\theta +\cos ^2\theta=1
)
=2+1
=3
The solution set of
(5+4\,cos\,\theta)\;(2\,cos\,\theta+1)=0
in the interval
[0,2\pi]
is
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\begin{Bmatrix}\displaystyle\frac{\pi}{3},\displaystyle\frac{2\pi}{3}\end{Bmatrix}
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\begin{Bmatrix}\displaystyle\frac{\pi}{3},{\pi}\end{Bmatrix}
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\begin{Bmatrix}\displaystyle\frac{2\pi}{3},\displaystyle\frac{4\pi}{3}\end{Bmatrix}
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\begin{Bmatrix}\displaystyle\frac{2\pi}{3},\displaystyle\frac{5\pi}{3}\end{Bmatrix}
Explanation
(5+4\cos\,\theta)\;(2\cos\,\theta+1)=0
\Rightarrow \cos\,\theta=-\displaystyle\frac{5}{4}
(not possible)
\cos\,\theta=-\displaystyle\frac{1}{2}
\Rightarrow \cos\theta=-\displaystyle\frac{1}{2}
\Rightarrow \cos \theta=-\cos {\dfrac{\pi}{3}}=\cos(\pi\pm\dfrac{\pi}{3})
\Rightarrow \theta=\displaystyle\frac{2\pi}{3},\displaystyle\frac{4\pi}{3}
(II and III quad, cosine is negaitve)
Which of the following is / are the value (S) of the expression?
sin A(1+ tan A) + cos A (1+ cot A) ?
1
. sec A + cosec A
2
. 2 cosec A ( sin A + cos A )
3
. tan A + cot A
Select the correct answer using the code given below.
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1 only
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1 and 2 only
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2 only
0%
1 and 3 only
Explanation
\sin A (1 + \tan A) + \cos A(1 + \cot A)
=
\sin A(1 + \frac{\sin A}{\cos A}) + \cos A (1 + \frac{\sin A}{\cos A})
=
(\sin A + \cos A) (\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})
=
(\sin A + \cos A) (\frac{\sin^2 A + \cos^2 A}{\sin A + \cos A}
=
\sec A + cosec A
Thus, 1 satisfies the given equation.
If
\displaystyle \sin x+\sin ^{2}x=1
then the value of
\displaystyle \cos ^{2}x+\cos ^{4}x
is equal to
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1
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\displaystyle \frac{1}{2}
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\displaystyle \frac{1}{3\sqrt{3}}
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\displaystyle \frac{3\sqrt{5}-5}{2}
Explanation
Given,
\sin x + \sin^2 x = 1
\Rightarrow \sin x = 1 - \sin^2 x
\Rightarrow \sin x = \cos^2x
Hence,
\cos^2 x + \cos^4x
=
\sin x + \sin^2 x
=
1
....(Given,
\sin x + \sin^2 x = 1
)
If
\displaystyle \cos \theta =\frac{5}{13}
, where
\theta
being an acute angle, then the value of
\dfrac{\cos \theta +5\cot \theta }{\text {cosec}\ \theta -\cos \theta }
will be
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\displaystyle \frac{169}{109}
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\displaystyle \frac{155}{109}
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\displaystyle \frac{385}{109}
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\displaystyle \frac{95}{109}
Explanation
Given,
\cos\theta=\dfrac{5}{13}
Then
\sin\theta=\dfrac{12}{13}
Hence,
\tan\theta=\dfrac{12}{5}
and
\text {cosec} \theta=\dfrac{13}{12}
Substituting in the equation, we get
=\dfrac{\frac{5}{13}+\frac{5.5}{12}}{\frac{13}{12}-\frac{5}{13}}
=\dfrac{5(12)+25(13)}{13(13)-5(12)}
=\dfrac{60+300+25}{169-60}
=\dfrac{385}{109}
Without using trigonometric tables evaluate:-
\displaystyle \frac{\cos ^{2}20^{\circ}+\cos ^{2}70^{\circ}}{\sec ^{2}50^{\circ}-\cot ^{2}40^{\circ}}+2\: cosec ^{2}58^{\circ}-2\cot 58^{\circ}\tan 32^{\circ}-4\tan 13^{\circ}\tan 37^{\circ}\tan 45^{\circ}\tan 53^{\circ}\tan 77^{\circ}
