Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\displaystyle {\text{cosec} }^{ 2 }{ 67 }^{ o }-{ \tan }^{ 2 }{ 23 }^{ o }$$ is :

A
$$2$$

B
$$0$$

C
$$1$$

D
$$3$$

Solution

$$\displaystyle { \text{cosec } }^{ 2 }{ 67 }^{ o }-{ \tan }^{ 2 }{ 23 }^{ o }=1+{ \cot }^{ 2 }{ 67 }^{ o }-{ \tan }^{ 2 }{ 23 }^{ o }$$
$$\displaystyle \left[ \because \quad { \text{cosec } }^{ 2 }\theta =1+{ \cot }^{ 2 }\theta \right] $$
$$\displaystyle =1+{ \cot }^{ 2 }\left( { 90 }^{ o }-{ 23 }^{ o } \right) -{ \tan }^{ 2 }{ 23 }^{ o }$$
$$\displaystyle =1+{ \tan }^{ 2 }{ 23 }^{ o }-{ \tan }^{ 2 }{ 23 }^{ o }=1+0=1$$
Question 2
1 / -0

The value of $$\displaystyle \frac { \cos { \left( { 90 }^{ o }-A \right) } }{ \text{cosec}\left( { 90 }^{ o }-A \right) } \times \frac { \cot{ \left( { 90 }^{ o }-A \right) } }{ \sin { A } } $$ is

A
$$\displaystyle \cos { A } $$

B
$$\displaystyle \sin { A } $$

C
$$\displaystyle \tan { A } $$

D
$$\displaystyle \sec { A } $$

Solution

The value of $$\displaystyle \frac { \cos { \left( { 90 }^{ o }-A \right) } }{ \text{cosec}\left( { 90 }^{ o }-A \right) } \times \frac { \cot{ \left( { 90 }^{ o }-A \right) } }{ \sin { A } } $$ is equal to

$$=\displaystyle \frac { \sin { A } }{ \sec { A } } \times \frac { \tan { A } }{ \sin { A } } $$

$$=\dfrac { \tan { A } }{ \sec { A } }$$

$$ =\dfrac { \frac { \sin { A } }{ \cos { A } } }{ \frac { 1 }{ \cos { A } } } $$

$$=\sin { A } $$
Question 3
1 / -0

The value of $$\displaystyle { \left( \frac { \sin { { 47 }^{ o } } }{ \cos { { 43 }^{ o } } } \right) }^{ 2 }+{ \left( \frac { \cos { { 43 }^{ o } } }{ \sin { { 47 }^{ o } } } \right) }^{ 2 }-4{ \cos }^{ 2 }{ 45 }^{ o }$$ is :

A
$$2$$

B
$$0$$

C
$$1$$

D
$$3$$

Solution

The value of $$\displaystyle { \left( \frac { \sin { { 47 }^{ o } } }{ \cos { { 43 }^{ o } } } \right) }^{ 2 }+{ \left( \frac { \cos { { 43 }^{ o } } }{ \sin { { 47 }^{ o } } } \right) }^{ 2 }-4{ \cos }^{ 2 }{ 45 }^{ o }$$ is
$$\displaystyle ={ \left[ \frac { \sin { { 47 }^{ o } } }{ { \cos\left( 90-47 \right) }^{ o } } \right] }^{ 2 }+{ \left[ \frac { \cos { { 43 }^{ o } } }{ { \sin\left( 90-43 \right) }^{ o } } \right] }^{ 2 }-{ 4\left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }$$
$$\displaystyle ={ \left( \frac { \sin { 47 } }{ \sin { 47 } } \right) }^{ 2 }+{ \left( \frac { \cos { 43 } }{ \cos { 43 } } \right) }^{ 2 }-4\left( \frac { 1 }{ 2 } \right) $$
$$\displaystyle ={ \left( 1 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 }-4\times \frac { 1 }{ 2 } $$
$$=1+1-2=0$$
Question 4
1 / -0

The value of $$\displaystyle \frac { 2\cos { { 67 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } } $$ is :

