Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 47

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Question 1
1 / -0

The value of $\displaystyle \sin^{3} \alpha \left ( 1+\cot \alpha \right )+\cos ^{3}\alpha \left ( 1+\tan \alpha \right )$ is equal to

A
$\displaystyle \sin \alpha -\cos \alpha$

B
$\displaystyle \sin \alpha +\cos \alpha$

C
$\displaystyle \sin \alpha \cos \alpha$

D
$\displaystyle co\sec \alpha \sec \alpha$

Solution

Given expression
$\displaystyle =\sin ^{3}\alpha \left ( 1+\frac{\cos \alpha }{\sin \alpha } \right )+\cos ^{3}\alpha \left ( 1+\frac{\sin \alpha }{\cos \alpha } \right )$

$\displaystyle =\sin ^{2}\alpha \left ( \sin \alpha +\cos \alpha \right )+\cos ^{2}\alpha$ $\displaystyle \left ( \cos \alpha +\sin \alpha \right )$

$\displaystyle =\left ( \cos \alpha +\sin \alpha \right )\left ( \sin ^{2}\alpha +\cos ^{2}\alpha \right )$

$\displaystyle =\sin \alpha +\cos \alpha$
Question 2
1 / -0

The value of $\displaystyle \sin ^{6}\theta +cos^{6}\theta+3\cos ^{2}\theta \sin ^{2}\theta$ is

A
$1$

B
$2$

C
$3$

D
$6$

Solution

$\displaystyle \sin ^{6}\theta +\cos ^{6}\theta+3\sin ^{2}\theta \cos ^{2}\theta$
$\displaystyle =\sin ^{6}\theta +\cos ^{6}\theta +3\sin ^{2}\theta \cos ^{2}\theta$
$\displaystyle =\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )^{3}=1$
Question 3
1 / -0

If $\displaystyle 7\sin ^{2}\theta +3\cos ^{2}\theta =4$ then the value of $\displaystyle \tan \theta$ is

A
$\displaystyle \pm \frac{1}{\sqrt{2}}$

B
$\displaystyle \pm \frac{1}{\sqrt{3}}$

C
$\displaystyle \pm \frac{1}{2}$

D
$\displaystyle \pm \frac{1}{3}$

Solution

Dividing both sides of the equation by $\displaystyle \cos ^{2}\theta$ we have:
$\displaystyle 7\tan ^{2}\theta +3=4\sec ^{2}\theta =4\left ( 1+\tan ^{2}\theta \right )$
$7\tan^2 \theta +3= 4 + 4\tan^2 \theta$
$\displaystyle \Rightarrow 3\tan ^{2}\theta=1$
$\tan\theta= \pm \frac{1}{\sqrt{3}}$
Question 4
1 / -0

If $\displaystyle \frac{\sin x}{1+\cos x}+\frac{\sin x}{1-\cos x}=4$ and $\displaystyle 0^{0}\leq x\leq 90^{0}$ , then the value of $x$ is

A
$\displaystyle 10^{0}$

B
$\displaystyle 15^{0}$

C
$\displaystyle 30^{0}$

D
$\displaystyle 45^{0}$

Solution

Given expression
$\displaystyle \Rightarrow \sin x\left [ \frac{\left ( 1-\cos x \right )+\left ( 1+\cos x \right )}{1-\cos ^{2}x} \right ]=4$
$\displaystyle \Rightarrow \sin x\times \frac{2}{\sin ^{2}x}=4$ or $\displaystyle \sin x=\frac{1}{2}$
$\displaystyle \Rightarrow x=30^{0}$
Question 5
1 / -0

If $\displaystyle x=r\cos \alpha \cos \beta$ and $y = r \cos \alpha \sin \beta$ , $z=r$ $\displaystyle \sin \alpha$ then $\displaystyle x^{2}+y^{2}+z^{2}$ is equal to

A
$\displaystyle r^{2}$

B
$\displaystyle r^{4}$

C
1

D
None of these

Solution

$\displaystyle x^{2}+y^{2}+z^{2}=r^{2}\left ( \cos ^{2}\alpha \right )\left ( \cos ^{2}\beta +\sin^2 \beta \right )+r^{2}\sin ^{2}\alpha$
$\displaystyle =r^{2}\cos ^{2}\alpha +r^{2}\sin ^{2}\alpha$
$\displaystyle =r^{2}\left ( \cos ^{2}\alpha +\sin ^{2}\alpha \right )=r^{2}$
Question 6
1 / -0

The value of $\displaystyle \sin ^{2}\alpha \cos ^{2}\beta +\cos ^{2}\alpha \sin ^{2}\beta +\sin ^{2}\alpha \sin ^{2}\beta +\cos ^{2}\alpha \cos ^{2}\beta$ is

A
$1$

B
$0$

C
$-1$

D
$3$

Solution

$\Rightarrow (\sin ^2\alpha \cos^2\beta+\sin ^2\alpha \sin ^2\beta)+(\displaystyle \cos ^{2}\alpha \sin ^{2}\beta +\cos ^{2}\alpha \cos ^{2}\beta)\\=\sin ^2\alpha(cos^2\beta+\sin ^2\beta)+cos^2\alpha(cos^2\beta+\sin ^2\beta)\\=\sin ^2\alpha+cos^2\alpha=1$

Therefore, Answer is $1$
Question 7
1 / -0

The value of $\displaystyle \sin ^{2}60^{\circ}.\cos ^{2}30^{\circ}+\tan 45^{\circ}.\cos 60^{\circ}\sin 30^{\circ}$ is

