Introduction to Trigonometry Test

Result

Introduction to Trigonometry Test - 30

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$ 3\cos\theta =2\sin\theta $$, then the value of $$ \dfrac{4\sin\theta -3\cos\theta }{2\sin\theta +6\cos\theta }$$ is:

A
$$\dfrac{1}{8}$$

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

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

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

Solution

Given that:
$$3\cos { \theta } =2\sin { \theta } $$
$$\dfrac { \cos { \theta } }{ \sin { \theta } } =\dfrac { 2 }{ 3 } $$
$$\cot { \theta } =\dfrac { 2 }{ 3 } $$
Now, $$\dfrac { 4\sin { \theta } -3\cos { \theta } }{ 2\sin { \theta } +6\cos { \theta } } =\dfrac { \sin { \theta } \left[ 4-3\dfrac { \cos { \theta } }{ \sin { \theta } } \right] }{ \sin { \theta } \left[ 2+6\dfrac { \cos { \theta } }{ \sin { \theta } } \right] } $$
Divided and multiplied the numerator and denominator by $$\sin \theta$$.
                                            $$=\dfrac { 4-3\cot { \theta } }{ 2+6\cot { \theta } } $$
                                            $$=\dfrac { 4-\left( 3\times \dfrac { 2 }{ 3 } \right) }{ 2+\left( 6\times \dfrac { 2 }{ 3 } \right) } $$
                                            $$=\dfrac { 4-2 }{ 2+4 } $$
                                            $$=\dfrac { 2 }{ 6 } $$
                                            $$=\dfrac { 1 }{ 3 }.$$
Hence, the answer is $$=\dfrac { 1 }{ 3 } .$$
Question 2
1 / -0

If $$\tan A =\cfrac{5}{12}$$, find the value of $$(\sin A+ \cos A) \times \sec A$$:

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

B
$$\dfrac{7}{12}$$

C
$$\dfrac{17}{12}$$

D
$$\dfrac{12}{17}$$

Solution

$$\tan { A } =\dfrac { 5 }{ 12 } $$

$$ \left( \sin { A } +\cos { A } \right) \times \sec { A }$$
$$ =\sin { A } .\sec { A } +\cos { A } .\sec { A } $$
$$=\tan { A } +1$$
$$=\cfrac { 5 }{ 12 } +1$$
$$=\cfrac { 17 }{ 12 } $$
Question 3
1 / -0

The value of tan$$1^{\circ}.tan2^{\circ}.tan3^{\circ}.......... tan89^{\circ}$$ is :

A
0

B
1

C
2

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

Solution

$$\tan \theta = \cot(90^{\circ}-\theta)$$
$$\therefore \tan1^{\circ}= \cot89^{\circ}$$
$$ \cot89^{\circ}=\dfrac{1}{\tan89^{\circ} }$$
Similarly all the term will cancel out each other, except $$\tan45^{\circ}$$, whose value is equal to 1.
Question 4
1 / -0

If sin$$\Theta$$ = cos$$\Theta$$, then the value of $$\Theta$$ is :

A
$$0^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$30^{\circ}$$

Solution

For $$\theta =45°,\sin { \theta } \& \cos { \theta } $$
$$\Rightarrow \sin { \theta } =\cos { \theta }$$
$$\Rightarrow \tan { \theta } =1$$
$$\Rightarrow \tan { \theta } =tan 45^0$$
$$\Rightarrow { \theta } =45^0$$
Question 5
1 / -0

Simplify:$$\displaystyle \dfrac {\cos 60^0+\sin 60^0}{\cos 60^0-\sin 60^0}$$

A
$$\sqrt 3-2$$

B
$$\sqrt 3+2$$

C
$$-(\sqrt 3+2)$$

D
$$1$$

E
$$0$$

Solution

$$\displaystyle \dfrac {\cos 60^0+\sin 60^0}{\cos 60^0-\sin 60^0}=\dfrac {\dfrac {1}{2}+\dfrac {\sqrt 3}{2}}{\dfrac {1}{2}-\dfrac {\sqrt 3}{2}}$$
$$=\dfrac {1+\sqrt 3}{1-\sqrt 3}\times \dfrac {1+\sqrt 3}{1+\sqrt 3}$$
$$=\dfrac {4+2\sqrt 3}{1-3}=-(2+\sqrt 3)$$
Question 6
1 / -0

The value of $$\cos 1^o.\cos 2^o.\cos 3^o.....\cos 179^o$$ is equal to

A
$$-1$$

B
$$0$$

C
$$1$$

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

Solution

$$A = \cos1^0\cdot\cos2^0\cdot\cos3^0\cdots\cos{90}^0\cdots\cos{179}^0$$
$$\because \cos{90}^0 = 0$$
$$\therefore A = 0$$

