Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 20

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Question 1
1 / -0

The given expression is $$\displaystyle \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } +\cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } +4 $$ equal to :

A
$$0$$

B
$$5$$

C
$$-5$$

D
$$4$$

Solution

Taking the given expression
$$\sin{\theta}\cos{(90-{\theta})} +\cos{\theta} sin{(90-{\theta})}$$
We know that $$\cos{(90-{\theta})}= sin {\theta} , sin{(90-{\theta})}=cos{\theta}$$
We get
$$\sin { \theta } \sin { \theta } +\cos { \theta } \cos { \theta } +4$$
$$\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +4 $$ we know $$ $$($$\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta }=1)$$
$$=1+4$$
$$=5$$
Question 2
1 / -0

If $$\displaystyle \theta ={ 30 }^{ o }$$ then $$\displaystyle \cos { 2\theta } $$ is :

A
$$0$$

B
$$1$$

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

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

Solution

Given : $$\theta=30^{o}$$
$$\cos 2\theta$$
$$=\cos{2} \left({30}^{o}\right) =\cos {{60}^{o}=\dfrac {1}{2}}$$
Question 3
1 / -0

Evaluate: $$\displaystyle \sqrt { \frac { 1+{ \cos60 }^{ o } }{ 2 } } =$$

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

B
$$1$$

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

D
$$\displaystyle \frac { 1 }{ 4 } $$

Solution

Given that; $$\sqrt { \dfrac { 1+\cos { { 60^{o} } } }{ 2 } } $$

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

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

$$=\sqrt { \dfrac { 3 }{ 4 } } =\dfrac { \sqrt { 3 } }{ 2 } $$
Question 4
1 / -0

Express $$\displaystyle \cos { { 79 }^{ o } } +\sec { { 79 }^{ o } } $$ in terms of angles between $$\displaystyle { 0 }^{ o }$$ and $$\displaystyle { 45 }^{ o }$$

A
$$1$$

B
$$2$$

C
$$\displaystyle \sin { { 11 }^{ o } } +\text{cosec }{ 11 }^{ o }$$

D
$$\displaystyle \cos { { 11 }^{ o } } +\sec { { 11 }^{ o } } $$

Solution

Given,
$$\displaystyle \cos { { 79 }^{ o } } +\sec { { 79 }^{ o } }$$
$$\displaystyle =\cos { \left( { 90 }^{ o }-{ 11 }^{ o } \right) } +\sec { \left( { 90 }^{ o }-{ 11 }^{ o } \right) }$$
$$=\displaystyle \sin{ { 11 }^{ o } }+\text{cosec }{ 11 }^{ o }$$
Question 5
1 / -0

$$\displaystyle \cos { { 45 }^{ o } } .\cos { { 30 }^{ o } } -\sin { { 45 }^{ o } } .\sin { { 30 }^{ o } } $$ is equal to :

A
$$\displaystyle \frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } $$

B
$$\displaystyle 0$$

C
$$\displaystyle \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } $$

D
$$\displaystyle \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } $$

Solution

Given that:
$$\cos { { 45 ^o } } .\cos { { 30^o-\sin { { 45 ^o } .\sin { { 30 ^o } } } } } $$
$$= \dfrac { 1 }{ \sqrt { 2 } } \times \dfrac { \sqrt { 3 } }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } \times \dfrac { 1 }{ 2 } $$(putting the values)

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

$$=\dfrac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } $$
Question 6
1 / -0

For an acute angle $$\theta$$ in a right angled triangle, $$\dfrac{1}{\sin{\theta}} =$$

A
$$\cos{\theta}$$

B
$$\cot{\theta}$$

C
$$\sec{\theta}$$

D
$$\text{cosec}\,{\theta}$$

Solution

$$\sin \theta=\dfrac{opposite}{hypotenuse}$$

$$\cos \theta = \dfrac{adjacent}{hypotenuse}$$

$$\dfrac{1}{\sin \theta}=\dfrac{1}{\dfrac{opposite}{hypotenuse}}=\dfrac{hypotenuse}{opposite}=\text{cosec}\ \theta$$
$$\therefore\ \dfrac{1}{\sin \theta}=\text{cosec}\ \theta$$
Question 7
1 / -0

What is the value of $$\displaystyle \sin { { 30 }^{ o } } \times co\sec{ 30 }^{ o }$$?

