Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \frac{\sin x}{a}= \frac{\cos x}{b}= \frac{\tan x}{c}= k,$$ then $$\displaystyle bc+\frac{1}{ck}+\frac{ak}{1+bk} $$ is equal to

A
$$\displaystyle k\left ( a+\frac{1}{a} \right )$$

B
$$\displaystyle \frac{1}{k}\left ( a+\frac{1}{a} \right )$$

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

D
$$\displaystyle \frac{a}{k}$$

Solution

$$bc+\cfrac { 1 }{ ck } +\cfrac { ak }{ 1+bk } $$
Substituting values from
$$\cfrac { sinx }{ a } =\cfrac { cosx }{ b } =\cfrac { tanx }{ c } =k$$
we get
$$\cfrac { cosx }{ k } .\cfrac { tanx }{ k } +\cfrac { k }{ tanxk } +\cfrac { \cfrac { sinx }{ k } k }{ 1+\cfrac { cosx }{ k } k } \\ =\cfrac { sinx }{ { k }^{ 2 } } +\cfrac { 1 }{ tanx } +\cfrac { sinx }{ 1+cosx } \\ =\cfrac { sinx }{ { k }^{ 2 } } +\cfrac { cosx }{ sinx } +\cfrac { sinx }{ 1+cosx } \\ =\cfrac { sinx }{ { k }^{ 2 } } +\cfrac { 1 }{ sinx } =\cfrac { ak }{ { k }^{ 2 } } +\cfrac { 1 }{ ak } \\ =\cfrac { 1 }{ k } \left( a+\cfrac { 1 }{ a } \right) $$
Question 2
1 / -0

If $$\displaystyle \sin \theta+\sin ^{2} \theta +\sin ^{3}\theta= 1$$ then the value of $$\displaystyle \cos ^{6}\theta-4\cos ^{4}\theta+8\cos ^{2}\theta$$ equals

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Given $$sin \theta+sin^{2} {\theta}+sin^{3}{\theta}=1$$

$$sin \theta+sin^{3}{\theta}=1-sin^{2} {\theta}$$

$$sin \theta \times (1+sin^{2} {\theta})=cos^{2}{\theta} $$

Squaring on both sides

$$\Rightarrow sin^{2}{\theta} \times (1+sin^{2} {\theta})^{2}=cos^{4}{\theta} $$

$$\Rightarrow (1-cos^{2}{\theta}) \times (1+1-cos^{2} {\theta})^{2}=cos^{4}{\theta} $$

$$\Rightarrow (1-cos^{2}{\theta}) \times (2-cos^{2} {\theta})^{2}=cos^{4}{\theta} $$

$$\Rightarrow (1-cos^{2}{\theta}) \times (4+cos^{4}{\theta}-4cos^{2} {\theta})=cos^{4}{\theta} $$

$$\Rightarrow 4+cos^{4}{\theta} -4 cos^{2}{\theta} -4 cos^{2}{\theta}-cos^{6}{\theta}+4cos^{4}{\theta}=cos^{4}{\theta} $$

$$\Rightarrow -cos^{6}{\theta}+4cos^{4}{\theta}-8cos^{2}{\theta}+4=0$$

$$\Rightarrow cos^{6}{\theta}-4cos^{4}{\theta}+8cos^{2}{\theta}=4$$
Question 3
1 / -0

If $$\displaystyle \sin x+ \sin^{2}x= 1,$$ then the value of $$\displaystyle \cos ^{12}x +3\cos ^{10}x+3\cos ^{8}x+\cos ^{6}x-2$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

$$sinx+{ sin }^{ 2 }x=1\Rightarrow sinx=1-{ sin }^{ 2 }x\\ \Rightarrow sinx={ cos }^{ 2 }x$$ ...(1)
$${ cos }^{ 12 }x+3{ cos }^{ 10 }x+3{ cos }^{ 8 }x+{ cos }^{ 6 }x-2$$
Substituting values from (1), we get
$${ sin }^{ 6 }x+3{ sin }^{ 5 }x+3{ sin }^{ 4 }x+{ sin }^{ 3 }x-2\\ ={ \left( { sin }^{ 2 }x+sinx \right) }^{ 3 }-2=1-2=-1$$
Question 4
1 / -0