A
$$1$$

B
$$0$$

C
$$4$$

D
$$2$$

Solution

$$\displaystyle \frac { 2\cos { { 67 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } }$$

$$\displaystyle =\frac { 2\cos { \left( { 90 }^{ o }-{ 23 }^{ o } \right) } }{ \sin { { 23 }^{ o } } } -\frac { \tan { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } }{ \cot { { 50 }^{ o } } } $$

$$\displaystyle =\frac { 2\sin { { 23 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \cot { { 50 }^{ o } } }{ \cot { { 50 }^{ o } } }$$

$$\displaystyle \left[ \because \quad \cos { \left( { 90 }^{ o }-\theta \right) } =\sin { \theta } ,\tan { \left( { 90 }^{ o }-\theta \right) =\cot { \theta } } \right] $$

$$\displaystyle =2-1=1$$
Question 5
1 / -0

The value of $$\displaystyle \frac { \cot { { 50 }^{ o } } }{ \tan { { 40 }^{ o } } } $$ is :

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

The value of $$\displaystyle \frac { \cot { { 50 }^{ o } } }{ \tan { { 40 }^{ o } } } $$
$$=\dfrac { \cot { \left( { 90 }^{ o }-{ 40 }^{ o } \right) } }{ \tan { { 40 }^{ o } } } $$
$$=\dfrac { \tan { { 40 }^{ o } } }{ \tan { { 40 }^{ o } } } =1$$
$$\displaystyle \left[ \because \quad \cot { \left( { 90 }^{ o }-\theta \right) } =\tan { \theta } \right] $$
Question 6
1 / -0

The value of $$\displaystyle \frac { \tan { { 49 }^{ o } } }{ \cot { { 41 }^{ o } } } $$ is :

A
$$1$$

B
$$2$$

C
$$0$$

D
$$3$$

Solution

The value of $$\displaystyle \frac { \tan { { 49 }^{ o } } }{ \cot { { 41 }^{ o } } } $$
$$=\dfrac { \tan { \left( { 90 }^{ o }-{ 41 }^{ o } \right) } }{ \cot { { 41 }^{ o } } }$$
$$ =\dfrac { \cot { { 41 }^{ o } } }{ \cot { { 41 }^{ o } } } =1$$
$$\displaystyle \left[ \because \quad \tan { \left( { 90 }^{ o }-\theta \right) =\cot { \theta } } \right] $$
Question 7
1 / -0

The value of $$\displaystyle \frac { \cot { { 40 }^{ o } } }{ \tan { { 50 }^{ o } } } -\frac { 1 }{ 2 } \left( \frac { \cos { { 35 }^{ o } } }{ \sin { { 55 }^{ o } } } \right) $$ is

A
0

B
1

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

D
2

Solution

The value of $$\displaystyle \frac { \cot { { 40 }^{ o } } }{ \tan { { 50 }^{ o } } } -\frac { 1 }{ 2 } \left( \frac { \cos { { 35 }^{ o } } }{ \sin { { 55 }^{ o } } } \right) $$ is

$$\displaystyle =\frac { \cot { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } }{ \tan { { 50 }^{ o } } } -\frac { 1 }{ 2 } \left[ \frac { \cos { \left( { 90 }^{ o }-{ 55 }^{ o } \right) } }{ \sin { { 55 }^{ o } } } \right] $$

$$\displaystyle =\frac { \tan { { 50 }^{ o } } }{ \tan { { 50 }^{ o } } } -\frac { 1 }{ 2 } \left( \frac { \sin { { 55 }^{ o } } }{ \sin { { 55 }^{ o } } } \right) $$

$$=1-\dfrac { 1 }{ 2 } \times 1=\dfrac { 1 }{ 2 } $$
Question 8
1 / -0

The value of $$\displaystyle \frac { \cos { \left( { 90 }^{ o }-A \right) } }{ 1+\sin { \left( { 90 }^{ o }-A \right) } } +\frac { 1+\sin { \left( { 90 }^{ o }-A \right) } }{ \cos { \left( { 90 }^{ o }-A \right) } } $$ is equal to :

A
$$\displaystyle \frac { 2 }{ \cos { A } } $$

B
$$\displaystyle \frac { 2 }{ \sin { A } } $$

C
$$\displaystyle \frac { 2 }{ \sec { A } } $$

D
$$\displaystyle \frac { 2 }{ \text{cosec }A } $$

Solution

Given that:
$$\displaystyle \frac { \cos { \left( { 90 }^{ o }-A \right) } }{ 1+\sin { \left( { 90 }^{ o }-A \right) } } +\frac { 1+\sin { \left( { 90 }^{ o }-A \right) } }{ \cos { \left( { 90 }^{ o }-A \right) } } $$

$$=\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}=\dfrac{\sin^2A+(1+\cos A)^2}{\sin A(1+\cos A)}$$
$$=\dfrac{\sin^2A+\cos^2A+2\cos A+1}{\sin A(1+\cos A)}=\dfrac{2+2\cos A}{\sin A(1+\cos A)}=\dfrac{2}{\sin A}$$
Question 9
1 / -0

