Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

If $$ \tan \theta = \dfrac {4}{7}$$, then $$\dfrac {7 \sin \theta - 3\cos \theta}{7\sin \theta + 3\cos \theta} = $$_____

A
$$\dfrac {1}{7}$$

B
$$\dfrac {5}{7}$$

C
$$\dfrac {3}{7}$$

D
$$\dfrac {5}{14}$$

Solution

$$\dfrac {7\sin \theta - 3\cos \theta}{7\sin \theta + 3\cos \theta} = \dfrac {7 \tan \theta - 3}{7 \tan \theta + 3} (\because$$ dividing numerator and denominator by $$\cos \theta)$$
$$= \dfrac {7\times \dfrac {4}{7} - 3}{7 \times \dfrac {4}{7} + 3} = \dfrac {1}{7}$$.
Question 2
1 / -0

$$\text{cosec } 69^{\circ} + \cot 69^{\circ}$$, when expressed in terms of angles between $$0^{\circ}$$ and $$45^{\circ}$$, becomes

A
$$\sec 21^{\circ} + \tan 21^{\circ}$$

B
$$\sin 21^{\circ} + \cot 21^{\circ}$$

C
$$\sin 21^{\circ} + \cos 21^{\circ}$$

D
$$\sec 21^{\circ} + \cot 21^{\circ}$$

Solution

We know that:
$$\cos A = \sin(90˚-A)$$
$$\cot A = \tan(90˚-A)$$
$$\text{cosec } A = \sec(90˚-A)$$

Using these, we get:
$$\text{cosec } 69^{\circ} = \text{cosec} (90˚ - 21˚) = \sec 21˚ $$
$$\cot 69^{\circ} = \cot (90˚ - 21˚) = \tan 21^{\circ}$$

$$\therefore \text{cosec } 69^{\circ} + \cot 69^{\circ} = \sec 21^{\circ} + \tan 21^{\circ}$$
Question 3
1 / -0

Evaluate $$: \dfrac {\cot 63˚}{\tan 27˚}$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-2$$

Solution

We know that
$$\tan A = \cot (90˚-A)$$

$$\dfrac {\cot 63˚}{\tan 27˚} = \dfrac {\tan (90˚ - 63˚)}{\tan 27˚} = \dfrac {\tan 27˚}{\tan 27˚} = 1$$
Question 4
1 / -0

Without trigonometric table, evaluate $$\dfrac {\sec 41^o}{\text{cosec } 49^o}$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-2$$

Solution

Given
$$\dfrac { \sec { { 41^o } } }{ \csc { { 49^o } } } $$
$$=\dfrac { \sec { \left( { 90^o- } { 49^o } \right) } }{ \csc { { 49^o } } } $$
$$=\dfrac { \csc { { 49^o } } }{ \csc { { 49^o } } } $$ as we know $$\sec { \left( 90^o-\theta \right) } =\csc { \theta } $$
$$=1$$
Hence $$1$$ is the answer
Question 5
1 / -0

$$\sin 84^{\circ} + \sec 84^{\circ}$$ expressed in terms of angles between $$0^{\circ}$$ and $$45^{\circ}$$ becomes

A
$$\cos 6^{\circ} + \text{cosec } 6^{\circ}$$

B
$$\sin 6^{\circ} + \cos 6^{\circ}$$

C
$$\sin 6^{\circ} + \text{cosec } 6^{\circ}$$

D
None of these

Solution

We know that
$$\sin A = \cos(90˚-A)$$
$$\tan A = \cot(90˚-A)$$
$$\sec A = \text{cosec }(90˚-A)$$

Thus, we get
$$\sin 84^{\circ} = \cos (90˚ - 84˚)\\ = \cos 6^{\circ}$$

$$\sec 84^{\circ} = \text{cosec } (90˚ - 84˚) \\=\text{cosec }6^{\circ}$$

$$\therefore \sin 84^{\circ} + \sec 84^{\circ} = \cos 6^{\circ} + \text{cosec } 6^{\circ}$$.
Question 6
1 / -0

$$\tan 68^{\circ} + \sec 68^{\circ}$$, when expressed in terms of angles between $$0^{\circ}$$ and $$45^{\circ}$$, becomes

A
$$\cot 22^{\circ} +\text{cosec } 22^{\circ}$$

B
$$\sin 22^{\circ} + \cos 22^{\circ}$$

C
$$\cos 22^{\circ} + \tan 22^{\circ}$$

D
None of these

Solution

Given trigonometric expression is $$\tan 68^o + \sec 68^o$$

We know that,
$$\sin A = \cos(90˚-A)$$
$$\tan A = \cot(90˚-A)$$
$$\sec A = \text{cosec }(90˚-A)$$

Using these, we get:
$$\tan 68^{\circ} = \tan (90˚-22˚) = \cot 22^{\circ}$$
$$\sec 68^{\circ} = \text{sec }(90˚ - 22˚) = \text{cosec } 22˚$$

$$\therefore \ \tan 68^{\circ} + \sec 68^{\circ} = \cot 22^{\circ} + \text{cosec } 22^{\circ}$$.
Question 7
1 / -0

$$\sin 81^{\circ} + \tan 81^{\circ}$$, when expressed in terms of angles between $$0^{\circ}$$ and $$45^{\circ}$$, becomes

