Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The expression $$ \displaystyle \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$$ can be written as:

A
$$\sec A \text{cosec} A + 1$$

B
$$ \tan A + \cot A$$

C
$$ \sec A + \text{cosec} A$$

D
$$ \sin A \cos A + 1$$

Solution

$$\cfrac{\tan A}{1-\cot A}+\cfrac{\cot A}{1-\tan A}$$

$$=\cfrac{\tan A}{\cfrac{\tan A-1}{\tan A}}+\cfrac{\cot A}{1-\tan A}$$

$$=\cfrac{\tan ^{2}A}{\tan A-1}-\cfrac{\cot A}{\tan A-1}$$

$$=\cfrac{\tan ^{2}A-\cot A}{\tan A-1}$$

$$=\cfrac{\tan ^{3}A-1}{\tan ^{2}A-\tan A}$$....................$$\because \left (\tan A=\dfrac {1}{\cot A}\right )$$

$$=\cfrac{(\tan A-1)(\tan ^{2}A+\tan A+1)}{\tan A(\tan A-1)}$$

$$=\cfrac{\tan ^{2}A+1+\tan A}{\tan A}$$

$$=\cfrac{\sec^{2}A+\tan A}{\tan A}$$

$$=\cfrac{\sec^{2}A}{\tan A}+1$$

$$=\cfrac{\cfrac{1}{\cos^{2}A}}{\cfrac{\sin A}{\cos A}}+1$$

$$=\cfrac{\cfrac{1}{\cos A}}{\sin A}+1$$

$$=\cfrac{1}{\sin A\cdot \cos A}+1$$

$$=\cfrac{1}{\cos A}.\cfrac{1}{\sin A}+1$$

$$=\sec A.\text{cosec}A+1$$.
Question 2
1 / -0

The value of tan $$1^o$$ tan $$2^o$$ tan $$3^o$$.... tan $$89^o$$ is.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

We have to find the value of $$tan 1^{0} tan 2^{0} tan 3^{0} .....tan 89^{0}$$
We know that $$tan 89^{0}=tan(90^{0}-1^{0})=cot 1^{0}$$
Similarly $$tan 88^{0}=cot 2^{0}$$ and so on
Finally we have $$tan46^{0}=cot44^{0}$$
So we get $$tan1^{0} tan 2^{0}........tan 89^{0}=(tan1^{0}cot1^{0})(tan2^{0}cot2^{0}).........(tan44^{0}cot44^{0})(tan45^{0})$$
$$=(1)(1).....(1)(1)=1$$
Question 3
1 / -0

If $$\tan \theta= \cot \theta$$, then the value of $$\sec \theta$$ is :

A
$$2$$

B
$$1$$

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

D
$$\sqrt{2}$$

Solution

$$\tan { \theta } =\cot { \theta } $$
$$\Rightarrow \tan ^{ 2 }{ \theta } =1$$
$$\Rightarrow \tan { \theta } =1$$
$$\Rightarrow \theta =45°$$
$$\therefore \sec { 45° } =\sqrt { 2 } $$
Hence, the answer is $$\sqrt { 2 }.$$
Question 4
1 / -0

The value of $$\displaystyle\,\frac{2}{tan\,30^{\circ}}$$ is ________

A
3.46

B
3.56

C
34.6

D
3.48

Solution

$$\dfrac{2}{\tan30^{\circ}} = \tfrac{2}{\frac{1}{\sqrt{3}}} = 2 \sqrt{3} = 3.46$$
Question 5
1 / -0

If $$\theta =45^{\circ},$$ then the value of $$cosec^{2}\theta$$ is

A
$$\dfrac{1}{\sqrt{2}}$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$2$$

Solution

$$\theta =45°$$
$$\Rightarrow cosec^{ 2 }45°={ \left( \sqrt { 2 } \right) }^{ 2 }=2$$
Hence, the answer is $$2.$$
Question 6
1 / -0

The value of $$\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }$$ is :

A
$$0$$

B
$$1$$

C
$$2$$

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

Solution

$$\displaystyle \tan { { 45 }^{ o } } \times \cot{ { 45 }^{ o } }=1\times 1=1$$
Question 7
1 / -0

$$\displaystyle \sin { { 90 }^{ o } } $$ is :

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$-2$$

Solution

$$\sin { 90° } =1$$
Hence, the answer is $$1.$$
Question 8
1 / -0

The value of $$\displaystyle \tan 45^{\circ} + \tan 60^{\circ}.$$

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

B
$$2+\sqrt{3}$$

C
$$3$$

D
None

Solution

Taking
$$tan 45^0+tan 60^0$$
substituting value of $$\tan { \mathring { 45 } }= 1 , \tan { \mathring { 60 } }=\sqrt{3} $$ we get
$$=1+\sqrt { 3 } $$
Question 9
1 / -0

If $$\displaystyle \theta =45$$ then $$\displaystyle \frac { 2\tan { \theta } }{ 1+{ \tan }^{ 2 }\theta } $$ is :

A
$$1$$

B
$$0$$

C
$$2$$

D
$$3$$

Solution

$$\dfrac{2 \tan \theta}{1+\tan \theta}$$

$$=\displaystyle \frac { 2\tan { { 45 }^{ o } } }{ 1+{ \tan }^{ 2 }{ 45 }^{ o } } $$

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

Hence, option A is correct.
Question 10
1 / -0

Find $$\tan\theta,$$ if $$\theta=45^o.$$

A
$$1$$

B
$$2$$

C
$$3$$

D
None of these

Solution

Given: $$\theta = 45^o$$
Therefore, $$ \tan \theta=\tan 45^o=1$$
Thus option A is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 19

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Introduction to Trigonometry Test - 19

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SHARING IS CARING

Question 1

$$\sec A \text{cosec} A + 1$$

$$ \tan A + \cot A$$

$$ \sec A + \text{cosec} A$$

$$ \sin A \cos A + 1$$

Solution

Question 2

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 3

$$2$$

$$1$$

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

$$\sqrt{2}$$

Solution

Question 4

3.46

3.56

34.6

3.48

Solution

Question 5

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

$$1$$

$$\dfrac{1}{2}$$

$$2$$

Solution

Question 6

$$0$$

$$1$$

$$2$$

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

Solution

Question 7

$$0$$

$$-1$$

$$1$$

$$-2$$

Solution

Question 8

$$1+\sqrt{3}$$

$$2+\sqrt{3}$$

$$3$$

None

Solution

Question 9

$$1$$

$$0$$

$$2$$

$$3$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

None of these

Solution

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