Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \sin \left ( A+B \right ) =\frac{\sqrt{3}}{2}$$ and $$\displaystyle \cot \left ( A-B \right )=1$$, then find $$A$$

A
$$\displaystyle 27\frac{1^{\circ}}{2}$$

B
$$\displaystyle 35\frac{1^{\circ}}{2}$$

C
$$\displaystyle 52\frac{1^{\circ}}{2}$$

D
$$\displaystyle 55\frac{1^{\circ}}{2}$$

Solution

Given, $$ \sin(A+B) = \frac {\sqrt{3}}{2} $$
$$ \Rightarrow A +B = 60^o $$ -(1)
And $$ \cot(A-B) = 1 $$
$$ \Rightarrow A-B = 45^o $$---- (2)
On ddding equations (1) and (2), we get
$$ \Rightarrow 2A = 105^0 $$
$$ \Rightarrow A = 52\dfrac{1}{2}^0 $$
Question 2
1 / -0

$$\displaystyle \sec ^{4}\theta -\sec ^{2}\theta $$ in terms of $$\displaystyle \tan \theta $$ is ___

A
$$\displaystyle \tan ^{4}\theta -\tan ^{2}\theta $$

B
$$\displaystyle \tan ^{4}\theta +\tan ^{2}\theta $$

C
$$\displaystyle \tan ^{2}\theta -\tan ^{4}\theta $$

D
None of these

Solution

$$\displaystyle \sec ^{4}\theta -\sec ^{2}\theta=(\sec^2\theta)^2-\sec^2\theta=(1+\tan ^2\theta)^2-(1+\tan ^2\theta)=1+\tan ^4\theta+2\tan ^2\theta-1-\tan ^2\theta\\=\tan ^4\theta+\tan ^2\theta$$

Therefore, Answer is $$\tan ^4\theta+\tan ^2\theta$$
Question 3
1 / -0

The value of $$\displaystyle \left ( \sin A-\cos A \right )^{2}+\left ( \sin A+\cos A \right )^{2}$$ is _____

A
$$1$$

B
$$3$$

C
$$2$$

D
$$4$$

Solution

$$ (\sin A - \cos A)^{2} + (\sin A + \cos A)^{2} = \sin ^{2} A + \cos ^{2} A -2\sin A \cos A + \sin ^{2} A + \cos ^{2} A + 2\sin A \cos A $$
$$ = \sin ^{2} A + \cos ^{2} A + \sin ^{2} A + \cos ^{2} A = 1 + 1 = 2$$
(since, $$ \sin ^{2} A + \cos ^{2} A = 1 $$)
Question 4
1 / -0

If $$\displaystyle \sin \left ( A+B \right )=\dfrac{\sqrt{3}}{2}$$ and $$\displaystyle \cot \left ( A-B \right )=\sqrt{3}$$ then find the value of $$B$$.

A
$$\displaystyle 15^{\circ}$$

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

C
$$\displaystyle 10^{\circ}$$

D
$$\displaystyle 20^{\circ}$$

Solution

Given, $$ \sin(A+B) = \dfrac {\sqrt{3}}{2} $$
$$ \Rightarrow A +B = 60^o $$ -(1)

And $$ \cot(A-B) = \sqrt {3} $$
$$ \Rightarrow A-B = 30^o $$---- (2)

Subtracting eqn 2 from 1
$$ \Rightarrow 2B= 30^0 $$
$$ \Rightarrow B= 15^0 $$
Question 5
1 / -0

The value of $$\displaystyle \sin 30^{\circ}\cdot \sin 45^{\circ} \cdot\text{cosec }45^{\circ}\cdot\cos 30^{\circ} $$ is ___

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

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

C
$$\displaystyle \frac{\sqrt{3}}{2} $$

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

Solution

The value of $$ \sin { { 30 }^{ 0 } } .\sin { { 45 }^{ 0 }. } \text {cosec} { 45 }^{ 0 }.\cos { { 30 }^{ 0 } } $$is
$$ = \dfrac {1}{2} \times \dfrac {1}{\sqrt {2}} \times \sqrt {2} \times \dfrac {\sqrt {3}}{2} $$
$$= \dfrac {\sqrt {3}}{4} $$
Question 6
1 / -0

