Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 43

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Question 1
1 / -0

$$\displaystyle \frac{4}{3}\cot ^{2}30^{\circ}+3\sin ^{2}60^{\circ}-2\text{cosec} ^{2}60^{\circ}-\frac{3}{4}\tan ^{2}30^{\circ}$$ is

A
$$1$$

B
$$\displaystyle -\frac{20}{3}$$

C
$$\displaystyle \frac{10}{3}$$

D
$$5$$

Solution

$$\displaystyle \frac{4}{3}\cot ^{2}30^{\circ}+3\sin

^{2}60^{\circ}-2\text{cosec}^{2}60^{\circ}-\frac{3}{4}\tan

^{2}30^{\circ}$$
$$= \displaystyle \frac{4}{3}(\sqrt 3)^2+3(\frac {\sqrt 3}{2})^2-2(\frac {2}{\sqrt 3})^2-\frac{3}{4}(\frac {1}{\sqrt 3})^2$$
$$=\displaystyle \frac{4}{3}\times 3+3\times \frac {3}{4}-2\times \frac {4}{3}-\frac{3}{4}\times \frac {1}{3}$$
$$=\displaystyle 4+ \frac {9}{4}- \frac {8}{3}-\frac{1}{4}$$
$$=\displaystyle \frac {10}{3}$$
Option C is correct.
Question 2
1 / -0

$$\displaystyle \frac{4}{\cot ^{2}30^{\circ}}+\frac{1}{\sin^{2}60^{\circ} }-\cos ^{2}45^{\circ}$$ is

A
$$\displaystyle \frac{12}{5}$$

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

C
$$\displaystyle \frac{13}{6}$$

D
1

Solution

$$\displaystyle \frac{4}{\cot ^{2}30^{\circ}}+\frac{1}{\sin^{2}60^{\circ} }-\cos ^{2}45^{\circ}$$
$$=\displaystyle \frac{4}{\left(\sqrt 3\right)^2}+\dfrac{1}{\left(\dfrac {\sqrt 3}{2}\right)^2 }- \left(\frac {1}{\sqrt 2}\right)^2$$

$$=\displaystyle \frac{4}{3}+\frac{4}{3}- \frac {1}{2}$$

$$=\displaystyle \frac{8}{3}- \frac {1}{2} = \frac {13}{6}$$

Option C is correct.
Question 3
1 / -0

If $$\displaystyle \tan 32^0.\cot (90^0-\theta )=1$$ find $$\theta $$.

A
$$\displaystyle 58^{\circ}$$

B
$$\displaystyle 122^{\circ}$$

C
$$\displaystyle 32^{\circ}$$

D
$$\displaystyle 158^{\circ}$$

Solution

Given, $$\tan  32^0.\cot (90^0-\theta )=1$$
$$\Rightarrow \tan  32^0=\dfrac1{\cot (90^0-\theta )}$$
$$\Rightarrow \tan  32^0=\tan  (90^0-\theta )$$
$$\Rightarrow (90^0-\theta )=32^0$$
$$\Rightarrow \theta=90^0-32^0$$
$$=58^0$$
Option A is correct.
Question 4
1 / -0

The value of the expression $$\displaystyle \frac{5\sin ^{2}30^{\circ}+\cos ^{2}45^{\circ}+4\tan ^{2}60^{\circ}}{2\sin 30^{\circ}\cos 60^{\circ}+\tan 45^{\circ}}$$ is

A
4

B
9

C
$$\displaystyle \frac{53}{12}$$

D
$$\displaystyle \frac{55}{6}$$

Solution

$$\displaystyle \frac{5\sin ^{2}30^{\circ}+\cos ^{2}45^{\circ}+4\tan

^{2}60^{\circ}}{2\sin 30^{\circ}\cos 60^{\circ}+\tan 45^{\circ}}$$

$$=\displaystyle \dfrac{5(\dfrac{1}{2})^2+(\dfrac {1}{\sqrt 2})^2+4\times (\sqrt 3)^2}{2\times \dfrac {1}{2}\times \dfrac {1}{2}+1}$$

$$=\displaystyle \dfrac {\dfrac 54+ \dfrac 12+ 4\times 3}{\dfrac 12+ 1}$$

$$=\displaystyle \dfrac {\dfrac 74+ 12}{\dfrac 32}$$

$$=\displaystyle \dfrac {\dfrac {55}{4}}{\dfrac 32}$$

$$=\displaystyle \dfrac {55}{6}$$

Option D is correct.
Question 5
1 / -0

The value of $$\displaystyle \csc (65^0+\theta )-\sec (25^0-\theta )-\tan (55^0-\theta )+\cot (35^0+\theta )$$ is

A
$$0$$

B
$$-1$$

C
$$\displaystyle \theta +90$$

D
None of these

Solution

Given, $$\displaystyle \csc (65^{\circ}+\theta )-\sec (25-\theta )-\tan (55-\theta )+\cot (35+\theta )$$
$$= \displaystyle \csc (90^{\circ}-25^{\circ}+\theta )-\sec (25^{\circ}-\theta )-\tan (55^{\circ}-\theta )+\cot (90^{\circ}-55^{\circ}+\theta )$$
$$= \displaystyle \csc (90^{\circ}-(25^{\circ}-\theta) )-\sec (25^{\circ}-\theta )-\tan (55^{\circ}-\theta )+\cot (90^{\circ}-(55^{\circ}-\theta) )$$
$$= \displaystyle \sec (25^{\circ}-\theta)-\sec (25^{\circ}-\theta )-\tan (55^{\circ}-\theta )+\tan (55^{\circ}-\theta )$$
$$=0$$
Question 6
1 / -0