A
$$\sin 9^{\circ} + \cos 9^{\circ}$$

B
$$\cos 9^{\circ} + \tan 9^{\circ}$$

C
$$\sin 9^{\circ} + \tan 9^{\circ}$$

D
$$\cos 9^{\circ} + \cot 9^{\circ}$$

Solution

We know that,
$$\sin A = \cos(90˚-A)$$
$$\tan A = \cot(90˚-A)$$
$$\sec A = \text{cosec }(90˚-A)$$

Using these, we get
$$\sin 81^{\circ} = \sin (90^{\circ} - 9^{\circ}) = \cos 9^{\circ}$$
$$\tan 81^{\circ} = \tan (90˚ - 9) = \cot 9^{\circ}$$
$$\therefore \sin 80^{\circ} + \tan 81^{\circ} = \cos 9^{\circ} + \cot 9^{\circ}$$
Question 8
1 / -0

If $$3\cot \theta = 4$$ then $$\dfrac {5 \sin \theta + 3 \cos \theta}{5 \sin \theta - 3 \cos \theta} = $$_____

A
$$\dfrac {1}{3}$$

B
$$3$$

C
$$\dfrac {1}{9}$$

D
$$9$$

Solution

We know that, $$cot\ \theta=\dfrac{sin\ \theta}{cos\ \theta}$$

Given:

$$\cot \theta = \dfrac {4}{3}$$

Thus, $$\dfrac {5 \sin \theta + 3 \cos \theta}{5\sin \theta - 3 \cos \theta} = \dfrac {5 + 3 \cot \theta}{5 - 3 \cot \theta}$$

($$\because$$ dividing the numerator and denominator by $$\sin \theta$$)

$$= \dfrac {5 + 3 \times \dfrac {4}{3}}{5 - 3 \times \dfrac {4}{3}} = 9$$.

Hence option (D) is coorect
Question 9
1 / -0

If $$16\cot \theta = 12$$, then $$\dfrac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = $$ _____

A
$$0$$

B
$$\dfrac {1}{7}$$

C
$$\dfrac {3}{7}$$

D
$$\dfrac {2}{7}$$

Solution

$$\cot \theta = \dfrac {12}{16} = \dfrac {3}{4}$$ and so $$\tan \theta = \dfrac {4}{3}$$ $$[\because \tan \theta=\dfrac{1}{\cot \theta}]$$

$$\dfrac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta} = \dfrac{\tan\theta -1}{\tan\theta+1}$$ (Dividing both the numerator and denominator by $$\cos\theta$$)

Putting values, $$\dfrac {\dfrac {4}{3} - 1}{\dfrac {4}{3} + 1} = \dfrac {\dfrac {1}{3}}{\dfrac {7}{3}} = \dfrac {1}{7}$$

So, $$\dfrac {1}{7}$$ is correct.
Question 10
1 / -0

Find the name of the person who first produce a table for solving a triangle's length and angles.

A
William Rowan Hamilton

B
Hipparchus

C
Euclid

D
Issac Newton

Solution

$$\text {Hipparchus}$$ gave the first table of chords analogus to modern table of sine values, and used them to solve trigonometric problems

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 57

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Introduction to Trigonometry Test - 57

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Question 1

$$\dfrac {1}{7}$$

$$\dfrac {5}{7}$$

$$\dfrac {3}{7}$$

$$\dfrac {5}{14}$$

Solution

Question 2

$$\sec 21^{\circ} + \tan 21^{\circ}$$

$$\sin 21^{\circ} + \cot 21^{\circ}$$

$$\sin 21^{\circ} + \cos 21^{\circ}$$

$$\sec 21^{\circ} + \cot 21^{\circ}$$

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$-2$$

Solution

Question 4

$$0$$

$$1$$

$$2$$

$$-2$$

Solution

Question 5

$$\cos 6^{\circ} + \text{cosec } 6^{\circ}$$

$$\sin 6^{\circ} + \cos 6^{\circ}$$

$$\sin 6^{\circ} + \text{cosec } 6^{\circ}$$

None of these

Solution

Question 6

$$\cot 22^{\circ} +\text{cosec } 22^{\circ}$$

$$\sin 22^{\circ} + \cos 22^{\circ}$$

$$\cos 22^{\circ} + \tan 22^{\circ}$$

None of these

Solution

Question 7

$$\sin 9^{\circ} + \cos 9^{\circ}$$

$$\cos 9^{\circ} + \tan 9^{\circ}$$

$$\sin 9^{\circ} + \tan 9^{\circ}$$

$$\cos 9^{\circ} + \cot 9^{\circ}$$

Solution

Question 8

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

$$3$$

$$\dfrac {1}{9}$$

$$9$$

Solution

Question 9

$$0$$

$$\dfrac {1}{7}$$

$$\dfrac {3}{7}$$

$$\dfrac {2}{7}$$

Solution

Question 10

William Rowan Hamilton

Hipparchus

Euclid

Issac Newton

Solution

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