Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 39

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Question 1
1 / -0

The value of $$\tan ^{ 2 }{ 60^0 } \text {cosec}^{ 2 }{45^0}+\sec ^{ 2 }{ 45^0 } \sin { 30 ^0}$$ is

A
$$-7$$

B
$$6$$

C
$$-6$$

D
$$7$$

Solution

Substituting the values of the trigonometric ratios, we get
$$(\sqrt{3})^{2}(\sqrt{2})^{2}+(\sqrt{2})^{2}.\dfrac{1}{2}$$
$$=3.2+1$$
$$=7$$
Question 2
1 / -0

Choose the correct option
$$\quad \displaystyle\frac{2\tan30^{\small\circ}}{1+(\tan30^{0})^{2}}$$

A
$$\sin60^{\small\circ}$$

B
$$\cos60^{\small\circ}$$

C
$$\tan60^{\small\circ}$$

D
$$\sin30^{\small\circ}$$

Solution

$$\quad \displaystyle\frac{2\tan30^{\small\circ}}{1+\tan30^{\small\circ}} = \displaystyle\frac{2\left(\displaystyle\frac{1}{\sqrt3}\right)}{1+\left(\displaystyle\frac{1}{\sqrt3}\right)^2} = \displaystyle\frac{\displaystyle\frac{2}{\sqrt3}}{1+\displaystyle\frac{1}{3}}$$

$$\quad = \displaystyle\frac{2}{\sqrt3}\times\displaystyle\frac{3}{4} = \displaystyle\frac{\sqrt3}{2} = \sin60^{\small\circ}$$
Question 3
1 / -0

Find the value of $$\displaystyle \left( \frac{3 \cos 43^{\circ}}{\sin 47^{\circ}} \right)^2 -\frac{\cos 37 ^{\circ}. \text{cosec} 53^{\circ}}{\tan 5^{\circ}. \tan 25^{\circ}. \tan 45^{\circ}. \tan 65^{\circ} \tan 85^{\circ}} $$

A
$$7$$

B
$$0$$

C
$$1$$

D
$$8$$

Solution

$$\left( \dfrac{3 \cos 43^{\circ}}{\sin 47^{\circ}} \right)^2 -\dfrac{\cos 37 ^{\circ}. \text{cosec} 53^{\circ}}{\tan 5^{\circ}. \tan 25^{\circ}. \tan 45^{\circ}. \tan 65^{\circ} \tan 85^{\circ}} = ?$$

$$=\left( \dfrac { 3\cos43^{ \circ } }{ \sin47^{ \circ } } \right) ^{ 2 }-\dfrac { \cos37^{ \circ }\times \text{cosec}53^{ \circ } }{ \tan5^{ \circ }\times \tan25^{ \circ }\times \tan45^{ \circ }\times \tan65^{ \circ }\times \tan85^{ \circ } } $$

$$=\left( \dfrac { 3\sin47^{ \circ } }{ \sin47^{ \circ } } \right) ^{ 2 }-\dfrac { \cos37^{ \circ } }{ \sin53^{ \circ } } \times \dfrac { 1 }{ \tan5^{ \circ }\times \tan25^{ \circ }\times (1)\times cot25^{ \circ }\times cot5^{ \circ } } $$

$$=3^2 - \dfrac{\sin 53^{\circ}}{\sin 53^{\circ}} \times \dfrac{1}{\displaystyle \dfrac{\tan 5^{\circ}}{\tan 5^{\circ}} \times \dfrac{\tan 25^{\circ}}{\tan 25^{\circ}}}$$

$$=9-1=8$$
Hence, option 'D' is correct.
Question 4
1 / -0

The value of $$\displaystyle\frac{3}{4}\tan^230^{\small\circ} - 3\sin^260^{\small\circ} + cosec^245^{\small\circ}$$ is

A
$$1$$

B
$$8$$

C
$$0$$

D
$$12$$

Solution

Substituting the corresponding trigonometric values, we get
$$\dfrac{3}{4}(\dfrac{1}{\sqrt{3}})^{2}-3(\dfrac{\sqrt{3}}{2})^{2}+(\sqrt{2})^{2}$$
$$=\dfrac{3}{4}.\dfrac{1}{3}-3.\frac{3}{4}+2$$
$$=\dfrac{1}{4}-\dfrac{9}{4}+2$$
$$=\dfrac{-8}{4}+2$$
$$=-2+2$$
$$=0$$
Question 5
1 / -0

If $$\sec 4A = cosec (A-20^{\small\circ})$$, where $$4A$$ is an acute angle, find the value of $$A$$.

