Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 60

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Question 1
1 / -0

Choose the correct option and justify your choice:
$$\displaystyle \sin { 2A } =2\sin { A } $$ is true when $$A =$$

A
$$\displaystyle { 0 }^{ \circ }$$

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

C
$$\displaystyle { 45 }^{ \circ }$$

D
$$\displaystyle { 60 }^{ \circ }$$

Solution

$$\sin 2A = 2\sin A$$ ....given

$$2\sin A \cos A=2\sin A$$ ($$\because \sin 2A =2\sin A \cos A$$)

$$2\sin A \cos A-2 \sin A=0$$

$$2\sin A (\cos A-1)=0$$

Either $$\sin A=0$$ or $$\cos A -1=0$$
$$A=0^\circ$$ or $$\cos A=1\implies A=0^\circ$$

Hence, option A is correct.
Question 2
1 / -0

Choose the correct option. Justify your choice.
$$\displaystyle \frac { 1+{ \tan }^{ 2 }A }{ 1+{ \cot }^{ 2 }A } =$$

A
$$\displaystyle { \sec }^{ 2 }A$$

B
$$\displaystyle -1$$

C
$$\displaystyle { \cot }^{ 2 }A$$

D
$$\displaystyle { \tan }^{ 2 }A$$

Solution

The value of $$\dfrac {1+\tan ^2A}{1+\cot ^2A}$$

$$=\dfrac{1+\dfrac{\sin^2A}{\cos^2A}}{1+\dfrac{\cos^2A}{\sin^2A}}$$

$$=\dfrac{\cos^2A\,+\sin^2A}{\sin^2A+\cos^2 A}\times \dfrac{\sin^2A}{\cos^2A}$$

$$=\tan^2A$$
Question 3
1 / -0

Choose the correct option and justify your choice:
$$\displaystyle \frac { 1-{ \tan }^{ 2 }{ 45 }^{ \circ } }{ 1+{ \tan }^{ 2 }{ 45 }^{ \circ } } =$$

A
$$\displaystyle \tan { { 90 }^{ \circ } } $$

B
$$\displaystyle 1$$

C
$$\displaystyle \sin { { 45 }^{ \circ } } $$

D
$$\displaystyle 0$$

Solution

The value of $$\dfrac {1- \tan ^2 45^o}{1+ \tan ^2 45^o}$$
$$=\displaystyle \frac{1-(1)^2}{1-(1)^2}$$
$$=\displaystyle \frac{1-1}{1+1}=\frac{0}{2}=0$$
Question 4
1 / -0

Choose the correct option and justify your choice:
$$\displaystyle \frac { 2\tan { { 30 }^{ \circ } } }{ 1+{ \tan }^{ 2 }{ 30 }^{ \circ } } =$$

A
$$\displaystyle \sin { { 60 }^{ \circ } } $$

B
$$\displaystyle \cos { { 60 }^{ \circ } } $$

C
$$\displaystyle \tan { { 60 }^{ \circ } } $$

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

Solution

Given trigonometric expression is $$\displaystyle \dfrac { 2\tan { { 30 }^{ \circ } } }{ 1+{ \tan }^{ 2 }{ 30 }^{ \circ } }$$

We know that, $$\tan30^\circ=\dfrac{1}{\sqrt3}$$

$$\therefore \ \displaystyle \frac { 2\tan { { 30 }^{ \circ } } }{ 1+{ \tan }^{ 2 }{ 30 }^{ \circ } } $$

$$= \displaystyle \frac{2\left(\dfrac{1}{\sqrt{3}}\right)}{1+\left(\dfrac{1}{\sqrt{3}}\right)^2}$$

$$=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}$$

$$=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$$

$$=\dfrac{6}{4\sqrt{3}}$$

$$=\dfrac{\sqrt{3}}{2}$$

We know that, $$\sin 60^\circ=\dfrac{\sqrt3}{2}$$

$$\therefore \ \displaystyle \dfrac { 2\tan { { 30 }^{ \circ } } }{ 1+{ \tan }^{ 2 }{ 30 }^{ \circ } }=\sin60^\circ$$

