Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 54

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Question 1
1 / -0

$$\dfrac { \tan ^{ 2 }{ 60^0 } +4\sin ^{ 2 }{ 45^0 } +3\sec ^{ 2 }{ 30 ^0} +5\cos ^{ 2 }{ 90^0 } }{ \csc { 30^0 } +\sec { 60 ^0} -\cot ^{ 2 }{ 30^0 } } =$$

A
$$\dfrac{9}{7}$$

B
$$9$$

C
$$\dfrac{7}{9}$$

D
$$6$$

Solution

Given that: $$\dfrac { \tan ^{ 2 }{ 60^0+4\sin ^{ 2 }{ 45^0+3\sec ^{ 2 }{ 30^0+5\cos ^{ 2 }{ 90^0 } } } } }{ \csc { 30^0+ } \sec { 60^0-\cot ^{ 2 }{ 30^0 } } } $$

$$=\dfrac { { \left( \sqrt { 3 } \right) }^{ 2 }+4\times { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+3\times { \left( \dfrac { 2 }{ \sqrt { 3 } } \right) }^{ 2 }+5\times 0 }{ 2+2-{ \left( \sqrt { 3 } \right) }^{ 2 } } $$

$$=\dfrac { 3+4\times \dfrac { 1 }{ 2 } +3\times \dfrac { 4 }{ 3 } +0 }{ 4-3 } $$

$$=\dfrac { 3+2+4+0 }{ 1 } $$

$$=9$$
Question 2
1 / -0

If $$(\sin \theta - \cos \theta) = 0$$, then $$(\sin^{4} \theta + \cos^{4} \theta) =$$

A
$$1$$

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

C
$$\dfrac {1}{4}$$

D
$$\dfrac {3}{4}$$

Solution

$$\sin { \theta - } \cos { \theta } =0$$
$$\sin { \theta =\cos { \theta } }$$
$$\dfrac { \sin { \theta } }{ \cos { \theta } } =1$$
$$\tan { \theta =1 } $$
We know that $$\tan { \mathring { 45=1 } } $$
$$so,\theta =\mathring { 45 } $$

$$\left( \sin ^{ 4 }{ \theta + } \cos ^{ 4 }{ \theta } \right) $$
$$\left( \sin ^{ 4 }{ \mathring { 45 } + } \cos ^{ 4 }{ \mathring { 45 } } \right) $$ $$=\left\{ { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 4 }+{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 4 } \right\} $$
$$=\left\{ \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 } \right\} $$
$$=\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } $$
Question 3
1 / -0

$$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 } \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 } \right] -5\cos ^{ 2 }{ 60^0 } =$$

A
$$\dfrac { 35 }{ 9 } $$

B
$$\dfrac { 45 }{ 8 } $$

C
$$\dfrac { 55 }{ 18 } $$

D
$$\dfrac { 49 }{ 36 } $$

Solution

$$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 } \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 } \right] -5\cos ^{ 2 }{ 60^0 } =$$
$$ =\dfrac { 1 }{ 4 } \left[ { \left( \sqrt { 3 } \right) }^{ 4 }-{ \left( \dfrac { 2 }{ \sqrt { 3 } } \right) }^{ 4 } \right] +\dfrac { 3 }{ 2 } \left[ { \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 } \right] -5\times { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$$, substitute the values from the table
$$ =\dfrac { 1 }{ 4 } \left( 9-\dfrac { 16 }{ 9 } \right) +\dfrac { 3 }{ 2 } \left( 2-\dfrac { 1 }{ 3 } \right) -\dfrac { 5 }{ 4 } $$
$$=\dfrac { 65 }{ 36 } +\dfrac { 5 }{ 2 } -\dfrac { 5 }{ 4 } =\dfrac { 110 }{ 36 } =\dfrac { 55 }{ 18 } $$
Question 4
1 / -0

$$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 } \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 } \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 } =$$

A
$$\dfrac { 8 }{ 3 } $$

B
$$\dfrac { 11 }{ 6 } $$

C
$$\dfrac { 13 }{ 6 } $$

D
$$\dfrac { 13 }{ 8 } $$

Solution

$$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 } \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 } \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 } $$
$$ =4\left[ { \left( \dfrac { 1 }{ 2 } \right) }^{ 4 }+{ \left( \dfrac { 1 }{ 2 } \right) }^{ 4 } \right] -\dfrac { 2 }{ 3 } \left[ { \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 } \right] +\dfrac { 1 }{ 2 } \times { \left( \sqrt { 3 } \right) }^{ 2 }$$
$$ =4\left( \dfrac { 1 }{ 16 } +\dfrac { 1 }{ 16 } \right) +\dfrac { 2 }{ 3 } \left( \dfrac { 3 }{ 4 } -\dfrac { 1 }{ 2 } \right) +\dfrac { 3 }{ 2 }$$
$$ =4\times \dfrac { 1 }{ 8 } -\dfrac { 2 }{ 3 } \times \dfrac { 1 }{ 4 } +\dfrac { 3 }{ 2 }$$
$$ =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 6 } +\dfrac { 3 }{ 2 }$$
$$ =\dfrac { 11 }{ 6 } $$
Question 5
1 / -0

$$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚ =$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution

$$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚$$
$$= 2\times \dfrac {1}{2} + 2\times 1 - 3\times \dfrac {1}{2} - 2\left (\dfrac {\sqrt {3}}{2}\right )^{2}$$
$$= 1 + 2 - \dfrac {3}{2} - \dfrac {3}{2}$$
$$= 0$$
Question 6
1 / -0

$$\sin 30^{\circ} \cos 60^{\circ} + \sin 60^{\circ} \cos 30^{\circ} $$ is equal to