A
$$A = 32^{\small\circ}$$

B
$$A = 22^{\small\circ}$$

C
$$A = 41^{\small\circ}$$

D
$$A = 16^{\small\circ}$$

Solution

Hence
$$\sin(A-20^{0})=\cos(4A)$$
$$\sin(A-20^{0})=\sin(90^{0}-4A)$$
Hence
$$A-20^{0}=90^{0}-4A$$
$$5A=110^{0}$$
$$A=22^{0}$$
Question 6
1 / -0

The value of $$2\tan^260^{\small\circ} - 4\cos^245^{\small\circ} - 3\sec^230^{\small\circ}$$ is

A
$$0$$

B
$$1$$

C
$$12$$

D
$$8$$

Solution

$$2\tan^260^{\small\circ} - 4\cos^245^{\small\circ} - 3\sec^230^{\small\circ}$$
$$=2(\sqrt3)^2-4(\dfrac{1}{\sqrt2})^2-3(\dfrac{2}{\sqrt 3})^2$$
$$=2(3)-\dfrac{4}{2}-4$$
$$=6-2-4$$
$$=0$$
Question 7
1 / -0

Say yes or no.

$$\tan48^{\small\circ}\tan23^{\small\circ}\tan42^{\small\circ}\tan67^{\small\circ} = 1$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

$$\quad LHS = \tan48^{\small\circ}\tan23^{\small\circ}\tan42^{\small\circ}\tan67^{\small\circ} $$

$$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times\tan(90^{\small\circ} - 48^{\small\circ})\times\tan(90^{\small\circ} - 23^{\small\circ})$$

$$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times cot 48^{\small\circ}\times cot 23^{\small\circ}$$

$$\quad = \tan48^{\small\circ}\times\tan23^{\small\circ}\times\displaystyle\frac{1}{\tan48^{\small\circ}}\times\displaystyle\frac{1}{\tan23^{\small\circ}} = 1$$

$$\therefore \quad LHS = RHS$$
Question 8
1 / -0

In $$\triangle ABC$$, $$\angle B = 90^{\small\circ}$$. If $$AB = 14\space cm$$ and $$AC = 50\space cm$$ then $$\tan A$$ equals:

A
$$\displaystyle\frac{24}{25}$$

B
$$\displaystyle\frac{24}{7}$$

C
$$\displaystyle\frac{7}{24}$$

D
$$\displaystyle\frac{25}{24}$$

Solution

In triangle ABC
$$B=90^{0}$$
Hence
$$AC=50cm$$ will be the hypotenuse.
And $$AB=14cm$$
Therefore
$$BC=\sqrt{AC^{2}-AB^{2}}cm$$
$$=48cm$$
Hence
$$tanA$$
$$=\dfrac{perpendicular}{base}$$

$$=\dfrac{BC}{AB}$$

$$=\dfrac{48}{14}$$

$$=\dfrac{24}{7}$$
Question 9
1 / -0

If $$\sin\theta - \cos\theta = 0$$, then the value of $$(\sin^4\theta + \cos^4\theta)$$ is

A
$$1$$

B
$$\displaystyle\frac{3}{4}$$

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

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

Solution

From the above equation, we get
$$\cos\theta=\sin\theta$$
Hence
$$\theta=45^{0}$$
Therefore
$$\sin^{4}\theta+\cos^{4}\theta$$
$$=\sin^{4}45^{0}+\cos^{4}45^{0}$$
$$=2\times(\dfrac{1}{\sqrt{2}})^{4}$$
$$=2\times\dfrac{1}{4}$$
$$=\dfrac{1}{2}$$
Question 10
1 / -0

$$\sin30^{\small\circ} + \cos60^{\small\circ}$$ equals

A
$$\displaystyle\frac{1+\sqrt3}{2}$$

B
$$\sqrt3$$

C
$$1$$

D
None of these

Solution

$$\sin30^{0}+\cos60^{0}$$
$$=\sin30^{0}+\sin(90^{0}-60^{0})$$
$$=2\sin30^{0}$$
$$=2\times\dfrac{1}{2}$$
$$=1$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 39

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Introduction to Trigonometry Test - 39

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SHARING IS CARING

Question 1

$$-7$$

$$6$$

$$-6$$

$$7$$

Solution

Question 2

$$\sin60^{\small\circ}$$

$$\cos60^{\small\circ}$$

$$\tan60^{\small\circ}$$

$$\sin30^{\small\circ}$$

Solution

Question 3

$$7$$

$$0$$

$$1$$

$$8$$

Solution

Question 4

$$1$$

$$8$$

$$0$$

$$12$$

Solution

Question 5

$$A = 32^{\small\circ}$$

$$A = 22^{\small\circ}$$

$$A = 41^{\small\circ}$$

$$A = 16^{\small\circ}$$

Solution

Question 6

$$0$$

$$1$$

$$12$$

$$8$$

Solution

Question 7

Yes

No

Ambiguous

Data insufficient

Solution

Question 8

$$\displaystyle\frac{24}{25}$$

$$\displaystyle\frac{24}{7}$$

$$\displaystyle\frac{7}{24}$$

$$\displaystyle\frac{25}{24}$$

Solution

Question 9

$$1$$

$$\displaystyle\frac{3}{4}$$

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

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

Solution

Question 10

$$\displaystyle\frac{1+\sqrt3}{2}$$

$$\sqrt3$$

$$1$$

None of these

Solution

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