Hence, option A is correct.
Question 5
1 / -0

If $$\sin \theta \, + \, \cos \theta \, = \, p, \, \sin^{3} \, \theta \, + \, \cos^{3} \, \theta \, = \, q$$, then $$ p(p^{2} \, - \, 3) \, = \, $$

A
$$q$$

B
$$2q$$

C
$$-q$$

D
$$-2q$$

Solution

Given, $$\sin\theta+\cos\theta=p$$, $$\sin^3\theta+\cos^3\theta=q$$

We know $$a^3+b^3=(a+b)(a^2+b^2-ab)$$

Therefore, $$\sin^3\theta+\cos^3\theta=(\sin\theta+\cos\theta)(\sin^2\theta+\cos^2\theta-\sin\theta\,\cos\theta)$$

$$q=p(1-\sin\theta\cos\theta)$$

$$\dfrac qp=1-\sin \theta \cos \theta$$

$$\because \sin\theta+\cos\theta=p$$

$$1+2\sin\theta\cos\theta=p^2$$

$$2\sin\theta\cos\theta=p^2-1$$

$$\sin\theta\cos\theta=\dfrac{p^2-1}{2}$$

$$\Rightarrow \dfrac qp=\left[1-\dfrac{(p^2-1)}{2}\right]$$

$$\Rightarrow 2q=p(2-p^2+1)$$

$$\Rightarrow 2q=p(3-p^2)$$

$$\Rightarrow p(p^2-3)=-2q$$
Question 6
1 / -0

If the area of $${\Delta}$$ $$ABC$$ is $$8\sqrt{3}$$ sq. units, then the length of $${AB}$$ is

A
$$4$$ units

B
$$6$$ units

C
$$7$$ units

D
$$8$$ units

Solution

Let the length of $$AB$$ be $$x$$.
Correspondingly, lengths of $$AC$$ and $$BC$$ would be $$x \sin(60^o)$$ and $$x \cos(60^o)$$ respectively.
i.e. $$AC = \cfrac{x\sqrt{3}}{2}$$ and $$AB = \cfrac{x}{2}$$
The area of the triangle $$=\dfrac{1}{2} \times AC \times BC$$
$$ 8 \sqrt {3}= \dfrac{(\frac{x\sqrt{3}}{2}) \times (\frac{x}{2})}{2}$$

$$8 \sqrt 3= \dfrac{\sqrt{3} x^2}{8}$$
Since the area is given to be $$8\sqrt{3}$$, we get
$$x^2 = 64$$
$$\therefore x=8$$
Question 7
1 / -0

$$\tan 10^{\circ} \tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ} $$ is equal to

A
$$0$$

B
$$-1$$

C
$$\dfrac {1}{\sqrt {3}}$$

D
$$1$$

Solution

$$\textbf{Step-1: Apply complementary angles of trigonometric function.}$$
$$\text{We have,}$$
$$\tan 10^{\circ} \tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}$$
$$\text{We know that}$$ $$\tan(90-\theta)=\cot\theta$$
$$= \tan 10^{\circ} . \tan 20^{\circ} . \tan 30^{\circ} . \tan 40^{\circ}. \cot 40^{\circ} . \cot 30^{\circ} . \cot 20^{\circ} . \cot 10^{\circ} $$
$$= 1\times1\times1\times1$$ $$[\because$$ $$\boldsymbol{\mathbf{\tan (90^o-\theta)=\cot\theta}}$$ $$\textbf{and}$$ $$\boldsymbol{\mathbf{\tan\theta\cot \theta=1]}}$$
$$=1$$
$$\textbf{Hence, option - D is correct option}$$
Question 8
1 / -0