Find the value of
$$\displaystyle 4\left( { \sin }^{ 4 }{ 30 }^{ o }+{ \cos }^{ 4 }{ 60 }^{ o } \right) -3\left( { \sin }^{ 2 }{ 45 }^{ o }-2{ \cos }^{ 2 }{ 45 }^{ o } \right) $$

A
$$1$$

B
$$2$$

C
$$0$$

D
$$3$$

Solution

$$4(sin^{4}30^{0}+cos^{4}60^{0})-3(sin^{2}45^{0}-2cos^{2}45^{0})$$
Put the value of $$\sin$$ and $$\cos$$ we get:
=$$\left [ \left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{4} \right ]-3\left [ \left ( \frac{1}{\sqrt{2}} \right )^{2}-2\left ( \frac{1}{\sqrt{2}} \right )^{2} \right ]$$
=$$4\left ( \frac{1}{16}+\frac{1}{16} \right )-3\left ( \frac{1}{2}-2(\frac{1}{2}) \right )$$
=$$\frac{1}{2}+\frac{3}{2}=2$$
Question 7
1 / -0

The value of $$\displaystyle \frac{2\sin 67^{\circ}}{\cos 23^{\circ}}-\frac{\cot 40^{\circ}}{\tan 50^{\circ}}$$

A
0

B
1

C
-1

D
None

Solution

$$\displaystyle \frac{2\sin 67^0}{\cos 23^0}-\frac{\cot 40^0}{\tan 50^0}=\displaystyle \frac{2\sin (90^0-23^0)}{\cos 23^0}-\frac{\cot (90^0-50^0)}{\tan50^0}$$

$$=\displaystyle \frac{2\cos 23^0}{\cos 23^0}-\frac{\tan50^0}{\tan 50^0}$$

$$=2-1=1$$

Option B is correct.
Question 8
1 / -0

The value of $$tan\;1^{\circ}\;tan\;2^{\circ}\;tan\;3^{\circ}....tan\;89^{\circ}$$ is

A
$$1$$

B
$$0$$

C
$$\infty$$

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

Solution

$$tan\;1^{\circ}\;tan\;2^{\circ}\;tan\;3^{\circ}....tan\;45^{\circ}.....tan\;87^{\circ}\;tan\;88^{\circ}\;tan\;89^{\circ}$$
$$=tan\;1^{\circ}\;tan\;2^{\circ}\;tan\;3^{\circ}.....tan\;45^{\circ}.....cot\;3^{\circ}\;cot\;2^{\circ}\;cot\;1^{\circ}=1$$
Question 9
1 / -0

The value of $$\displaystyle { \text{cosec} }^{ 2 }\left( { 90 }^{ o }-\theta \right) -{ \tan }^{ 2 }\theta $$ is :

A
$$2$$

B
$$3$$

C
$$0$$

D
$$1$$

Solution

The value of $$\text{cosec} ^2(90-\theta)-\tan ^2\theta$$ is
$$=\displaystyle { \sec }^{ 2 }\theta -{ \tan }^{ 2 }\theta $$

$$=\dfrac { 1 }{ { \cos }^{ 2 }\theta } -\dfrac { { \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =\dfrac { 1-{ \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta }$$

$$ =\dfrac { { \cos }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } $$

$$=1$$
Question 10
1 / -0

$$\displaystyle \tan { { 5 }^{ o } } .\tan { { 40 }^{ o } } .\tan { 4{ 5 }^{ o } } .\tan { { 50 }^{ o } } .\tan { 8{ 5 }^{ o } } $$ is equal to :

A
$$1$$

B
$$0$$

C
$$2$$

D
$$-1$$

Solution

The value of $$\tan 5^o. \tan 40^o. \tan 45^o. \tan 50^o. \tan 85^o$$ is
$$=\displaystyle \tan { { 5 }^{ o } } .\tan { \left( { 90 }^{ o }-{ 5 }^{ o } \right) . } \tan { { 40 }^{ o } } .\tan { \left( { 90 }^{ o }-{ 40 }^{ o } \right) } . \tan 45^o$$
$$\displaystyle =\tan { { 5 }^{ o } } .\cot { { 5 }^{ o } } .\tan { { 40 }^{ o } } .\cot{ { 40 }^{ o } }.1$$
$$\displaystyle =1\times 1\times 1$$
$$=1$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 43

Result

Introduction to Trigonometry Test - 43

Score

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Rank

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SHARING IS CARING

Question 1

$$1$$

$$\displaystyle -\frac{20}{3}$$

$$\displaystyle \frac{10}{3}$$

$$5$$

Solution

Question 2

$$\displaystyle \frac{12}{5}$$

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

$$\displaystyle \frac{13}{6}$$

1

Solution

Question 3

$$\displaystyle 58^{\circ}$$

$$\displaystyle 122^{\circ}$$

$$\displaystyle 32^{\circ}$$

$$\displaystyle 158^{\circ}$$

Solution

Question 4

4

9

$$\displaystyle \frac{53}{12}$$

$$\displaystyle \frac{55}{6}$$

Solution

Question 5

$$0$$

$$-1$$

$$\displaystyle \theta +90$$

None of these

Solution

Question 6

$$1$$

$$2$$

$$0$$

$$3$$

Solution

Question 7

0

1

-1

None

Solution

Question 8

$$1$$

$$0$$

$$\infty$$

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

Solution

Question 9

$$2$$

$$3$$

$$0$$

$$1$$

Solution

Question 10

$$1$$

$$0$$

$$2$$

$$-1$$

Solution

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