Use the information given to find the length of $$AB$$ in cm

A
$$11.74$$

B
$$26.48$$

C
$$19.37$$

D
$$25.32$$

Solution

In $$\triangle APQ$$,
$$AQ = 10$$, $$\angle APQ = 30^{\circ}$$
$$\tan \angle APQ = \dfrac{P}{B} = \dfrac{AQ}{AP}$$
$$\dfrac{1}{\sqrt{3}} = \dfrac{AQ}{AP}$$
$$AP = AQ \sqrt{3}$$
$$AP = 10 \sqrt{3}$$
$$AP = 17.32$$

In $$\triangle PBR$$,
$$RB = 8$$, $$\angle PRB = 45^{\circ}$$
$$\tan \angle PRB = \dfrac{P}{B} = \dfrac{PB}{RB}$$
$$1 = \dfrac{PB}{8}$$
$$PB = 8$$

Now, $$AB = AP + PB$$
$$AB = 17.32 + 8 $$
$$AB = 25.32$$ cm
Question 5
1 / -0

If $$\displaystyle \cos A= \frac{1}{2}$$ and $$\displaystyle \sin B= \frac{1}{\sqrt{2}}$$ , find the value of $$\displaystyle \frac{\tan A\, -\, \tan B}{1\, +\, \tan A\tan B}$$

A
$$\displaystyle 3\, -\, \sqrt{3}$$

B
$$\displaystyle 2\, -\, \sqrt{3}$$

C
$$\displaystyle 2\, +\, \sqrt{3}$$

D
$$\displaystyle 2\, -\, \sqrt{2}$$

Solution

$$\cos A = \dfrac{1}{2}$$
$$\cos A = \cos 60^0$$
$$A = 60^{\circ}$$

$$\sin B = \dfrac{1}{\sqrt{2}}$$
$$\sin B = \sin 45^0$$
$$B = 45^{\circ}$$

Now, $$\displaystyle \frac{\tan A\, -\, \tan B}{1\, +\, \tan A\tan B}$$

= $$\displaystyle \frac{\tan 60^0\, -\, \tan 45^0}{1\, +\, \tan 60^0\tan 45^0}$$

= $$\displaystyle \frac{\sqrt{3}\, -\, 1}{1\, +\, (\sqrt{3})(1)}$$

= $$\displaystyle \frac{(\sqrt{3}\, -\, 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$$

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

= $$\displaystyle 2 - \sqrt{3}$$
Question 6
1 / -0

If $$\displaystyle \frac{\cos ^{4}\theta} {\theta_{1}}+\frac{\sin ^{4}\theta} {\theta_{2}}= \frac{1}{\theta_{1}+\theta_{2}}$$ ,then $$\displaystyle \frac{\theta_{2}+\theta_{1}}{\theta_{1}}$$ equals

A
$$\sin ^{2}\theta $$

B
$$\sec ^{2}\theta $$

C
$$\cos ^{2}\theta $$

D
$$\tan ^{2}\theta $$

Solution

$$\displaystyle \dfrac { \cos ^{ 4 }{ \theta } }{ { \theta }_{ 1 } } +\dfrac { \sin ^{ 4 }{ \theta } }{ { \theta }_{ 2 } } =\dfrac { 1 }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \\ \Rightarrow \dfrac { \cos ^{ 4 }{ \theta } }{ { \theta }_{ 1 } } +\dfrac { \sin ^{ 4 }{ \theta } }{ { \theta }_{ 2 } } =\dfrac { { \left( \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right) }^{ 2 } }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \\ \Rightarrow \dfrac { \cos ^{ 4 }{ \theta } }{ { \theta }_{ 1 } } +\dfrac { \sin ^{ 4 }{ \theta } }{ { \theta }_{ 2 } } =\dfrac { \cos ^{ 4 }{ \theta } +\sin ^{ 4 }{ \theta } +2\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \\ \Rightarrow \cos ^{ 4 }{ \theta } \left( \dfrac { 1 }{ { \theta }_{ 1 } } -\dfrac { 1 }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \right) +\sin ^{ 4 }{ \theta } \left( \dfrac { 1 }{ { \theta }_{ 2 } } -\dfrac { 1 }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \right) =\dfrac { 2\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \\ \Rightarrow \dfrac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } \cos ^{ 4 }{ \theta } +\dfrac { { \theta }_{ 1 } }{ { \theta }_{ 2 } } \sin ^{ 4 }{ \theta } -2\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } =0\\ \Rightarrow { { \theta }_{ 2 } }^{ 2 }\cos ^{ 4 }{ \theta } +{ { \theta }_{ 1 } }^{ 2 }\sin ^{ 4 }{ \theta } -2{ \theta }_{ 1 }{ \theta }_{ 2 }\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } =0\\ \Rightarrow { \left( { \theta }_{ 2 }\cos ^{ 2 }{ \theta } -{ \theta }_{ 1 }\sin ^{ 2 }{ \theta } \right) }^{ 2 }=0\\ \Rightarrow \tan ^{ 2 }{ \theta } =\dfrac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } $$