The simplified value of
$$\displaystyle \text{cosec} ^{2}\alpha \left ( 1+\frac{1}{\sec \alpha } \right )\left (1-\frac{1}{\sec \alpha } \right )$$ is _____

A
$$1$$

B
$$0$$

C
$$2$$

D
$$-1$$

Solution

We know that $$ \text{cosec}\, \alpha = \dfrac {1}{\sin \alpha} $$ and $$ \sec \alpha = \dfrac {1}{\cos \alpha} $$

So, $$ \text{cosec}\, ^2 \alpha \left(1 + \dfrac {1}{\sec \alpha}\right)\left(1 - \dfrac {1}{\sec \alpha}\right) = \dfrac {1}{\sin ^2 \alpha}\left(1+\cos \alpha\right)\left(1-\cos \alpha\right) $$

$$ = \dfrac {1}{\sin ^2 \alpha}\left(1-\cos ^2\alpha\right) = \dfrac {1}{\sin ^2 \alpha}\left(\sin ^2\alpha\right) = 1 $$
Question 7
1 / -0

If $$\displaystyle\tan \theta +\cot \theta =2 $$ find the value of $$\displaystyle \tan ^{1025}\theta +\cot ^{1025}\theta $$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$\dfrac12$$

Solution

Given $$\tan\theta+\cot\theta=2$$
$$\Rightarrow \tan\theta+\dfrac{1}{\tan\theta}=2$$
$$\Rightarrow \tan^2\theta+1=2\tan\theta$$
$$\Rightarrow \tan^2\theta-2\tan\theta+1=0$$
$$\Rightarrow (\tan\theta-1)^2=0$$
$$\Rightarrow \tan\theta=1\therefore \cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{1}=1$$

$$\therefore \tan^{1025}\theta+\cot^{1025}\theta=1^{1025}+1^{1025}=1+1=2$$
Question 8
1 / -0

If $$\displaystyle \tan \theta -\cot \theta =7 $$ find the value of $$\displaystyle \tan ^{3}\theta -\cot^{3} \theta $$

A
$$284$$

B
$$296$$

C
$$345$$

D
$$364$$

Solution

$$\displaystyle \tan ^{3}\theta -\cot^{3} \theta=( \tan\theta- \cot\theta)(\tan^2\theta+\cot^2\theta+\tan\theta \cot\theta)\\=( \tan\theta- \cot\theta)((\tan\theta-\cot \theta)^2+2\tan\theta cot \theta+\tan \theta \cot \theta)\\=7(7^2+3\tan\theta \cot \theta)=7(49+3)\\=7\times52=364$$

Therefore, Answer is $$364$$
Question 9
1 / -0

If $$\displaystyle \text{cosec}\, \theta -\cot \theta =2,$$ then find the value of $$\displaystyle \text{cosec} ^{2}\theta +\cot^{2} \theta.$$