A
$\displaystyle \frac{13}{16}$

B
5

C
1

D
$\displaystyle -\frac{1}{2}$

Solution

$\displaystyle \sin ^{2}60^{\circ}.\cos ^{2}30^{\circ}+\tan 45^{\circ}.\cos 60^{\circ}\sin 30^{\circ}$
$\displaystyle =\frac{3}{4}\times \frac{3}{4}+1\times \frac{1}{2}\times \frac{1}{2}=\frac{9}{16}+\frac{1}{4}=\frac{13}{16}$
Question 8
1 / -0

If $\displaystyle \sin \theta +co\sec \theta =2$ , then $\displaystyle \sin^n\theta +co\sec ^{n}\theta$ is equal to

A
$2$

B
$2n$

C
$\displaystyle 2^{n-1}$

D
$\displaystyle 2^{n-2}$

Solution

$\displaystyle \sin \theta +co\sec \theta =2\Rightarrow \left ( \sin \theta -1 \right )^{2}=0$
$\displaystyle \therefore \sin \theta =1$
$\displaystyle \therefore \sin ^{n}\theta co\sec ^{n}\theta =1+1=2$
Question 9
1 / -0

If $\displaystyle \sin \theta =\dfrac12$ and $\displaystyle 0^{\circ}< \theta < 90^{\circ}$ then $\displaystyle \cos 2\theta =$ _____

A
$0$

B
$1$

C
$\displaystyle \frac{1}{2}$

D
$\displaystyle \frac{\sqrt{3}}{2}$

Solution

Given, $\sin { \theta } = \dfrac {1}{2}$
$=> \theta = {30}^{0}$ (Since, $\sin { {30}^{0} } = \dfrac {1}{2}$ and ${0}^{0}<\theta< {90}^{0})$

So, $\cos { 2\theta } = \cos ({60}^{0}) = \dfrac {1}{2}$
Question 10
1 / -0

If co $\displaystyle \sec \left ( 20^{\circ} + x \right )=\sec \left ( 50^{\circ}+x \right )$ the value of x is

A
$\displaystyle 10^{\circ}$

B
$\displaystyle 20^{\circ}$

C
$\displaystyle 30^{\circ}$

D
$\displaystyle 40^{\circ}$

Solution

We know that $cosec (A) = sec(90-A)$

This means, if ${20}^{0} + x = A$ ,
then ${50}^{0} + x = {90}^{0} - ({20}^{0} + x)$
$=> {50}^{0} + x = {90}^{0} - {20}^{0}-x$
$=> 2 x = {20}^{0}$
$=> x = {10}^{0}$

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Introduction to Trigonometry Test - 47

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Introduction to Trigonometry Test - 47

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Question 1

sin⁡α−cos⁡α\displaystyle \sin \alpha -\cos \alpha sinα−cosα

sin⁡α+cos⁡α\displaystyle \sin \alpha +\cos \alpha sinα+cosα

sin⁡αcos⁡α\displaystyle \sin \alpha \cos \alpha sinαcosα

cosec⁡αsec⁡α\displaystyle co\sec \alpha \sec \alpha cosecαsecα

Solution

Question 2

111

222

333

666

Solution

Question 3

±12\displaystyle \pm \frac{1}{\sqrt{2}}±2​1​

±13\displaystyle \pm \frac{1}{\sqrt{3}}±3​1​

±12\displaystyle \pm \frac{1}{2}±21​

±13\displaystyle \pm \frac{1}{3}±31​

Solution

Question 4

100\displaystyle 10^{0}100

150\displaystyle 15^{0}150

300\displaystyle 30^{0}300

450\displaystyle 45^{0}450

Solution

Question 5

r2\displaystyle r^{2}r2

r4\displaystyle r^{4}r4

1

None of these

Solution

Question 6

111

000

−1-1−1

333

Solution

Question 7

1316 \displaystyle \frac{13}{16} 1613​

5

1

−12 \displaystyle -\frac{1}{2} −21​

Solution

Question 8

222

2n2n2n

2n−1\displaystyle 2^{n-1}2n−1

2n−2\displaystyle 2^{n-2}2n−2

Solution

Question 9

000

111

12\displaystyle \frac{1}{2}21​

32\displaystyle \frac{\sqrt{3}}{2}23​​

Solution

Question 10

10∘\displaystyle 10^{\circ}10∘

20∘\displaystyle 20^{\circ}20∘

30∘\displaystyle 30^{\circ}30∘

40∘\displaystyle 40^{\circ}40∘

Solution

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$\displaystyle \sin \alpha -\cos \alpha$

$\displaystyle \sin \alpha +\cos \alpha$

$\displaystyle \sin \alpha \cos \alpha$

$\displaystyle co\sec \alpha \sec \alpha$

$1$

$2$

$3$

$6$

$\displaystyle \pm \frac{1}{\sqrt{2}}$

$\displaystyle \pm \frac{1}{\sqrt{3}}$

$\displaystyle \pm \frac{1}{2}$

$\displaystyle \pm \frac{1}{3}$

$\displaystyle 10^{0}$

$\displaystyle 15^{0}$

$\displaystyle 30^{0}$

$\displaystyle 45^{0}$

$\displaystyle r^{2}$

$\displaystyle r^{4}$

$1$

$0$

$-1$

$3$

$\displaystyle \frac{13}{16}$

$\displaystyle -\frac{1}{2}$

$2$

$2n$

$\displaystyle 2^{n-1}$

$\displaystyle 2^{n-2}$

$0$

$1$

$\displaystyle \frac{1}{2}$

$\displaystyle \frac{\sqrt{3}}{2}$

$\displaystyle 10^{\circ}$

$\displaystyle 20^{\circ}$

$\displaystyle 30^{\circ}$

$\displaystyle 40^{\circ}$