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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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$6$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$\displaystyle 10^{0}$$

$$\displaystyle 15^{0}$$

$$\displaystyle 30^{0}$$

$$\displaystyle 45^{0}$$

Solution

Question 5

$$\displaystyle r^{2}$$

$$\displaystyle r^{4}$$

1

None of these

Solution

Question 6

$$1$$

$$0$$

$$-1$$

$$3$$

Solution

Question 7

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

5

1

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

Solution

Question 8

$$2$$

$$2n$$

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

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

Solution

Question 9

$$0$$

$$1$$

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

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

Solution

Question 10

$$\displaystyle 10^{\circ}$$

$$\displaystyle 20^{\circ}$$

$$\displaystyle 30^{\circ}$$

$$\displaystyle 40^{\circ}$$

Solution

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