Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \tan ^{2}\theta =\left ( 1-e^{2} \right )$$ then $$\displaystyle \sec \theta +\tan ^{3}\theta\, \text{cosec}\, \theta $$ is equal to

A
$$\displaystyle \left ( 2+e^{2} \right )^{\tfrac32}$$

B
$$\displaystyle \left ( 2-e^{2} \right )^{\tfrac32}$$

C
$$\displaystyle \left ( 1-e^{2} \right )^{\tfrac32}$$

D
$$\displaystyle \left ( 1+e^{2} \right )^{\tfrac32}$$

Solution

$$\displaystyle \sec \theta +\tan ^{3}\theta \,\text{cosec}\, \theta $$
$$\displaystyle = \frac{1}{\cos \theta }+\frac{\sin ^{2}\theta }{\cos ^{3}\theta }=\frac{\cos ^{2}\theta+\sin ^{2}\theta}{\cos ^{3}\theta}$$
$$\displaystyle =\sec ^{3}\theta $$
Now $$\displaystyle \tan ^{2}\theta =\left ( 1-e^{2} \right )\Rightarrow \sec ^{2}\theta = 2-e^{2}$$
$$\displaystyle \therefore \sec ^{3}\theta =\left ( 2-e^{2} \right )^{\tfrac32}$$
Question 2
1 / -0

If $$\displaystyle a\cos ^{3}\theta +3a\cos \theta \sin ^{2}\theta =m$$ and $$\displaystyle a\sin ^{3}\theta +3a\cos ^{2}\theta \sin \theta=n$$ then $$\displaystyle \left ( m+n \right )^{\tfrac23}+\left ( m-n \right )^{\tfrac23}$$ is equal to

A
$$\displaystyle a^{\tfrac23}$$

B
$$\displaystyle a^{\tfrac43}$$

C
$$\displaystyle 2a^{\tfrac23}$$

D
$$\displaystyle 2a^{\tfrac43}$$

Solution

Given : $$\displaystyle a\cos ^{3}\theta +3a\cos \theta \sin ^{2}\theta =m$$.....(1)
$$\displaystyle a\sin ^{3}\theta +3a\cos ^{2}\theta \sin \theta=n$$.....(2)

Adding equation (1) and (2) we get,

$$ \therefore \displaystyle m+n=a\left ( \cos ^{3}\theta +\sin ^{3}\theta \right )+3a\cos \theta \sin \theta \left ( \cos \theta +\sin \theta \right )=a\left ( \cos \theta +\sin \theta \right )^{3}$$
Similarly, we get
$$\displaystyle m-n=a\left ( \cos \theta -\sin \theta \right )^{3}$$

$$=\displaystyle \left ( m+n \right )^{\tfrac23}+\left ( m-n \right )^{\tfrac23}$$

$$=\displaystyle a^{{2}/{3}}[((cos \theta + sin \theta)^3)^{2/3}+((cos \theta-sin\theta)^3)^{2/3}]$$

$$=\displaystyle a^{2/3}[(cos\theta+sin\theta)^2+(cos\theta-sin\theta)^2]$$

After solving the above two brackets we get,

$$=a^{\tfrac23}\left [ \left ( 2\left ( \cos ^{2}\theta +\sin ^{2}\theta \right ) \right ) \right ]=2a^{\tfrac23}$$

Hence option $$'C'$$ is the answer.
Question 3
1 / -0

The value of $$\displaystyle 4\left ( \sin ^{4}30^{\circ}+\cos ^{4}30^{\circ} \right )-3\left ( \cos ^{2}45^{\circ}+\sin ^{2}90^{\circ} \right ) $$ is:

A
$$-\dfrac12$$

B
$$-2$$

C
$$2$$

D
$$\dfrac12$$

Solution

Given: $$4\left( \sin ^{ 4 }{ 30° } +\cos ^{ 4 }{ 30° } \right) -3\left( \cos ^{ 2 }{ 45° } +\sin ^{ 2 }{ 90° } \right) $$

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

$$=4\left( \dfrac { 1 }{ 16 } +\dfrac { 9 }{ 16 } \right) -3\left( \dfrac { 1 }{ 2 } +1 \right) $$

$$=\left( 4\times \dfrac { 10 }{ 16 } \right) -\left( 3\times \dfrac { 3 }{ 2 } \right) $$

$$=\dfrac { 5 }{ 2 } -\dfrac { 9 }{ 2 } $$

$$=\dfrac { -4 }{ 2 } $$

$$=-2$$

Hence, the answer is $$-2.$$
Question 4
1 / -0

$$\displaystyle \sin^{2} 85^{\circ} + \sin ^{2}5^{\circ} =$$ ______

A
$$0$$

B
$$1$$

C
$$\dfrac12$$

D
$$\dfrac23$$

Solution

Given, $$ \sin ^{ 2 }{ { 85 }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } $$