If $$\displaystyle \sin { x } +{ \sin }^{ 2 }x=1$$, then $$\displaystyle { \cos }^{ 12 }x+3{ \cos }^{ 10 }x+3{ \cos }^{ 8 }x+{ \cos }^{ 6 }x$$ is equal to:

A
1

B
2

C
3

D
0

Solution

We have $$\sin x+\sin^2x=1\Rightarrow \sin x=1-\sin^2x=\cos^2x$$

$$\therefore \cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x$$
$$=\sin^6x+3\sin^5x+3\sin^4x+\sin^3x$$ ......... $$[\because\sin x = \cos ^2 x]$$
$$=\sin^3x(\sin^3x+3\sin^2x+3\sin x+1)$$
$$=\sin^3x(\sin x+1)^3$$
$$=(\sin x+\sin^2x)^3=1$$
Question 9
1 / -0

Simplify $$\sin^3\theta +\sin\theta -\sin\theta \cos^2\theta $$.

A
$$0$$

B
$$\sin \theta$$

C
$$\sin 2\theta$$

D
$$2\sin \theta$$

E
$$2\sin^3 \theta$$

Solution

$${ \sin }^{ 3 }\theta + \sin\theta- \sin\theta { \cos }^{ 2 }\theta $$
$$= { \sin }^{ 3 }\theta + \sin\theta (1-{ \cos }^{ 2 }\theta )$$
$$ = { \sin }^{ 3 }\theta + \sin\theta \times { \sin }^{ 2 }\theta $$
$$= 2{ \sin }^{ 3 }\theta$$
Question 10
1 / -0

Calculate the value of $$\sqrt {\dfrac {9\sin^{2} \theta + 9\cos^{2} \theta}{4}} $$.

A
$$1.5$$

B
$$2.25$$

C
$$8.9$$

D
$$23.1$$

Solution

Trigonometric identity $$:{\sin}^2\theta +{\cos}^2\theta =1$$

$$\sqrt{\dfrac {9{\sin}^2\theta +9{\cos}^2\theta} {4}}= \sqrt { \dfrac { 9{ (\sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta ) }{ 4 } } $$
$$ = \sqrt { \dfrac { 9\times 1 }{ 4 } } $$
$$=\dfrac { 3 }{ 2 } $$
$$=1.5$$

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Introduction to Trigonometry Test - 60

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Introduction to Trigonometry Test - 60

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Question 1

$$\displaystyle { 0 }^{ \circ }$$

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

$$\displaystyle { 45 }^{ \circ }$$

$$\displaystyle { 60 }^{ \circ }$$

Solution

Question 2

$$\displaystyle { \sec }^{ 2 }A$$

$$\displaystyle -1$$

$$\displaystyle { \cot }^{ 2 }A$$

$$\displaystyle { \tan }^{ 2 }A$$

Solution

Question 3

$$\displaystyle \tan { { 90 }^{ \circ } } $$

$$\displaystyle 1$$

$$\displaystyle \sin { { 45 }^{ \circ } } $$

$$\displaystyle 0$$

Solution

Question 4

$$\displaystyle \sin { { 60 }^{ \circ } } $$

$$\displaystyle \cos { { 60 }^{ \circ } } $$

$$\displaystyle \tan { { 60 }^{ \circ } } $$

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

Solution

Question 5

$$q$$

$$2q$$

$$-q$$

$$-2q$$

Solution

Question 6

$$4$$ units

$$6$$ units

$$7$$ units

$$8$$ units

Solution

Question 7

$$0$$

$$-1$$

$$\dfrac {1}{\sqrt {3}}$$

$$1$$

Solution

Question 8

1

2

3

0

Solution

Question 9

$$0$$

$$\sin \theta$$

$$\sin 2\theta$$

$$2\sin \theta$$

$$2\sin^3 \theta$$

Solution

Question 10

$$1.5$$

$$2.25$$

$$8.9$$

$$23.1$$

Solution

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