Introduction to Trigonometry Test

Result

Introduction to Trigonometry Test - 41

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

Find $$\theta$$, if $$\displaystyle\frac{2\tan\displaystyle\frac{\theta}{2}}{1 + \tan^2\displaystyle\frac{\theta}{2}} = 1,\quad 0^{\small\circ} < \theta \le 90^{\small\circ}$$

A
$$\theta = 60^{\small\circ}$$

B
$$\theta = 40^{\small\circ}$$

C
$$\theta = 90^{\small\circ}$$

D
$$\theta = 70^{\small\circ}$$

Solution

Put $$ \displaystyle\frac{\theta}{2} = A$$, then we must have $$\displaystyle\frac{2\tan A}{1+\tan^2A} = 1$$
$$ \Rightarrow 1+\tan^2A - 2\tan A = 0$$
$$ \Rightarrow(1-\tan A)^2 = 0$$
$$ \Rightarrow\tan A = 1 = \tan45^{\small\circ}$$
$$A = 45^{\small\circ}$$
We know $$\displaystyle\frac{\theta}{2} = A = 45^{\small\circ} $$
$$\Rightarrow \displaystyle\frac{\theta}{2} = 45^{\small\circ} $$
$$ \Rightarrow \theta = 90^{\small\circ}$$
Question 2
1 / -0

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

A
1

B
$$\displaystyle 9\frac{1}{6}$$

C
$$\displaystyle 7\frac{3}{7}$$

D
$$\displaystyle \frac{47}{12}$$

Solution

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

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

= $$\displaystyle \dfrac{\dfrac{5}{4} + \dfrac{1}{2} + 12}{\dfrac{1}{2} + 1}$$

= $$\displaystyle \dfrac{\dfrac{5 + 48 + 2}{4}}{\dfrac{3}{2}}$$

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

= $$\displaystyle 9\dfrac{1}{6}$$
Question 3
1 / -0

$$2(\sin^{6}{\theta}+\cos^{6}{\theta})-3(\sin^{4}{\theta}+\cos^{4}{\theta})+1$$ is equal to

A
$$2$$

B
$$0$$

C
$$4$$

D
$$6$$

Solution

$$2\left( \sin ^{ 6 }{ \theta } +\cos ^{ 6 }{ \theta } \right) -3\left( \sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } \right) +1$$

Using $$(a^3+b^3=(a+b)^3-3\cdot a\cdot b(a+b)) \ and \ (a^2+b^2=(a+b)^2-2ab) $$, we get

$$ =2\left( { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 3 }-3\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) \right) -3\left( { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 2 }-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } \right) +1$$

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

Value of $$\displaystyle\frac{\sin^{2}{{20}^{o}}+\cos^{4}{{20}^{o}}}{\sin^{4}{{20}^{o}}+\cos^{2}{{20}^{o}}}$$ is

A
$$1$$

B
$$2$$

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

D
None of these

Solution

$$\displaystyle\frac{\sin^{2}{{20}^{o}}+\cos^{4}{{20}^{o}}}{\sin^{4}{{20}^{o}}+\cos^{2}{{20}^{o}}}$$

$$\displaystyle=\frac{\sin^{2}{20^0}+\cos^{2}{20^0}.\cos^{2}{20^0}}{\sin^{2}{20^0}.\sin^{2}{20^0}+\cos^{2}{20^0}}$$

$$\displaystyle=\frac{\sin^{2}{20^0}+\cos^{2}{20^0}(1-\sin^{2}{20^0})}{\sin^{2}{20^0}(1-\cos^{2}{20^0})+\cos^{2}{20^0}}$$

$$=\displaystyle\frac{\sin^{2}{20^0}+\cos^{2}{20^0}-\sin^{2}{20^0}\cos^{2}{20^0}}{\sin^{2}{20^0}-\sin^{2}{20^0}\cos^{2}{20^0}+\cos^{2}{20^0}}$$

$$=\displaystyle\frac{1-\sin^{2}{20^0}\cos^{2}{20^0}}{\sin^{2}{20^0}-\sin^{2}{20^0}\cos^{2}{20^0}+\cos^{2}{20^0}}$$

$$=\displaystyle\frac{1-\sin^{2}{20^0}\cos^{2}{20^0}}{1-\sin^{2}{20^0}\cos^{2}{20^0}}=1$$
Question 5
1 / -0