We know that $$ \sin ^{ 2 }{ { A }^{ o } } = \cos ^{ 2 }{ { (90-A) }^{ o } } $$

$$ \Rightarrow \sin ^{ 2 }{ { 85 }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } = \cos ^{ 2 }{ {(90-85) }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } = \cos ^{ 2 }{{ 5 }^{ o } } + \sin ^{ 2 }{ { 5 }^{ o } } = 1 $$

(Since, $$ \cos ^{ 2 }{ {A }^{ o } } + \sin ^{ 2 }{ { A }^{ o } } = 1 $$)
Question 5
1 / -0

If $$\displaystyle \sin ^{4}\theta +\cos ^{4}\theta =\dfrac{1}{2} $$ then the value of $$\displaystyle \sin \theta \cos \theta $$ is

A
$$\displaystyle \pm \frac{1}{8} $$

B
$$\displaystyle \pm \frac{1}{4} $$

C
$$\displaystyle \pm 1 $$

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

Solution

We know that $$ { a }^{ 4 }+{ b }^{ 4 }=\quad \left( { a }^{ 2 }+b^{ 2 } \right) ^{ 2 }-2a^{ 2 }{ b }^{ 2 } $$

So, $$ \sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ 0 } =\quad \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) ^{ 2 }-2(\sin ^{ 2 }{ \theta } )^{ 2 }{ (\cos ^{ 2 }{ \theta } ) }^{ 2 } $$

But $$ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } = 1 $$

So, $$ \sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta }= 1^{2} -2(\sin \theta \cos \theta) ^ {2} $$
$$ \Rightarrow \dfrac {1}{2} = 1 - 2(\sin \theta \cos \theta) ^ {2} $$
$$ \Rightarrow 2(\sin \theta \cos \theta) ^ {2}= \dfrac {1}{2} $$
$$ \Rightarrow (\sin \theta \cos \theta) ^ {2}= \dfrac {1}{4} $$
$$ \Rightarrow \sin \theta \cos \theta = \pm \dfrac {1}{2} $$
Question 6
1 / -0

If $$\displaystyle \sqrt{3}\tan \theta =3\sin \theta $$ then the value of $$\displaystyle \sin ^{2}\theta -\cos ^{2}\theta $$ is

A
2

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

C
$$\displaystyle \frac{1}{3}$$

D
3

Solution

Given $$\sqrt{3}\tan\theta=3\sin \theta$$

$$\sqrt 3\dfrac{\sin \theta}{\cos \theta}=3\sin \theta$$

$$\sqrt 3\dfrac{\sin \theta}{\cos \theta}-3\sin \theta=0$$

$$\sqrt3\sin \theta \big(\dfrac{1}{\cos \theta}-\sqrt3\big)=0$$

Now either $$\sin \theta=0$$ or $$\cos \theta=\dfrac{1}{\sqrt3}$$

But $$\sin \theta \neq 0$$

Hence $$ \cos \theta =\dfrac{1}{\sqrt3}$$

Now u\sin g $$\sin ^2\theta+\cos ^2\theta=1$$,we will get $$\sin \theta=\sqrt{\dfrac{2}{3}}$$

Hence $$\sin ^2\theta-\cos ^2\theta=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}$$
Question 7
1 / -0

If $$\displaystyle a=\cos \theta +b\sin \theta =c$$, then the value of $$\displaystyle \left ( a\sin \theta -b\cos \theta \right )$$ is

A
$$\displaystyle \pm \sqrt{a^{2}+b^{2}-c^{2}}$$

B
$$\displaystyle \pm \sqrt{a^{2}-b^{2}+c^{2}}$$

C
$$\displaystyle \pm \sqrt{a^{2}+b^{2}+c^{2}}$$

D
$$\displaystyle \pm \sqrt{b^{2}+c^{2}-a^{2}}$$

Solution

$$a\cos { \theta } +b\sin { \theta } =c$$

Squaring both sides, we get
$$\Rightarrow { \left( a\cos { \theta } +b\sin { \theta } \right) }^{ 2 }={ c }^{ 2 }$$

$$\Rightarrow { a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } +2ab\sin { \theta } \cos { \theta } ={ c }^{ 2 }$$

$$\Rightarrow { a }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) +{ b }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) +2ab\sin { \theta } \cos { \theta } ={ c }^{ 2 }$$

$$\Rightarrow { a }^{ 2 }-{ a }^{ 2 }\sin ^{ 2 }{ \theta } +{ b }^{ 2 }-{ b }^{ 2 }\cos ^{ 2 }{ \theta } +2ab\sin { \theta } \cos { \theta } ={ c }^{ 2 }$$

$$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }-\left[ { a }^{ 2 }\sin ^{ 2 }{ \theta } +{ b }^{ 2 }\cos ^{ 2 }{ \theta } -2ab\sin { \theta } \cos { \theta } \right] ={ c }^{ 2 }$$

$$\Rightarrow { a }^{ 2 }\sin ^{ 2 }{ \theta } +{ b }^{ 2 }\cos ^{ 2 }{ \theta } -2ab\sin { \theta } \cos { \theta } ={ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }$$

$$\Rightarrow { \left( a\cos { \theta } -b\sin { \theta } \right) }^{ 2 }={ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } $$

$$\Rightarrow { \left( a\cos { \theta } -b\sin { \theta } \right) }=\pm \sqrt { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } $$