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1
0%
2
0%
-1
0%
-2
Explanation
\dfrac{\cos ^2 20^{\circ}+\cos ^2 70^{\circ}}{\sec ^2 50^{\circ}-\cot ^2 40^{\circ}}+2\csc ^2 58^{\circ}-2\cot 58^{\circ}\tan 32^{\circ}-4\tan 13^{\circ}\tan 37^{\circ}\tan 45^{\circ}\tan 53^{\circ}\tan 77^{\circ}
=\dfrac{\cos ^2 20^{\circ}+\sin ^2 20^{\circ}}{\sec ^2 50^{\circ}-\tan ^2 50^{\circ}}+2\csc ^2 58^{\circ}-2\cot 58^{\circ}\cot 58^{\circ}-4\tan 13^{\circ}\cot 13^{\circ} \cot 53^{\circ} \tan 53^{\circ} \tan 45^{\circ}
=\dfrac{1}{1}+2\csc ^2 58^{\circ}-2\cot 58^{\circ}\cot 58^{\circ}-4\times 1\times 1\times 1
=1+2(\csc ^2 58^{\circ}-\cot ^258^{\circ})-4
=1+2-4
=-1
\displaystyle \frac{\tan ^{2}\theta }{1+\sec \theta }+1
equals to
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\displaystyle \tan \theta
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\displaystyle \frac{1}{\cos \theta }
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\displaystyle \sec \theta -1
0%
\displaystyle \sec \theta +\tan \theta
Explanation
\displaystyle \frac{\tan ^{2}\theta }{1+\sec \theta }+1=\frac{\sec^2\theta-1 }{1+\sec \theta }+1
=\dfrac{(\sec\theta+1)(\sec\theta-1) }{1+\sec \theta }+1
=\sec\theta-1 +1
=\sec\theta
=\dfrac1{\cos\theta}
Option B is correct.
\displaystyle \sqrt{\frac{1-\sin \theta }{1+\sin \theta}}
is equal to ............
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\displaystyle cosec\ \theta -\cot \theta
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\displaystyle \tan \theta -\sec \theta
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\displaystyle\sec \theta -\tan \theta
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\displaystyle \cot \theta -cosec\ \theta
Explanation
\sqrt { \dfrac { 1-\sin { \theta } }{ 1+\sin { \theta } } }
=\sqrt { \dfrac { \left( 1-\sin { \theta } \right) \left( 1-\sin { \theta } \right) }{ \left( 1+\sin { \theta } \right) \left( 1-\sin { \theta } \right) } }
=\sqrt { \dfrac { { \left( 1-\sin { \theta } \right) }^{ 2 } }{ 1-\sin ^{ 2 }{ \theta } } }
=\sqrt { \dfrac { { \left( 1-\sin { \theta } \right) }^{ 2 } }{ \cos ^{ 2 }{ \theta } } }
=\dfrac { 1-\sin { \theta } }{ \cos { \theta } }
=\dfrac { 1 }{ \cos { \theta } } -\dfrac { \sin { \theta } }{ \cos { \theta } }
=\sec { \theta } -\tan { \theta }
Hence, the answer is
\sec { \theta } -\tan { \theta }.
The value of
\displaystyle \frac{(1+\tan ^{2}\theta )}{(1+\cot ^{2}\theta )}
is
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\displaystyle \tan ^{2}\theta
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\displaystyle \cot ^{2}\theta
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\displaystyle \sec ^{2}\theta
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\displaystyle \text{cosec } ^{2}\theta
Explanation
The value of
\displaystyle \dfrac {(1+\tan ^{2}\theta )}{(1+\cot ^{2}\theta )}
is
=\displaystyle \dfrac {\left (1+\frac {\sin ^{2}\theta}{\cos ^{2}\theta} \right)}{\left (1+\frac {\cos ^{2}\theta}{\sin ^{2}\theta} \right)}
=\displaystyle \dfrac {\left (\frac {\cos ^{2}\theta + \sin ^{2}\theta}{\cos ^{2}\theta} \right)}{\left (\frac {\sin ^{2}\theta + \cos ^{2}\theta}{\sin ^{2}\theta}\right )}
=\displaystyle \dfrac {\left (\frac {1}{\cos ^{2}\theta}\right )}{\left (\frac {1}{\sin ^{2}\theta} \right)}
=\displaystyle \frac {\sin ^{2}\theta}{\cos ^{2}\theta}
=\displaystyle \tan^{2}\theta
Hence, option A is correct.
