Mathematics Test

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Mathematics Test - 17

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Question 1
1 / -0

$\left(\frac{\tan 25^{\circ}}{\operatorname{cosec} 65^{\circ}}\right)^{2}+\left(\frac{\cot 25^{\circ}}{\sec 65^{\circ}}\right)^{2}=$

A
1

B
2

C
3/4

D
8

Solution

$\left(\frac{\tan 25^{\circ}}{\operatorname{cosec} 65^{\circ}}\right)^{2}+\left(\frac{\cot 25^{\circ}}{\sec 65^{\circ}}\right)^{2}$
$=\left(\tan 25^{\circ} \cdot \frac{1}{\operatorname{cosec} 65^{\circ}}\right)^{2}+\left(\cot 25^{\circ} \cdot \frac{1}{\sec 65^{\circ}}\right)^{2}$
$=\left(\frac{\sin 25^{\circ}}{\cos 25^{\circ}} \cdot \sin 65^{\circ}\right)^{2}+\left(\frac{\cos 25^{\circ}}{\sin 25^{\circ}} \cdot \cos 65^{\circ}\right)^{2}$
$=\left(\frac{\sin 25^{\circ}}{\cos 25^{\circ}} \cdot \cos 25^{\circ}\right)^{2}+\left(\frac{\cos 25^{\circ}}{\sin 25^{\circ}} \cdot \sin 25^{\circ}\right)^{2}$
$=\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}=1,$ Pythagorean identity
Hence, the correct option is A.
Question 2
1 / -0

A circle is drawn in a sector of a larger circle of radius r as shown in figure. The smaller circle is tangent to the two bounding radii and the arc of the sector. The radius of the small circle is

A
r/2

B
r/3

C
[2√(3r)]/5

D
r/√2

Solution

From the given figure. Radius of the larger circle $=A C=r$ Radius of the smaller circle $=O B=B C=R$ $A B=A C-B C=r-R$
$\angle O A B=\frac{\angle O A D}{2}=30$
In right angled triangle $\mathrm{ABO} \sin (\angle O A B)=\frac{O B}{A B}$
$\sin 30^{\circ}=\frac{R}{r-R}$
$\Rightarrow R=\frac{r}{3}$
Hence, the correct option is B.
Question 3
1 / -0

A wheel makes 240 revolutions in one minute The measure of the angle it describes at the centre in 15 seconds is,

A
60π

B
120π

C
8π

D
π

Solution

In 60 seconds, 240 revolutions are made This means, in 15 seconds, $\frac{15 \times 240}{60}=60$ revolutions are made.
In one revolution, angle described at centre $=2 \pi$ $\Rightarrow$ In 60 revolutions, angle described at centre $=60 \times 2 \pi=120 \pi$
Hence, the correct option is B.
Question 4
1 / -0

Find the value of $\frac{\left(\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}\right)}{\left(\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}\right)}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}=$

A
0

B
1

C
2

D
3

Solution

Given $\frac{\left(\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}\right)}{\left(\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}\right)}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$
$=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2} 22^{\circ}+\cos ^{2}\left(90^{\circ}-22^{\circ}\right)}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \times \sin \left(90^{\circ}-63^{\circ}\right)$
$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\cos ^{2} 22^{\circ}+\sin ^{2} 22^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \times \cos 63^{\circ}$
(as we know $\sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\left.\cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$
$=\frac{1}{1}+\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}$
$=1+1$
$=2$
Hence, the correct option is C.
Question 5
1 / -0

Find the value of $\sin \theta,$ if $\tan \theta=\frac{3}{4}$ and $180^{\circ}<\theta<270^{\circ}$

A
4/3

B
-4/3

C
3/5

D
-3/5

Solution

$1+\tan ^{2} \theta=\sec ^{2} \theta$
Hence $\sec ^{2} \theta=1+\frac{9}{16}=\frac{16+9}{16}=\frac{25}{16}$
Hence sec $\theta=\frac{-5}{4}$ since $\theta$ is in $3^{\text {rd }}$ quadrant and in $3^{r d}$ quadrant only tan and cot are positive. Therefore $\cos \theta=\frac{-4}{5}$
Hence $\sin \theta=-\sqrt{1-\cos ^{2} \theta}$
$=-\sqrt{1-\frac{16}{25}}$
$=-\sqrt{\frac{25-16}{25}}$
$=-\sqrt{\frac{9}{25}}$
$=\frac{-3}{5}$
Hence, the correct option is D.
Question 6
1 / -0

In a triangle ABC, angle A is greater than B. If the measure of angles A and B satisfy the equation 3sinx−4sin³x−k=0,0

A
π/3

B
π/2

C
2π/3

D
5π/6

Solution

We have the expression $3 \sin x-4 \sin ^{3} x-k=0 \Rightarrow \sin 3 x-k=0$
since, $A$ and $B$ satisfy (1), therefore $\sin 3 A-k=0 ; \sin 3 B-k=0 \Rightarrow \sin 3 A=\sin 3 B$
since $A>B$, therefore $\sin 3 A=\sin 3 B \Rightarrow 3 A=180^{\circ}-3 B$
$\Rightarrow 3 A+3 B=180^{\circ} \Rightarrow A+B=60^{\circ}$
$\Rightarrow 180^{\circ}-C=60^{\circ} \Rightarrow C=120^{\circ}=\frac{2 \pi}{3}$
Hence, the correct option is C.
Question 7
1 / -0