A
$$\displaystyle \frac{8}{15}$$

B
$$\displaystyle \frac{15}{8}$$

C
$$\displaystyle \frac{8}{17}$$

D
$$\displaystyle \frac{17}{8}$$

Solution

$$cosec\;\theta-\cot\theta=2 \longrightarrow \left( 1 \right) $$
Now, $$\left( cosec^{ 2 }\theta -\cot ^{ 2 }{ \theta } \right) =1$$
$$\Rightarrow \left( cosec\;\theta -\cot { \theta } \right) \left( cosec\;\theta +\cot { \theta } \right) =1$$
$$\Rightarrow 2\left( cosec\;\theta +\cot { \theta } \right) =1$$
$$\Rightarrow \left( cosec\;\theta +\cot { \theta } \right) =\dfrac { 1 }{ 2 } \longrightarrow \left( 2 \right) $$
Adding $$\left( 1 \right) \;\&\;\left( 2 \right) ,$$ we get
$$\Rightarrow 2\;cosec\;\theta =2+\dfrac { 1 }{ 2 } =\dfrac { 5 }{ 2 } $$
$$\Rightarrow cosec\;\theta =\dfrac { 5 }{ 4 } $$
$$\Rightarrow cosec^{ 2 }\theta =\dfrac { 25 }{ 16 } $$
Again $$\left( 2 \right) \;-\;\left( 1 \right) ,$$ we get
$$\Rightarrow 2\cot { \theta } =\dfrac { 1 }{ 2 } -2=-\dfrac { 3 }{ 2 } $$
$$\Rightarrow \cot { \theta } =-\dfrac { 3 }{ 4 } $$
$$\Rightarrow \cot ^{ 2 }{ \theta } =\dfrac { 9 }{ 16 } $$
$$\Rightarrow cosec^{ 2 }\theta +\cot ^{ 2 }{ \theta } =\dfrac { 34 }{ 16 } =\dfrac { 17 }{ 8 } $$
Question 10
1 / -0

The simplified value of $$\displaystyle \left ( \frac{1-\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{1+\sin \alpha } \right )\left ( \sec \alpha +\frac{1}{\cot \alpha } \right )$$ is _____

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$ \left( \dfrac { 1-\sin \alpha }{ \cos \alpha } +\dfrac { \cos \alpha }{ 1+\sin \alpha } \right) \left( \sec \alpha \quad +\quad \dfrac { 1 }{ \cot \alpha } \right) \\= \left( \dfrac { (1-\sin \alpha )(1+\sin \alpha )+\cos ^ 2 \alpha \quad }{ (1+\sin \alpha )\cos \alpha } \right) \left( \dfrac { 1 }{ \cos \alpha } +\quad \dfrac { \sin \alpha }{ \cos \alpha } \right) $$
$$\\ = \left( \dfrac { (1-\sin ^{ 2 }\alpha +\cos ^{ 2 }\alpha \quad }{ (1+\sin \alpha )\cos \alpha } \right) \left( \dfrac { 1+\sin \alpha }{ \cos \alpha } \right) \\= \left( \dfrac { \cos ^{ 2 }\alpha +\cos ^{ 2 }\alpha \quad }{ \cos \alpha } \right) \left( \dfrac { 1 }{ \cos \alpha } \right) \\= 2 $$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 49

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Introduction to Trigonometry Test - 49

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SHARING IS CARING

Question 1

$$\displaystyle 27\frac{1^{\circ}}{2}$$

$$\displaystyle 35\frac{1^{\circ}}{2}$$

$$\displaystyle 52\frac{1^{\circ}}{2}$$

$$\displaystyle 55\frac{1^{\circ}}{2}$$

Solution

Question 2

$$\displaystyle \tan ^{4}\theta -\tan ^{2}\theta $$

$$\displaystyle \tan ^{4}\theta +\tan ^{2}\theta $$

$$\displaystyle \tan ^{2}\theta -\tan ^{4}\theta $$

None of these

Solution

Question 3

$$1$$

$$3$$

$$2$$

$$4$$

Solution

Question 4

$$\displaystyle 15^{\circ}$$

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

$$\displaystyle 10^{\circ}$$

$$\displaystyle 20^{\circ}$$

Solution

Question 5

$$\displaystyle \sqrt{3} $$

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

$$\displaystyle \frac{\sqrt{3}}{2} $$

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

Solution

Question 6

$$1$$

$$0$$

$$2$$

$$-1$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

$$\dfrac12$$

Solution

Question 8

$$284$$

$$296$$

$$345$$

$$364$$

Solution

Question 9

$$\displaystyle \frac{8}{15}$$

$$\displaystyle \frac{15}{8}$$

$$\displaystyle \frac{8}{17}$$

$$\displaystyle \frac{17}{8}$$

Solution

Question 10

$$0$$

$$1$$

$$2$$

$$3$$

Solution

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