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

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

C
$$1$$

D
$$\dfrac {1}{4}$$

Solution

Given: $$\sin { { 30^o } } \cos { { 60^o } } +\sin { { 60 ^o } } \cos { { 30^o } } $$
(Putting the values as per trigonometry table)
$$=\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } $$

$$=\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } =\dfrac { 4 }{ 4 } =1$$
Question 7
1 / -0

If $$\cos ^{ 2 }{ 45^o- } \cos ^{ 2 }{ 30^o=x\cos { { 45^o\sin { { 45^o } } } } } $$, then $$x = $$

A
$$2$$

B
$$\dfrac {3}{2}$$

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

D
$$\dfrac {-3}{4}$$

Solution

Given that: $$\cos ^{ 2 }{ 45^o- } \cos ^{ 2 }{ 30^o=x\cos { { 45^o\sin { { 45^o } } } } } $$
$$\therefore \quad x=\dfrac { \cos ^{ 2 }{ 45^o- } \cos ^{ 2 }{ 30^o } }{ \cos { { 45^o\sin { { 45^o } } } } } $$

$$x=\dfrac { { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }-{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } }{ \dfrac { 1 }{ \sqrt { 2 } } \times \dfrac { 1 }{ \sqrt { 2 } } } =\dfrac { \dfrac { 1 }{ 2 } -\dfrac { 3 }{ 4 } }{ \dfrac { 1 }{ 2 } } =\dfrac { \dfrac { 2-3 }{ 4 } }{ \dfrac { 1 }{ 2 } } =\dfrac { -1 }{ 4 } \times 2=-\dfrac { 1 }{ 2 } $$
Question 8
1 / -0

$$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0 = $$

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

B
$$\dfrac {3}{2}$$

C
$$\dfrac {5}{2}$$

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

Solution

Taking
$$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0 $$
Substituting the values of sin $$30 ^0$$ , sin $$60 ^0$$ , cos $$45 ^0$$
$$={ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$$
$$=\dfrac { 1 }{ 2 } +\dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 } $$

$$=\dfrac { 6 }{ 4} $$

$$=\dfrac { 3 }{ 2 } $$
Question 9
1 / -0

Solve: $$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec } 70^{\circ} $$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

Given:
$$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec} 70^{\circ}$$

$$= \sec 70^{\circ} \sin (90^{\circ} - 70^{\circ}) + \cos 20^{\circ} \text{cosec} (90^{\circ} - 20^{\circ})$$

$$= \sec 70^{\circ} \cos 70^{\circ} + \cos 20^{\circ} \sec 20^{\circ}$$ $$[\because \sin(90^{\circ}-\theta)=\cos \theta$$ and $$\text{cosec}(90^{\circ}-\theta)=\sec \theta]$$

$$= 1 + 1$$ $$[\because ~\sec \theta.\cos \theta=1]$$

$$= 2$$

So, $$2$$ is correct.
Question 10
1 / -0

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

A
$$cosec \theta$$

B
$$\sin \theta$$

C
$$\cos \theta$$

D
None of these

Solution

In the figure,
$$\sec (90 - \theta) = \dfrac {r}{y} = \csc \theta$$

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

So, $$\csc \theta$$ is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 54

Result

Introduction to Trigonometry Test - 54

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SHARING IS CARING

Question 1

$$\dfrac{9}{7}$$

$$9$$

$$\dfrac{7}{9}$$

$$6$$

Solution

Question 2

$$1$$

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

$$\dfrac {1}{4}$$

$$\dfrac {3}{4}$$

Solution

Question 3

$$\dfrac { 35 }{ 9 } $$

$$\dfrac { 45 }{ 8 } $$

$$\dfrac { 55 }{ 18 } $$

$$\dfrac { 49 }{ 36 } $$

Solution

Question 4

$$\dfrac { 8 }{ 3 } $$

$$\dfrac { 11 }{ 6 } $$

$$\dfrac { 13 }{ 6 } $$

$$\dfrac { 13 }{ 8 } $$

Solution

Question 5

$$0$$

$$1$$

$$-1$$

None of these

Solution

Question 6

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

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

$$1$$

$$\dfrac {1}{4}$$

Solution

Question 7

$$2$$

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

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

$$\dfrac {-3}{4}$$

Solution

Question 8

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

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

$$\dfrac {5}{2}$$

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

Solution

Question 9

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 10

$$cosec \theta$$

$$\sin \theta$$

$$\cos \theta$$

None of these

Solution

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