Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

If $$A$$ and $$B$$ are complementary angles, then $$\sin A \times \sec B =$$

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$2$$

Solution

Given that both the angles are complementary i.e., $$A+B=90^{\circ}$$
So, $$A=90^{\circ}-B$$
We have :
$$=\sin A\times \sec B$$
$$=\sin(90^o-B)\times \sec B$$
$$= \cos B\times \sec B$$ [using $$\cos \theta = \sin (90^o - \theta)$$]
$$=\cos B \times \dfrac{1}{\cos B}$$
$$=1$$
Question 2
1 / -0

$$\cot (90 - \theta) = $$

A
$$\sin \theta$$

B
$$\cos \theta$$

C
$$\tan \theta$$

D
None of these

Solution

$$\cot (90 - \theta) = \dfrac {y}{x} = \tan \theta$$

So, $$\cot (90 - \theta) = \tan \theta$$

So, $$\tan \theta$$ is correct.
Question 3
1 / -0

If $$A$$ and $$B$$ are acute angle such that $$\sin A = \cos B$$, then $$(A + B) = .....$$

A
$$45^{\circ}$$

B
$$60^{\circ}$$

C
$$90^{\circ}$$

D
$$180^{\circ}$$

Solution

$$\sin A=\cos B$$

$$\Rightarrow \sin A=\sin(90^o-B)$$

$$\Rightarrow A=90^o-B$$

$$\Rightarrow A+B=90^o$$

Therefore, Answer is $$90^o$$
Question 4
1 / -0

If $$\cos 9\alpha = \sin \alpha$$, and $$9\alpha < 90^{\circ}$$, then $$\tan 5\alpha = .....$$

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

B
$$\sqrt {3}$$

C
$$1$$

D
$$0$$

Solution

$$\cos 9\alpha = \sin \alpha$$,
So, $$\cos 9\alpha = \cos (90 - \alpha)\\\therefore 9\alpha = 90 - \alpha\Rightarrow 9\alpha + \alpha = 90^{\circ}\Rightarrow 10\alpha=90^o\Rightarrow \alpha=9^o$$

Now, $$\tan5\alpha=\tan(5*9^o)=\tan45^o=1$$

Therefore, Answer is $$1$$
Question 5
1 / -0

$$\tan (90 - \theta) $$ is equivalent to

A
$$\cos \theta$$

B
$$\sin \theta$$

C
$$(\sin \theta + \cos \theta)$$

D
$$\cot \theta$$

Solution

In the figure,
$$\tan (90^{\circ} - \theta) = \dfrac {x}{y}$$ and $$\cot \theta = \dfrac {x}{y}$$

So $$\tan (90 - \theta) = \cot \theta$$

So $$\cot \theta$$ is correct.
Question 6
1 / -0

$$\cot (90^{\circ} - \theta) $$ is equivalent to

A
$$\cot \theta$$

B
$$\cos \theta$$

C
$$\tan \theta$$

D
$$-\tan \theta$$

Solution

we know
$$\cos (90 - \theta)= \tan \theta$$

So, $$\tan \theta$$ is correct.
Question 7
1 / -0

$$\csc (90 - \theta) = $$ is equivalent to

A
$$\sec \theta$$

B
$$\sin \theta$$

C
$$\cos \theta$$

D
None of these

Solution

In the figure,
$$cosec (90 - \theta) = \dfrac {r}{x}$$ and $$\sec \theta = \dfrac {r}{x}$$

So, $$cosec (90 - \theta) = \sec \theta$$

Hence, $$\sec \theta$$ is correct.
Question 8
1 / -0

$$\sin^{2} 20^{\circ} + \sin^{2} 70^\circ=$$?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$\sin^{2} 20^{\circ} = \cos^{2} (90^{\circ} - 20^{\circ}) = \cos^{2} 70^\circ$$

Substituting $$\sin^{2} 20^{\circ} = \cos^{2} 70^\circ$$

$$\sin^220^o+\sin^270^o=\cos^{2} 70^{\circ} + \sin^{2} 70^{\circ} = 1$$ $$[\because \sin^2\theta+\cos^2\theta=1]$$
Question 9
1 / -0

Evaluate: $$\dfrac {\tan 35^{\circ}}{\cot 55^{\circ}} + \dfrac {\cot 78^{\circ}}{\tan 12^{\circ}} = $$

A
$$0$$

B
$$1$$

C
$$2$$

D
None of these

Solution

Given expression is $$\dfrac { \tan { { 35^o } } }{ \cot { { 55^o } } } +\dfrac { \cot { { 78^o } } }{ \tan { { 12^o } } } $$

We can rewrite the expression as:

$$\dfrac { \tan { { 35^o } } }{ \cot { \left( { 90^o- } { 35^o } \right) } } +\dfrac { \cot { { 78^o } } }{ \tan { \left( { 90^o- } { 78 } \right) } } $$

We know that, $$\cot { \left( 90^o-\theta \right) } =\tan { \theta } $$ and $$\tan { \left( 90^o-\theta \right) } =\cot { \theta } $$

$$\therefore \ $$ The expression becomes,

$$\dfrac { \tan { { 35^o } } }{ \tan { { 35^o } } } +\dfrac { \cot { { 78^o } } }{ \cot { { 78^o } } } $$

$$\Rightarrow 1+1$$

$$\Rightarrow 2$$

Hence, $$\dfrac { \tan { { 35^o } } }{ \cot { { 55^o } } } +\dfrac { \cot { { 78^o } } }{ \tan { { 12^o } } } =2$$
Question 10
1 / -0

If $$\sin 15^{\circ} = \cos (n\times15^{\circ})$$, then $$n = .....$$

A
$$1$$

B
$$2$$

C
$$5$$

D
$$0$$

Solution

As we know that
$$\cos (90^{\circ}-\theta)=\sin \theta$$

$$\sin 15^{\circ}=\cos (90^{\circ}-15^{\circ})=\cos 75^{\circ}=\cos (5\times 15^{\circ})$$

So $$n=5$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 55

Result

Introduction to Trigonometry Test - 55

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SHARING IS CARING

Question 1

$$1$$

$$0$$

$$-1$$

$$2$$

Solution

Question 2

$$\sin \theta$$

$$\cos \theta$$

$$\tan \theta$$

None of these

Solution

Question 3

$$45^{\circ}$$

$$60^{\circ}$$

$$90^{\circ}$$

$$180^{\circ}$$

Solution

Question 4

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

$$\sqrt {3}$$

$$1$$

$$0$$

Solution

Question 5

$$\cos \theta$$

$$\sin \theta$$

$$(\sin \theta + \cos \theta)$$

$$\cot \theta$$

Solution

Question 6

$$\cot \theta$$

$$\cos \theta$$

$$\tan \theta$$

$$-\tan \theta$$

Solution

Question 7

$$\sec \theta$$

$$\sin \theta$$

$$\cos \theta$$

None of these

Solution

Question 8

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

None of these

Solution

Question 10

$$1$$

$$2$$

$$5$$

$$0$$

Solution

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