If sinθ+cosecθ=2, then [sin⁸θ+cosec⁸θ] will have the value of

A
2

B
2⁴

C
2⁶

D
2⁸

Solution

Given, $\sin \theta+\operatorname{cosec} \theta=2$
On squaring both sides, we get, $(\sin \theta+\operatorname{cosec} \theta)^{2}=2^{2}$
$\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2 \sin \theta \operatorname{cosec} \theta=4$
$\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2=4$
$\because \sin \theta \operatorname{cosec} \theta=1)$
$\Rightarrow \sin ^{2} \theta+\operatorname{cosec}^{2} \theta=2$
Again, On squaring both sides, we get, $\sin ^{4} \theta+\operatorname{cosec}^{4} \theta+2 \sin ^{2} \theta \operatorname{cosec}^{2} \theta=4$
$\Rightarrow \sin ^{4} \theta+\operatorname{cosec}^{4} \theta+2=4$
$\Rightarrow \sin ^{4} \theta+\operatorname{cosec}^{4} \theta=2$
Again, squaring both sides. $\Rightarrow \sin ^{8} \theta+\operatorname{cosec}^{8} \theta+2 \sin ^{4} \theta \operatorname{cosec}^{4} \theta=4$
$\Rightarrow \sin ^{8} \theta+\operatorname{cosec}^{8} \theta+2=4$
$\Rightarrow \sin ^{8} \theta+\operatorname{cosec}^{8} \theta=2$
Hence, the correct option is A.
Question 8
1 / -0

If acosθ+bsinθ=m and asinθ−bcosθ=n, then the value of a²+b² is,

A
m + n

B
mn

C
mn

D
m² + n²

Solution

$a \cos \theta+b \sin \theta=m$
$a \sin \theta-b \cos \theta=n \quad \ldots(2)$
Squaring (1) and (2), we get $a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cos \theta=m^{2}$
$-(3)$
and $a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta=n^{2}$
(4)
Adding (3) and (4)
$a^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=m^{2}+n^{2}$
or $a^{2}+b^{2}=m^{2}+n^{2}$
Hence, the correct option is D.
Question 9
1 / -0

If $\phi$ is an acute angle such that $\tan \phi=\frac{2}{3},$ then evaluate
$$
\left(\frac{1+\tan \phi}{\sin \phi+\cos \phi}\right)\left(\frac{1-\cot \phi}{\sec \phi+\operatorname{cosec} \phi}\right)
$$

A
-1/5

B
-4/√13

C
1/5

D
4/√13

Solution

Given $: \tan \phi=\frac{2}{3}$
Consider, $\left(\frac{1+\tan \phi}{\sin \phi+\cos \phi}\right)\left(\frac{1-\cot \phi}{\sec \phi+\operatorname{cosec} \phi}\right)$
$=\left(\frac{1+\tan \phi}{\sin \phi+\cos \phi}\right)\left(\frac{1-\frac{1}{\tan \phi}}{\frac{1}{\cos \phi}+\frac{1}{\sin \phi}}\right)$
$=-\left(\frac{(1+\tan \phi)(1-\tan \phi)}{\left(\frac{\sin \phi}{\cos \phi}+\frac{\cos \phi}{\sin \phi}+1+1\right) \tan \phi}\right)$
$=-\left(\frac{(1+\tan \phi)(1-\tan \phi)}{\left(\tan \phi+\frac{1}{\tan \phi}+2\right) \tan \phi}\right)$
$=-\left(\frac{(1+\tan \phi)(1-\tan \phi)}{\frac{(\tan \phi+1)^{2}}{\tan \phi} \tan \phi}\right)$
$=-\frac{(1-\tan \phi)}{(\tan \phi+1)}$
$=-\frac{\frac{2}{3}-1}{\frac{2}{3}+1}$
=-1/5
Hence, the correct option is A.
Question 10
1 / -0

Find the value of:
$\sin ^{2} 30 \cos ^{2} 45+4 \tan ^{2} 30+\frac{1}{2} \sin ^{2} 90+\frac{1}{2} \cot ^{2} 60$

A
2

B
3

C
4

D
17/8

Solution

$\sin ^{2} 30 \cos ^{2} 45+4 \tan ^{2} 30+\frac{1}{2} \sin ^{2} 90+\frac{1}{2} \cot ^{2} 60$
$=\left(\frac{1}{2}\right)^{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}+4 \times\left(\frac{1}{\sqrt{3}}\right)^{2}+\frac{1}{2} \times(1)^{2}+\frac{1}{2} \times\left(\frac{1}{\sqrt{3}}\right)^{2}$
$=\frac{1}{4} \times \frac{1}{2}+4 \times \frac{1}{3}+\frac{1}{2}+\frac{1}{2} \times \frac{1}{3}$
$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{6}$
$=\frac{3+32+12+4}{24}$
$=\frac{17}{8}$
Hence, the correct option is D.

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Mathematics Test - 17

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Mathematics Test - 17

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SHARING IS CARING

Question 1

1

2

3/4

8

Solution

Question 2

r/2

r/3

[2√(3r)]/5

r/√2

Solution

Question 3

60π

120π

8π

π

Solution

Question 4

0

1

2

3

Solution

Question 5

4/3

-4/3

3/5

-3/5

Solution

Question 6

π/3

π/2

2π/3

5π/6

Solution

Question 7

2

24

26

28

Solution

Question 8

m + n

mn

mn

m2 + n2

Solution

Question 9

-1/5

-4/√13

1/5

4/√13

Solution

Question 10

2

3

4

17/8

Solution

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2⁴

2⁶

2⁸

m² + n²