$$\dfrac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ { \theta }_{ 1 } } =\dfrac { { \theta }_{ 1 }+{ \theta }_{ 1 }\tan ^{ 2 }{ \theta } }{ { \theta }_{ 1 } } =\sec ^{ 2 }{ \theta } $$

Ans: B
Question 7
1 / -0

The value of the expression $$\displaystyle \frac{\cos ^{2}\alpha \sin ^{2}\theta +\sin ^{2}\alpha }{\tan ^{2}\theta+\sin ^{2}\alpha }$$ equals

A
$$\sin ^{2}\alpha$$

B
$$\cos ^{2}\alpha$$

C
$$\sin ^{2}\theta$$

D
$$\cos ^{2}\theta$$

Solution

$$\displaystyle a=\frac { \cos ^{ 2 }{ \alpha } \sin ^{ 2 }{ \theta } +\sin ^{ 2 }{ \alpha } }{ \tan ^{ 2 }{ \theta } +\sin ^{ 2 }{ \alpha } } $$

$$\displaystyle \Rightarrow a=\frac { \cos ^{ 2 }{ \theta } \left[ \sin ^{ 2 }{ \theta } \left( 1-\sin ^{ 2 }{ \alpha } \right) +\sin ^{ 2 }{ \alpha } \right] }{ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \alpha } } $$

$$\displaystyle \Rightarrow a=\frac { \cos ^{ 2 }{ \theta } \left[ \sin ^{ 2 }{ \theta } +\sin ^{ 2 }{ \alpha } \left( 1-\sin ^{ 2 }{ \theta } \right) \right] }{ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \alpha } } $$

$$\displaystyle \Rightarrow a=\frac { \cos ^{ 2 }{ \theta } \left[ \sin ^{ 2 }{ \theta } +\sin ^{ 2 }{ \alpha } \cos ^{ 2 }{ \theta } \right] }{ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \alpha } } $$

$$\therefore \quad a=\cos ^{ 2 }{ \theta } $$

Ans: D
Question 8
1 / -0

If $$\displaystyle\frac{\cos^{4}x }{\theta _{1}}+\displaystyle\frac{\sin^{4}x}{\theta _{2}}=\frac{1}{\theta _{1}+\theta _{2}},$$ then $$\displaystyle\frac{\theta _{2}}{\theta _{1}}$$ equals

