Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

$${cos}^{2}0+{cos}^{2}10+{cos}^{2}20+{cos}^{2}30+{cos}^{2}40+{cos}^{2}50+{cos}^{2}60+{cos}^{2}70+{cos}^{2}80+{cos}^{2}90=$$

A
$$0$$

B
$$1$$

C
$$5$$

D
$$10$$

Solution
Question 2
1 / -0

The value of $$\cos{\cfrac{\pi}{10}}\cos{\cfrac{2\pi}{10}}\cos{\cfrac{4\pi}{10}}\cos{\cfrac{8\pi}{10}}\cos{\cfrac{16\pi}{10}}$$ is:

A
$$\cfrac{\sqrt{10+2\sqrt{5}}}{64}$$

B
$$-\cfrac{\cos{(\pi/10)}}{16}$$

C
$$\cfrac{\cos{(\pi/10)}}{16}$$

D
$$\cfrac{\sqrt{10+2\sqrt{5}}}{16}$$

Solution

$$\cos\left({\cfrac{\pi}{10}}\right)\cos\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{4\pi}{10}}\right)\cos\left({\cfrac{8\pi}{10}}\right)\cos\left({\cfrac{16\pi}{10}}\right)$$

Multiplying numerators and denominator by $$2\sin\left(\pi/10\right)$$

$$\dfrac{1}{2\sin(\pi/10)}\times 2\sin\left({\cfrac{\pi}{10}}\right)\cos\left({\cfrac{\pi}{10}}\right)\cos\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{4\pi}{10}}\right)\cos\left({\cfrac{8\pi}{10}}\right)\cos\left(\dfrac{16\pi}{10}\right)$$

$$\dfrac{1}{2\sin(\pi/10)}\sin\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{4\pi}{10}}\right)\cos\left({\cfrac{8\pi}{10}}\right)\cos\left({\cfrac{16\pi}{10}}\right)$$

Multiplying numerators and denominator by $$2$$, we get

$$=\dfrac{1}{4\sin(\pi/10)}\times 2\sin\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{4\pi}{10}}\right)\cos\left({\cfrac{8\pi}{10}}\right)\cos\left({\cfrac{16\pi}{10}}\right)$$

$$=\dfrac{1}{4\sin(\pi/10)}\times \sin\left({\cfrac{4\pi}{10}}\right)\cos\left({\cfrac{4\pi}{10}}\right)\cos\left({\cfrac{8\pi}{10}}\right)\cos\left({\cfrac{16\pi}{10}}\right)$$

$$[As\, ,2\sin\left({\cfrac{2\pi}{10}}\right)\cos\left({\cfrac{2\pi}{10}}\right)=\sin\left({\cfrac{4\pi}{10}}\right)]$$

$$=\dfrac{1}{32\sin(\pi/10)} \sin\left( \dfrac{32\pi}{10}\right)$$

$$=\dfrac{\sin(3\pi+\dfrac{2\pi}{10})}{32\sin(\pi/10)}$$

$$=\dfrac{-\sin(2\pi/10)}{32\sin(\pi/10)}$$ $$[As,\sin(3\pi+\theta)=-\sin\theta]$$

$$\dfrac{-2\sin(\pi/10)\cos(\pi/10)}{32\sin(\pi/10)}$$ [As, $$\sin(2\pi/10)=2\sin(\pi/10)\cos(\pi/10)$$]

$$=-\dfrac{-1}{16}\cos(\pi/10)$$
Question 3
1 / -0

Evaluate: $$\dfrac{{\sin}^{4}{\theta}-{\cos}^{4}{\theta}}{{\sin}^{2}{\theta}-{\cos}^{2}{\theta}}$$

A
$$3$$

B
$$2$$

C
$$0$$

D
$$1$$

Solution

$$\dfrac{{\sin}^{4}{\theta}-{\cos}^{4}{\theta}}{{\sin}^{2}{\theta}-{\cos}^{2}{\theta}}$$

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

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

If $$\sin \theta = \cos \theta$$, then $$\theta = ?$$

A
$$45^{o}$$

B
$$90^{o}$$

C
$$0^{o}$$

D
$$30^{o}$$

Solution

$$\sin\theta =\cos \theta$$

$$\Rightarrow \dfrac {\sin\theta}{\cos\theta}=1$$

$$\Rightarrow \tan\theta=1$$

$$\Rightarrow \theta=\tan^{-1}1$$

$$\therefore \theta= 45^{\circ}$$
Question 5
1 / -0

$$\sin { 2A } =2\sin { A } $$ is true then A = _____

A
0

B
30

C
45

D
60

Solution

$$\sin{2A}=2\sin{A}$$
$$\Rightarrow 2\sin{A}\cos{A}=2\sin{A}$$
$$\Rightarrow 2\sin{A}\left(\cos{A}-1\right)=0$$
$$\Rightarrow \sin{A}=0,\cos{A}-1=0$$
$$\Rightarrow \sin{A}=0,\cos{A}=1$$
$$\sin{A}=0\Rightarrow A=0,\pi,2\pi$$
$$\cos{A}=1\Rightarrow A=0,2\pi$$
$$\therefore A=\left\{0,\pi,2\pi\right\}\cap\left\{0,2\pi\right\}$$
$$=\left\{0,2\pi\right\}$$
Hence $$A=0$$
Question 6
1 / -0

$$\cot{1^0}\cot{2^0}\cot{3^0}\cot{4^0}...\cot{89^0}$$.