The
\triangle ABC
has a right angle at C. If
\displaystyle \sin A=\frac{2}{3}
then
\displaystyle \tan B
is
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\displaystyle \frac{3}{5}
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\displaystyle \frac{\sqrt{5}}{3}
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\displaystyle \frac{2}{\sqrt{5}}
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\displaystyle \frac{\sqrt{5}}{2}
Explanation
\Rightarrow \sin { A } =\dfrac { 2 }{ 3 } =\dfrac { BC }{ AB }
In right angled
\triangle ABC,
\Rightarrow AC^2+BC^2=AB^2
\Rightarrow AC^2=AB^2-BC^2
=9-4
\Rightarrow AC=\sqrt {5}
\Rightarrow \tan B=\dfrac{AC}{BC}=\dfrac{\sqrt {5}}{2}
Hence, the answer is
\dfrac{\sqrt {5}}{2}.
The simplification of
\displaystyle \sqrt{\frac{1+\cos A}{1-\cos A}}
gives
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\displaystyle \text{cosec } A+\cot A
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\displaystyle \text{cosec } A-\cot A
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\displaystyle \frac{1+\cos A}{\sin A}
0%
Both A and C.
Explanation
\displaystyle \sqrt{\dfrac{1+\cos A}{1-\cos A}}=\sqrt{\dfrac{1+\cos A}{1-\cos A}\times\dfrac{1+\cos A}{1+\cos A}}
=\displaystyle \dfrac{1+\cos A}{\sqrt{1-\cos ^2 A}}
=\displaystyle \dfrac{1+\cos A}{\sqrt{\sin^2 A}}
=\displaystyle \dfrac{1+\cos A}{\sin A}
...(i)
=\displaystyle \dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}
=\displaystyle \text{cosec} A+\cot A
...(ii)
From (i) and (ii) , Option D is correct.
The expression
\displaystyle (1-\tan A+\sec A)(1-\cot A+\cos \sec A)
has value
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-1
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0
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+1
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+2
Explanation
\displaystyle \left(1-\tan A+\sec A\right)\left(1-\cot A+\text{cosec } A\right)
=\displaystyle \left(1-\frac {\sin A}{\cos A}+\frac {1}{cos A}\right)\left(1-\frac {\cos A}{\sin A}+\frac {1}{\sin A}\right)
=\displaystyle \left(\frac {\cos A- \sin A + 1}{\cos A}\right)\left(\frac {\sin A -\cos A + 1}{\sin A}\right)
=\displaystyle \left(\frac {\cos A- \sin A + 1}{\cos A}\right)\left(\frac {-\left(\cos A - \sin A -1\right)}{\sin A}\right)
=\displaystyle - \left(\frac {\left(\cos A- \sin A\right) + 1}{\cos A}\right)\left(\frac {\left(\cos A - \sin A\right) -1}{\sin A}\right)
=\displaystyle - \frac {\left(\cos A- \sin A\right)^2 - 1}{\sin A \cos A}
=\displaystyle - \frac {\left(\cos^{2} A+ \sin^{2} A- 2\sin A \cos A\right) - 1}{\sin A \cos A}
=\displaystyle - \frac {\left(1- 2\sin A \cos A\right) - 1}{\sin A \cos A}
=\displaystyle - \frac {- 2\sin A \cos A }{\sin A \cos A} = 2
Option D is correct.