If $$\displaystyle \tan \theta =\frac{x}{y},$$ then $$\displaystyle \frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$$ is equal to

A
$$\displaystyle \frac{x^{2}+y^{2}}{x^{2}-y^{2}}$$

B
$$\displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$$

C
$$\displaystyle \frac{x}{\sqrt{x^{2}+y^{2}}}$$

D
$$\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}$$

Solution

$$\displaystyle \cfrac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$$

Multiply and divide by $$\cos \theta$$
$$=\cfrac{x\cfrac{\sin \theta }{\cos \theta }+y\cfrac{\cos \theta }{\cos \theta }}{x\cfrac{\sin \theta }{\cos \theta }-y\cfrac{\cos \theta}{\cos \theta }}$$

$$=\cfrac{x\times\cfrac{x}{y}+y}{x\times\cfrac{x}{y}-y}$$
$$=\cfrac{\cfrac{x^{2}+y^{2}}{y}}{\cfrac{x^{2}-y^{2}}{y}}$$
$$=\cfrac{x^{2}+y^{2}}{x^{2}-y^{2}}$$
Question 6
1 / -0

If $$\theta$$ is an acute angle such that $$\tan^{2}{\theta}=\dfrac{8}{7}$$,then the value of $$\displaystyle\frac{(1+\sin{\theta})(1-\sin{\theta})}{(1+\cos{\theta})(1-\cos{\theta})}$$ is

A
$$\displaystyle\frac{7}{8}$$

B
$$\displaystyle\frac{8}{7}$$

C
$$\displaystyle\frac{7}{4}$$

D
$$\displaystyle\frac{64}{49}$$

Solution

$$\displaystyle \frac { \left( 1+\sin { \theta } \right) \left( 1-\sin { \theta } \right) }{ \left( 1+\cos { \theta } \right) \left( 1-\cos { \theta } \right) } =\frac { 1-sin^{ 2 }{ \theta } }{ 1-cos^{ 2 }{ \theta } } =\frac { \cos ^{ 2 }{ \theta } }{ sin^{ 2 }{ \theta } } =\frac { 1 }{ \tan ^{ 2 }{ \theta } } =\frac { 7 }{ 8 } $$
Question 7
1 / -0

If $$\sin{x}+\sin^{2}{x}=1$$, then $$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$$ is equals to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

Given $$\sin{x}+\sin^{2}{x}=1$$
$$\Rightarrow \sin x=\cos^2 x$$

Now,$$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$$
$$= \sin^{6}{x}+3\sin^{5}{x}+3\sin^{4}{x}+\sin^{3}{x}-2$$
$$={ (\sin ^{ 2 }{ x } ) }^{ 3 }+3{ (\sin ^{ 2 }{ x } ) }^{ 2 }\sin { x } +3{ (\sin ^{ 2 }{ x } ) }^{ 2 }+\sin ^{ 2 }{ x } \sin { x } -2$$
$$={ (\sin ^{ 2 }{ x } ) }^{ 3 }+3{ (\sin ^{ 2 }{ x } ) }^{ 2 }\sin { x } +3\sin ^{ 2 }{ x } { (\sin { x } ) }^{ 2 }+\sin ^{ 3 }{ x } -2$$
$$={(\sin^{2}{x}+\sin{x})}^{3}-2$$
$$=1-2=-1$$
Question 8
1 / -0

If $$\displaystyle \tan 2A= \cot (A-60^{\circ}),$$ where 2A is an acute angle, then the value of A is

A
$$\displaystyle 30^{\circ}$$

B
$$\displaystyle 60^{\circ}$$

C
$$\displaystyle 50^{\circ}$$

D
$$\displaystyle 24^{\circ}$$

Solution

$$\displaystyle \tan 2A= \cot (A-60^{\circ})$$
$$\cot (90 - 2A) = \cot (A - 60)$$
$$90 - 2 A = A - 60$$
$$3A = 150$$
$$A = 50^{\circ}$$
Question 9
1 / -0

If $$(\sec{A}-\tan{A})(\sec{B}-\tan{B})(\sec{C}-\tan{C})=(\sec{A}+\tan{A})(\sec{B}+\tan{B})(\sec{C}+\tan{C})$$ represents each side of a equilateral triangle, then each side is equal to -