A
$$1$$

B
$$0$$

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

D
$$2$$

Solution

$$\displaystyle \sin { { 30 }^{ o }\times co\sec{ 30 }^{ o } } =\cfrac { 1 }{ 2 } \times \cfrac { 2 }{ 1 } =\cfrac { 1 }{ 1 } =1$$
Question 8
1 / -0

$$\dfrac{1}{\cos{\theta}} =$$_____ , for an acute angle $$\theta$$ in a right angled triangle.

A
$$\text{cosec}\,{\theta}$$

B
$$\sec{\theta}$$

C
$$\cot{\theta}$$

D
None of these

Solution

$$\sin \theta=\dfrac{opposite}{hypotenuse}$$

$$\cos \theta = \dfrac{adjacent}{hypotenuse}$$

$$\dfrac{1}{\cos \theta}=\dfrac{1}{\dfrac{adjacent}{hypotenuse}}=\dfrac{hypotenuse}{adjacent}=\sec \theta$$
$$\therefore\ \dfrac{1}{\cos \theta}=\sec \theta$$
Question 9
1 / -0

The value of $$\tan{\theta}\cdot\tan{(90-\theta)}$$ is equal to

A
$$\sin^2{\theta}$$

B
$$1$$

C
$$\cos^2{\theta}$$

D
$$0$$

Solution

$$ \tan\theta \cdot \tan(90-\theta) = \tan\theta \times \cot\theta = 1 $$
$$\because \tan(90-\theta) =\cot \theta $$
Question 10
1 / -0

$$\dfrac{1}{\sec{\theta}} =$$ _____ for an acute angle $$\theta$$ in a right angled triangle.

A
$$\cos{\theta}$$

B
$$\sin{\theta}$$

C
$$\tan{\theta}$$

D
None of these

Solution

$$\sin \theta=\dfrac{opposite}{hypotenuse}$$

$$\cos \theta = \dfrac{adjacent}{hypotenuse}$$

$$\text{sec}\ \theta = \dfrac{hypotenuse}{adjacent}$$
$$\dfrac{1}{\text{sec}\ \theta}=\dfrac{1}{\dfrac{hypotenuse}{adjacent}}=\dfrac{adjacent}{hypotenuse}=\cos\ \theta$$
$$\therefore\ \dfrac{1}{\text{sec}\ \theta}=\cos\ \theta$$

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Introduction to Trigonometry Test - 20

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Introduction to Trigonometry Test - 20

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Question 1

$$0$$

$$5$$

$$-5$$

$$4$$

Solution

Question 2

$$0$$

$$1$$

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

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

Solution

Question 3

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

$$1$$

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

$$\displaystyle \frac { 1 }{ 4 } $$

Solution

Question 4

$$1$$

$$2$$

$$\displaystyle \sin { { 11 }^{ o } } +\text{cosec }{ 11 }^{ o }$$

$$\displaystyle \cos { { 11 }^{ o } } +\sec { { 11 }^{ o } } $$

Solution

Question 5

$$\displaystyle \frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } $$

$$\displaystyle 0$$

$$\displaystyle \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } $$

$$\displaystyle \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } $$

Solution

Question 6

$$\cos{\theta}$$

$$\cot{\theta}$$

$$\sec{\theta}$$

$$\text{cosec}\,{\theta}$$

Solution

Question 7

$$1$$

$$0$$

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

$$2$$

Solution

Question 8

$$\text{cosec}\,{\theta}$$

$$\sec{\theta}$$

$$\cot{\theta}$$

None of these

Solution

Question 9

$$\sin^2{\theta}$$

$$1$$

$$\cos^2{\theta}$$

$$0$$

Solution

Question 10

$$\cos{\theta}$$

$$\sin{\theta}$$

$$\tan{\theta}$$

None of these

Solution

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