In a right angled
\displaystyle \Delta ABC
right angled at
B
the ratio of
AB
to
AC
is
\displaystyle 1:\sqrt{5}
then
\displaystyle 3\tan \theta +5\sec ^{2}\theta
is
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\displaystyle \frac{2}{\sqrt{5}}
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\displaystyle 3+\sqrt{5}
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\displaystyle \dfrac{25}4+\frac{3}{2}
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\displaystyle \frac{\sqrt{5+1}}{2}
Explanation
h=\sqrt { 5 } x
p=x
b=\sqrt { (\sqrt { 5 } )^ 2-x^ 2 } =\sqrt { 4{ x }^{ 2 } } =2x
\tan \theta =\dfrac { p }{ b } =\dfrac { 1x }{ 2x } =\dfrac { 1 }{ 2 }
\sec { \theta } =\dfrac { h }{ b } =\dfrac { \sqrt { 5 } x }{ 2x } =\dfrac { \sqrt { 5 } }{ 2 }
3\tan \theta +5\sec ^{ 2 }{ \theta =3\times } \dfrac { 1 }{ 2 } +5{ \times \left(\dfrac { \sqrt { 5 } }{ 2 } \right) }^{ 2 }=\dfrac { 3 }{ 2 } +\dfrac { 25 }{ 4 }
The value of
\displaystyle \frac{\sin ^{2}53+\cos ^{2}53}{\sec ^{2}37-\tan ^{2}37 }
is
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1
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2
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\displaystyle \frac{1}{4}
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\displaystyle \frac{3}{2}
Explanation
Given,
\displaystyle \frac{\sin ^{2}53+\cos ^{2}53}{\sec ^{2}37-\tan ^{2}37 }
We know,
\sin^2 \theta+\cos^2 \theta=1
and
\sec^2 \theta- \tan^2 \theta =1
\therefore \dfrac{\sin^2 53+ \cos^2 53}{\sec^2 37- \tan^2 37}=1
Value of
(1+\tan {\theta}+\sec {\theta})(1+\cot {\theta}-co\sec {\theta})
is:
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1
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-1
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2
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-4
Explanation
(1+\tan {\theta}+\sec {\theta})(1+\cot {\theta}-co\sec {\theta})
\left( 1+\cfrac { \sin { \theta } }{ \cos { \theta } } +\cfrac { 1 }{ \cos { \theta } } \right) \left( 1+\cfrac { \cos { \theta } }{ \sin { \theta } } -\cfrac { 1 }{ \sin { \theta } } \right)
\left( \cfrac { \cos { \theta } +\sin { \theta } +1 }{ \cos { \theta } } \right) \left( \cfrac { \sin { \theta } +\cos { \theta } -1 }{ \sin { \theta } } \right)
=\cfrac { { \left( \cos { \theta } +\sin { \theta } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } }{ \cos { \theta } \sin { \theta } }
=\cfrac { \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } +2\sin { \theta } \cos { \theta } -1 }{ \cos { \theta } \sin { \theta } }
=\cfrac { 1+2\sin { \theta } \cos { \theta } -1 }{ \cos { \theta } \sin { \theta } } \quad =\cfrac { 2\sin { \theta \cos { \theta } } }{ \cos { \theta } \sin { \theta } } =2
If
\displaystyle \sec \theta =2,
evaluating
\displaystyle \frac{1-\tan \theta }{1+\tan \theta }
gives
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\displaystyle -\sqrt{3}
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\displaystyle \sqrt{3}+1
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\displaystyle2 -\sqrt{3}
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\displaystyle\frac{ \sqrt{3}+1}{2}
Explanation
we know that,
\sec^2\theta-\tan^2\theta=1
\Rightarrow \tan^2\theta=\sec^2\theta-1
\Rightarrow \tan^2\theta=4-1
[given\,\sec\theta=2]
\Rightarrow \tan\theta=\sqrt3
Now,
\displaystyle \dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{1-\sqrt3}{1+\sqrt3}\times\dfrac{1-\sqrt3}{1-\sqrt3}
\displaystyle=\frac{1+3-2\sqrt3}{1-3}
\displaystyle=\frac{4-2\sqrt3}{-2}
\displaystyle={\sqrt3-2}
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