Hence, option B.
Question 7
1 / -0

The value of
$$32 { \cot }^{ 2 }{ 45 }^{ \circ }-8 { \sec }^{ 2 }{ 60 }^{ \circ }+8 { \cos }^{ 3 }{ 30 }^{ \circ }\\$$

A
$$3\sqrt{3}$$

B
$$2\sqrt{3}$$

C
$$\sqrt{3}$$

D
$$6\sqrt{3}$$

Solution

$$32 { \cot }^{ 2 }{ 45 }^{ \circ }-8 { \sec }^{ 2 }{ 60 }^{ \circ }+8 { \cos }^{ 3 }{ 30 }^{ \circ }\\ =32\times 1-8\times { 2 }^{ 2 }+8\times \dfrac { 3\sqrt { 3 } }{ 8 } \\ =32-32+3\sqrt { 3 } \\ =3\sqrt { 3 } $$
Question 8
1 / -0

$$\tan 9^o\times \tan 27^o\times \tan 63^o\times \tan 81^o=$$

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$\tan { { 9 }^{ \circ }.\tan { { 27 }^{ \circ }.\tan { { 63 }^{ \circ }.\tan { { 81 }^{ \circ } } } } } \\ =\tan { \left( { 90 }^{ \circ }-{ 81 }^{ \circ } \right) .\tan { \left( { 90 }^{ \circ }-{ 63 }^{ \circ } \right) .\tan { { 63 }^{ \circ }.\tan { { 81 }^{ \circ } } } } } \\ =\cot { { 81 }^{ \circ }.\cot { { 63 }^{ \circ }. } \tan { { 63 }^{ \circ }.\tan { { 81 }^{ \circ } } } } \\ =1$$
Question 9
1 / -0

Which one of the following is correct ?

A
$$sec^2\alpha=1-tan^2 \alpha$$

B
$$sin^2\alpha=1+cos^2\alpha$$

C
$$tan \alpha cot \alpha=1$$

D
None of these

Solution

Only statement $$(C)$$ is correct.
$$\Rightarrow \tan { \alpha } \cot { \alpha } =1$$
$$\Rightarrow L.H.S=\tan { \alpha } \cot { \alpha } $$
$$=\dfrac { 1 }{ \cot { \alpha } } \cot { \alpha } $$
$$=1=R.H.S.$$
Hence, the answer is $$ \tan { \alpha } \cot { \alpha }=1.$$
Question 10
1 / -0

If $$\tan30^0=x, \tan 45^0=y, \tan 60^0=z,$$ then which of the following is/are correct?

A
$$x+y=z$$

B
$$xy=z$$

C
$$xz=y$$

D
$$y+z=x$$

Solution

$$\displaystyle x=\tan { { 30 }^{ \circ }=\frac { 1 }{ \sqrt { 3 } }, \\ y=\tan { { 45 }^{ \circ }=1,\\ z=\tan { { 60 }^{ \circ }=\sqrt { 3 } } } }$$

$$xz=\dfrac { \sqrt { 3 } }{ \sqrt { 3 } } =1=y$$

$$x + y = \dfrac1{\sqrt3} + 1 = \dfrac{1+\sqrt3}{\sqrt3} ≠ \sqrt3 = z$$

$$y+z = 1+\sqrt3 ≠\dfrac1{\sqrt3} = x$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 30

Result

Introduction to Trigonometry Test - 30

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac{1}{8}$$

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

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

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

Solution

Question 2

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

$$\dfrac{7}{12}$$

$$\dfrac{17}{12}$$

$$\dfrac{12}{17}$$

Solution

Question 3

0

1

2

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

Solution

Question 4

$$0^{\circ}$$

$$45^{\circ}$$

$$60^{\circ}$$

$$30^{\circ}$$

Solution

Question 5

$$\sqrt 3-2$$

$$\sqrt 3+2$$

$$-(\sqrt 3+2)$$

$$1$$

$$0$$

Solution

Question 6

$$-1$$

$$0$$

$$1$$

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

Solution

Question 7

$$3\sqrt{3}$$

$$2\sqrt{3}$$

$$\sqrt{3}$$

$$6\sqrt{3}$$

Solution

Question 8

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 9

$$sec^2\alpha=1-tan^2 \alpha$$

$$sin^2\alpha=1+cos^2\alpha$$

$$tan \alpha cot \alpha=1$$

None of these

Solution

Question 10

$$x+y=z$$

$$xy=z$$

$$xz=y$$

$$y+z=x$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.