Hence, the answer is $$\pm \sqrt { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } .$$
Question 8
1 / -0

If $$\displaystyle \left ( \tan x+\sec x \right )\left ( \tan y+\sec y \right )\left ( \tan z+\sec z \right )=\left ( \sec x-\tan x \right )\left ( \sec y-\tan z \right )=K$$, then the value of $$K$$ is

A
$$\displaystyle \pm \sqrt{2}$$

B
$$\displaystyle \pm \sqrt{3}$$

C
$$\displaystyle \pm 2$$

D
$$\displaystyle \pm 1$$

Solution

$$\displaystyle \left ( \tan x+\sec x \right )^{2}\left ( \tan y+\sec y \right )^{2}\left ( \tan z+\sec z \right )^{2}$$
$$\displaystyle =\left ( \sec ^{2}x-\tan ^{2}x \right )\left ( \sec ^{2}y-\tan ^{2}y \right )\left ( \sec ^{2}z-\tan ^{2}z \right )$$
$$\displaystyle = 1\times 1\times 1=1=k^{2}$$
$$\displaystyle \therefore K=\pm 1$$
Question 9
1 / -0

If $$\displaystyle \sin \Theta +\text{cosec }\Theta =2 $$ find the value of $$\displaystyle \cot \Theta +\cos \Theta $$

A
$$0$$

B
$$\dfrac12$$

C
$$1$$

D
$$2$$

Solution

Given, $$ \sin \Theta + \text{cosec } \Theta = 2 $$

This is possible only when $$ \Theta = 90^o $$ as $$ \sin 90^0 = \text{cosec } 90^o = 1 $$

So, $$ \cot \Theta + \cos \Theta = \cot 90^o + \cos 90^ 0 = 0 + 0 = 0 $$
Question 10
1 / -0

If $$\displaystyle \text{cosec}\, \theta -\cot \theta =x $$ then $$\displaystyle \text{cosec}\, \theta +\cot \theta =$$ _____

A
$$x$$

B
$$2x$$

C
$$\dfrac 1x$$

D
$$\dfrac1{2x}$$

Solution

$$cosec\;\theta -\cot { \theta } =x$$
Now, $$cosec^{ 2 }\theta -\cot ^{ 2 }{ \theta } =1$$
$$\Rightarrow \left( cosec\;\theta -\cot { \theta } \right) \left( cosec\;\theta +\cot { \theta } \right) =1$$
$$\Rightarrow x\left( cosec\;\theta +\cot { \theta } \right) =1$$
$$\Rightarrow cosec\;\theta +\cot { \theta } =\dfrac { 1 }{ x } $$
Hence, the answer is $$\dfrac { 1 }{ x } .$$

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Introduction to Trigonometry Test - 48

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Introduction to Trigonometry Test - 48

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SHARING IS CARING

Question 1

$$\displaystyle \left ( 2+e^{2} \right )^{\tfrac32}$$

$$\displaystyle \left ( 2-e^{2} \right )^{\tfrac32}$$

$$\displaystyle \left ( 1-e^{2} \right )^{\tfrac32}$$

$$\displaystyle \left ( 1+e^{2} \right )^{\tfrac32}$$

Solution

Question 2

$$\displaystyle a^{\tfrac23}$$

$$\displaystyle a^{\tfrac43}$$

$$\displaystyle 2a^{\tfrac23}$$

$$\displaystyle 2a^{\tfrac43}$$

Solution

Question 3

$$-\dfrac12$$

$$-2$$

$$2$$

$$\dfrac12$$

Solution

Question 4

$$0$$

$$1$$

$$\dfrac12$$

$$\dfrac23$$

Solution

Question 5

$$\displaystyle \pm \frac{1}{8} $$

$$\displaystyle \pm \frac{1}{4} $$

$$\displaystyle \pm 1 $$

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

Solution

Question 6

2

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

$$\displaystyle \frac{1}{3}$$

3

Solution

Question 7

$$\displaystyle \pm \sqrt{a^{2}+b^{2}-c^{2}}$$

$$\displaystyle \pm \sqrt{a^{2}-b^{2}+c^{2}}$$

$$\displaystyle \pm \sqrt{a^{2}+b^{2}+c^{2}}$$

$$\displaystyle \pm \sqrt{b^{2}+c^{2}-a^{2}}$$

Solution

Question 8

$$\displaystyle \pm \sqrt{2}$$

$$\displaystyle \pm \sqrt{3}$$

$$\displaystyle \pm 2$$

$$\displaystyle \pm 1$$

Solution

Question 9

$$0$$

$$\dfrac12$$

$$1$$

$$2$$

Solution

Question 10

$$x$$

$$2x$$

$$\dfrac 1x$$

$$\dfrac1{2x}$$

Solution

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