A
$$\cos^{2}x$$

B
$$\sin^{2}x$$

C
$$\tan^{2}x$$

D
None of these

Solution

$$\cfrac { { cos }^{ 4 }x }{ { \theta }_{ 1 } } +\cfrac { { sin }^{ 4 }x }{ { \theta }_{ 2 } } =\cfrac { 1 }{ { \theta }_{ 1 }+{ \theta }_{ 2 } } \\ \Rightarrow \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) \left( \cfrac { { cos }^{ 4 }x }{ { \theta }_{ 1 } } +\cfrac { { sin }^{ 4 }x }{ { \theta }_{ 2 } } \right) =1\\ \Rightarrow \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) \left( \cfrac { { cos }^{ 4 }x }{ { \theta }_{ 1 } } +\cfrac { { sin }^{ 4 }x }{ { \theta }_{ 2 } } \right) ={ \left( { sin }^{ 2 }x+{ cos }^{ 2 }x \right) }^{ 2 }\\ \Rightarrow \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) \left( \cfrac { { cos }^{ 4 }x }{ { \theta }_{ 1 } } +\cfrac { { sin }^{ 4 }x }{ { \theta }_{ 2 } } \right) ={ sin }^{ 4 }x+{ cos }^{ 4 }x+2{ sin }^{ 2 }x{ cos }^{ 2 }x\\ \Rightarrow \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) { \theta }_{ 2 }{ cos }^{ 4 }x+\left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) { \theta }_{ 1 }{ sin }^{ 4 }x={ { \theta }_{ 1 }{ \theta }_{ 2 }sin }^{ 4 }x+{ { \theta }_{ 1 }{ \theta }_{ 2 }cos }^{ 4 }x+{ \theta }_{ 1 }{ \theta }_{ 2 }2{ sin }^{ 2 }x{ cos }^{ 2 }x\\ \Rightarrow { { \theta }_{ 2 } }^{ 2 }{ cos }^{ 4 }x+{ { \theta }_{ 1 } }^{ 2 }{ sin }^{ 4 }x-{ \theta }_{ 1 }{ \theta }_{ 2 }2{ sin }^{ 2 }x{ cos }^{ 2 }x=0\\ \Rightarrow \cfrac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } { cos }^{ 4 }x+\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 2 } } { sin }^{ 4 }x-2{ sin }^{ 2 }x{ cos }^{ 2 }x=0\\ \Rightarrow { \left( \sqrt { \cfrac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } { cos }^{ 2 }x } -\sqrt { \cfrac { { \theta }_{ 1 } }{ { \theta }_{ 2 } } { sin }^{ 2 }x } \right) }^{ 2 }=0\\ \Rightarrow { tan }^{ 2 }x=\cfrac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } $$
Question 9
1 / -0

$$\cfrac { \cos { \theta } }{ 1-\sin { \theta } } -\cfrac { \cos { \theta } }{ 1+\sin { \theta } } =2$$ is satisfied by which one of the following values of $$\theta$$?

A
$$\cfrac { \pi }{ 2 } $$

B
$$\cfrac { \pi }{ 3 } $$

C
$$\cfrac { \pi }{ 4 } $$

D
$$\cfrac { \pi }{ 6 } $$
Question 10
1 / -0

Evaluate :$$\displaystyle \frac{5\cos ^{2}60^{\circ}+4\sec ^{2}30^{\circ}-\tan ^{2}45^{\circ}}{\sin ^{2}30^{\circ}+\cos ^{2}30^{\circ}}$$

A
$$\displaystyle 2\frac{5}{16}$$

B
$$\displaystyle \frac{67}{12}$$

C
0

D
1

Solution

$$\displaystyle \frac{5\cos ^{2}60^{\circ}+4\sec ^{2}30^{\circ}-\tan ^{2}45^{\circ}}{\sin ^{2}30^{\circ}+\cos ^{2}30^{\circ}}$$

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

= $$\displaystyle \dfrac{\dfrac{5}{4} + \dfrac{16}{3} - 1}{\dfrac{1}{4} + \dfrac{3}{4}}$$

= $$\displaystyle \dfrac{15 + 64 - 12}{12}$$

= $$\displaystyle \dfrac{67}{12}$$

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Introduction to Trigonometry Test - 69

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Introduction to Trigonometry Test - 69

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Question 1

$$\displaystyle k\left ( a+\frac{1}{a} \right )$$

$$\displaystyle \frac{1}{k}\left ( a+\frac{1}{a} \right )$$

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

$$\displaystyle \frac{a}{k}$$

Solution

Question 2

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 3

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 4

$$11.74$$

$$26.48$$

$$19.37$$

$$25.32$$

Solution

Question 5

$$\displaystyle 3\, -\, \sqrt{3}$$

$$\displaystyle 2\, -\, \sqrt{3}$$

$$\displaystyle 2\, +\, \sqrt{3}$$

$$\displaystyle 2\, -\, \sqrt{2}$$

Solution

Question 6

$$\sin ^{2}\theta $$

$$\sec ^{2}\theta $$

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

$$\tan ^{2}\theta $$

Solution

Question 7

$$\sin ^{2}\alpha$$

$$\cos ^{2}\alpha$$

$$\sin ^{2}\theta$$

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

Solution

Question 8

$$\cos^{2}x$$

$$\sin^{2}x$$

$$\tan^{2}x$$

None of these

Solution

Question 9

$$\cfrac { \pi }{ 2 } $$

$$\cfrac { \pi }{ 3 } $$

$$\cfrac { \pi }{ 4 } $$

$$\cfrac { \pi }{ 6 } $$

Question 10

$$\displaystyle 2\frac{5}{16}$$

$$\displaystyle \frac{67}{12}$$

0

1

Solution

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