A
1

B
0

C
-1

D
NONE OF THE GIVEN

Solution

$$\cot{1^0}\cot{2^0}\cot{3^0}\cot{4^0}...\cot{89^0}$$

$$=\cot{\left(90^0-89^0\right)}\cot{\left(90^0-88^0\right)}\cot{\left(90^0-87^0\right)}\cot{\left(90^0-86^0\right)}...\cot{45^0}.....\cot{86^0}\cot{87^0}\cot{88^0}\cot{89^0}$$

There are $$\dfrac{89}{2}=44$$ terms $$+1$$ term
That $$1$$ term is the middle term$$=\cot{45^0}=1$$
and other terms will form the reciprocals

$$=\tan{89^0}\tan{88^0}\tan{87^0}\tan{86^0}...\cot{45^0}.....\cot{86^0}\cot{87^0}\cot{88^0}\cot{89^0}$$

$$=\left(\tan{89^0}\cot{89^0}\right)\times \left(\tan{88^0}\cot{88^0}\right)\times \left(\tan{87^0}\cot{87^0}\right)\times \left(\tan{86^0}\cot{86^0}\right)\times ...\tan{45^0}$$ by re-arranging

$$=1\times 1\times 1\times 1...1=1$$
Question 7
1 / -0

Choose the correct alternative:
$$\ cosec \theta = \dfrac { 1 } { \dots \ldots }$$

A
$$\cos \theta$$

B
$$\sin \theta$$

C
$$\tan \theta$$

D
$$\cot \theta$$

Solution

By using trigonometric formula,
$$cosec{\theta}$$ $$=1/sin {\theta}$$
Question 8
1 / -0

If $$\sin A+\cos A= {\sqrt2}$$ and $$\tan A+\cot A=2$$, then the value of $$\sec A\cdot \csc A$$ is equal to

A
$$2$$

B
$$3$$

C
$$4$$

D
$$\dfrac{1}{2}$$
Question 9
1 / -0

If $$\displaystyle \tan \theta + \cot \theta = 2$$, then $$\displaystyle \sin \theta = $$?

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

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

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

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

Solution

$$\displaystyle \tan \theta + \cot \theta = 2 \Rightarrow \tan \theta + \frac{1}{\tan \theta} = 2$$
$$\displaystyle \Rightarrow \tan^2 \theta + 1 = 2 \tan \theta \Rightarrow 1 + \tan^2 \theta - 2\tan \theta = 0$$
$$\displaystyle \Rightarrow ( 1 - \tan \theta )^2 = 0 \Rightarrow 1 - \tan \theta = 0 \Rightarrow \tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4}$$.
$$\displaystyle \therefore \ \ \ \sin \theta = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$$
Question 10
1 / -0

Consider the following statements :
1. $$\cos\theta+\sec\theta$$ can never be equal to $$1.5$$
2. $$\tan\theta+\cot\theta$$ can never be less than to $$2$$
Which of the above statements is/are correct ?

A
$$1$$ only

B
$$2$$ only

C
Both $$1$$ and $$2$$

D
Neither $$1$$ nor $$2$$

Solution

$$\cos\theta=\dfrac{1}{\sec\theta}$$

$$\tan\theta=\dfrac{1}{cot\theta}$$

$$\Rightarrow A.M\ge G.M$$
$$\Rightarrow x+\dfrac{1}{x}\ge 2, x>0$$
From above inequality, we can write $$ x+\dfrac{1}{x}\le -2, x<0$$

So $$\tan\theta+\dfrac{1}{\tan\theta}\ge 2$$ (when $$\tan \theta > 0$$)
and $$\cos\theta+\dfrac{1}{\cos\theta}\ge 2$$ (when $$\cos \theta > 0$$)

Similarly $$\tan\theta+\dfrac{1}{\tan\theta}\le -2$$ (when $$\tan \theta < 0$$)
and $$\cos\theta+\dfrac{1}{\cos\theta}\le -2$$ (when $$\cos \theta < 0$$)

Hence $$\cos \theta +\sec \theta $$ can never be $$1.5$$.

But $$\tan \theta + \cot \theta $$ will always be less than $$2$$ when they are negative.

Hence only (1) is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 65

Result

Introduction to Trigonometry Test - 65

Score

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$5$$

$$10$$

Solution

Question 2

$$\cfrac{\sqrt{10+2\sqrt{5}}}{64}$$

$$-\cfrac{\cos{(\pi/10)}}{16}$$

$$\cfrac{\cos{(\pi/10)}}{16}$$

$$\cfrac{\sqrt{10+2\sqrt{5}}}{16}$$

Solution

Question 3

$$3$$

$$2$$

$$0$$

$$1$$

Solution

Question 4

$$45^{o}$$

$$90^{o}$$

$$0^{o}$$

$$30^{o}$$

Solution

Question 5

0

30

45

60

Solution

Question 6

1

0

-1

NONE OF THE GIVEN

Solution

Question 7

$$\cos \theta$$

$$\sin \theta$$

$$\tan \theta$$

$$\cot \theta$$

Solution

Question 8

$$2$$

$$3$$

$$4$$

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

Question 9

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

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

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

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

Solution

Question 10

$$1$$ only

$$2$$ only

Both $$1$$ and $$2$$

Neither $$1$$ nor $$2$$

Solution

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