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$\pm 1$$

Solution

Let $$\left ( \sec A+\tan A \right )\left ( \sec B+\tan B \right )\left ( \sec C+\tan C \right )=x$$............(i)

    $$\therefore \left ( \sec A-\tan A \right )\left ( \sec B-\tan B \right )\left ( \sec C-\tan C \right )=x$$.............(ii)

Multiplying Eqs. (i) and (ii), we get

    $$\left ( \sec ^{2}A-\tan ^{2}A \right )\left ( \sec ^{2}B-\tan ^{2}B \right )\left ( \sec ^{2}C-\tan ^{2}C \right )=x^{2}$$...........$$\left ( \sec ^{2}A-\tan ^{2}A \right )=1$$

or   $$x^{2}=1$$ $$\therefore $$   $$x=\pm 1$$

Hence, each side is equal to $$\pm 1$$.
Question 10
1 / -0

If $$\mathrm{cosec} {A}+\cot{A}=\cfrac{11}{2}$$, then $$\tan{A}$$ is equal to

A
$$\displaystyle\frac{22}{117}$$

B
$$\displaystyle\frac{2}{11}$$

C
$$\displaystyle\frac{117}{22}$$

D
$$\displaystyle\frac{44}{117}$$

Solution

Given, $$\displaystyle cosec { A } +\cot { A } =\frac { 11 }{ 2 } $$

$$\displaystyle \Rightarrow { \left( cosec { A } +\cot { A } \right) }^{ 2 }=\frac { 121 }{ 4 } $$

$$\displaystyle \Rightarrow cosec ^{ 2 }{ A } +\cot ^{ 2 }{ A } +2 cosec { A } \cot { A } =\frac { 121 }{ 4 } $$

$$\displaystyle \Rightarrow \frac { 2\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } +2\cos { A } }{ \sin ^{ 2 }{ A } } =\frac { 121 }{ 4 } $$

Applying componendo and dividendo
$$\displaystyle \frac { 2\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } +2\cos { A } -\sin ^{ 2 }{ A } }{ 2\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } +2\cos { A } +\sin ^{ 2 }{ A } } =\frac { 121-4 }{ 121+4 } $$

$$\displaystyle \Rightarrow \frac { 2\cos ^{ 2 }{ A } +2\cos { A } }{ 2+2\cos { A } } =\frac { 117 }{ 125 } $$

$$\displaystyle \Rightarrow \cos { A } =\frac { 117 }{ 125 } \Rightarrow \tan { A } =\frac { 44 }{ 117 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 41

Result

Introduction to Trigonometry Test - 41

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\theta = 60^{\small\circ}$$

$$\theta = 40^{\small\circ}$$

$$\theta = 90^{\small\circ}$$

$$\theta = 70^{\small\circ}$$

Solution

Question 2

1

$$\displaystyle 9\frac{1}{6}$$

$$\displaystyle 7\frac{3}{7}$$

$$\displaystyle \frac{47}{12}$$

Solution

Question 3

$$2$$

$$0$$

$$4$$

$$6$$

Solution

Question 4

$$1$$

$$2$$

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

None of these

Solution

Question 5

$$\displaystyle \frac{x^{2}+y^{2}}{x^{2}-y^{2}}$$

$$\displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$$

$$\displaystyle \frac{x}{\sqrt{x^{2}+y^{2}}}$$

$$\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}$$

Solution

Question 6

$$\displaystyle\frac{7}{8}$$

$$\displaystyle\frac{8}{7}$$

$$\displaystyle\frac{7}{4}$$

$$\displaystyle\frac{64}{49}$$

Solution

Question 7

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 8

$$\displaystyle 30^{\circ}$$

$$\displaystyle 60^{\circ}$$

$$\displaystyle 50^{\circ}$$

$$\displaystyle 24^{\circ}$$

Solution

Question 9

$$0$$

$$1$$

$$-1$$

$$\pm 1$$

Solution

Question 10

$$\displaystyle\frac{22}{117}$$

$$\displaystyle\frac{2}{11}$$

$$\displaystyle\frac{117}{22}$$

$$\displaystyle\frac{44}{117}$$

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.