The value of $$\displaystyle \frac { \cos { { 75 }^{ o } } }{ \sin { { 15 }^{ o } } } +\frac { \sin { { 12 }^{ o } } }{ \cos { { 78 }^{ o } } } -\frac { \cos { { 18 }^{ o } } }{ \sin { { 72 }^{ o } } } $$ is :

A
0

B
2

C
3

D
1

Solution

we have, $$\displaystyle \cos { { 75 }^{ o } } =\cos { \left( { 90 }^{ o }-{ 15 }^{ o } \right) = } \sin { { 15 }^{ o } } $$
$$\displaystyle \sin { { 12 }^{ o } } =\sin { \left( { 90 }^{ o }-{ 78 }^{ o } \right) } =\cos { { 78 }^{ o } } $$
$$\displaystyle \cos { { 18 }^{ o } } =\cos { \left( { 90 }^{ o }-{ 72 }^{ o } \right) } =\sin { { 72 }^{ o } } $$

$$\displaystyle \therefore \quad \frac { \cos { { 75 }^{ o } } }{ \sin { { 15 }^{ o } } } +\frac { \sin { { 12 }^{ o } } }{ \cos { { 78 }^{ o } } } -\frac { \cos { { 18 }^{ o } } }{ \sin { { 72 }^{ o } } } $$
$$\displaystyle =\frac { \sin { { 15 }^{ o } } }{ \sin { { 15 }^{ o } } } +\frac { \cos { { 78 }^{ o } } }{ \cos { { 78 }^{ o } } } -\frac { \sin { { 72 }^{ o } } }{ \sin { { 72 }^{ o } } } $$
$$\displaystyle =1+1-1=1$$
Hence, option D is correct.
Question 10
1 / -0

The value of $$\displaystyle \frac { \cos { { 70 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \cos { { 59 }^{ o } } }{ \sin { { 31 }^{ o } } } -8{ \sin }^{ 2 }{ 30 }^{ o }$$ is :

A
$$1$$

B
$$2$$

C
$$0$$

D
$$3$$

Solution

The value of $$\displaystyle \frac { \cos { { 70 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \cos { { 59 }^{ o } } }{ \sin { { 31 }^{ o } } } -8{ \sin }^{ 2 }{ 30 }^{ o }$$ is
$$\displaystyle =\frac { \cos { \left( { 90 }^{ o }-{ 20 }^{ o } \right) } }{ \sin { { 20 }^{ o } } } +\frac { \cos { \left( { 90 }^{ o }-{ 31 }^{ o } \right) } }{ \sin { { 31 }^{ o } } } -{ 8\left( \frac { 1 }{ 2 } \right) }^{ 2 }$$
$$\displaystyle =\frac { \sin { { 20 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \sin { { 31 }^{ o } } }{ \sin { { 31 }^{ o } } } -8\times \frac { 1 }{ 4 } $$
$$\displaystyle =1+1-2=0$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 45

Result

Introduction to Trigonometry Test - 45

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2$$

$$0$$

$$1$$

$$3$$

Solution

Question 2

$$\displaystyle \cos { A } $$

$$\displaystyle \sin { A } $$

$$\displaystyle \tan { A } $$

$$\displaystyle \sec { A } $$

Solution

Question 3

$$2$$

$$0$$

$$1$$

$$3$$

Solution

Question 4

$$1$$

$$0$$

$$4$$

$$2$$

Solution

Question 5

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 6

$$1$$

$$2$$

$$0$$

$$3$$

Solution

Question 7

0

1

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

2

Solution

Question 8

$$\displaystyle \frac { 2 }{ \cos { A } } $$

$$\displaystyle \frac { 2 }{ \sin { A } } $$

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

$$\displaystyle \frac { 2 }{ \text{cosec }A } $$

Solution

Question 9

0

2

3

1

Solution

Question 10

$$1$$

$$2$$

$$0$$

$$3$$

Solution

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Question Analysis

My Perfomance

Score

-

Rank

-

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