Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

sin$$^{2}\theta$$ + cos$$^{2}\theta$$ = _____

A
$$1$$

B
$$-1$$

C
$$0$$

D
$$-\sqrt{2}$$

Solution

From Pythagoras theorem, we write
$$a^{2}+b^{2}= c^{2}$$

Dividing through by $$c^{2}$$ gives,
$$\dfrac{a^{2}}{c^{2}}+\dfrac{b^{2}}{c^{2}} = \dfrac{c^{2}}{c^{2}}$$

This can be simplified to:
$$\left(\dfrac{a}{c}\right)^{2}+\left(\dfrac{b}{c}\right)^{2} = \left(\dfrac{c}{c}\right)^{2}$$

Now, $$\dfrac{a}{c}$$ is $$\dfrac{\text{opposite}}{\text{hypotenuse}}$$, which is $$\sin \theta$$

$$\dfrac{b}{c}$$ is $$\dfrac{\text{adjacent}}{\text{hypotenuse}}$$, which is $$\cos \theta$$

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

So, option $$A$$ is correct.
Question 2
1 / -0

The inverse of $$\sin\theta$$ is

A
$$\cos\theta$$

B
$$\sec\theta$$

C
$$\text{cosec}\,\theta$$

D
$$\tan\theta$$

Solution

inverse of $$\sin\theta\rightarrow \dfrac{1}{\sin\theta}=\text{cosec}\theta$$

Therefore, Answer is $$\text{cosec}\theta$$
Question 3
1 / -0

If $$\tan^{2} 45^{\circ} - \cos^{2} 30^{\circ} = x \sin 45^{\circ} \cos 45^{\circ}$$, then $$x = $$

A
$$2$$

B
$$-2$$

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

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

Solution

We know, $$\tan 45^{\circ} = 1, \cos 30 = \dfrac {\sqrt {3}}{2}, \sin 45^{\circ} = \dfrac {1}{\sqrt {2}}, \cos 45^{\circ} = \dfrac {1}{\sqrt {2}}$$
Now, $$ \tan^245^o-\cos^230^o= x \sin 45^0\cos 45^o$$
$$\Rightarrow 1 - \dfrac {3}{4} = x\times \dfrac {1}{\sqrt {2}} \times \dfrac {1}{\sqrt {2}} = >1 - \dfrac {3}{4} = x \times \dfrac {1}{2}$$
$$\Rightarrow \dfrac {1}{4} = \dfrac {x}{2}$$, $$\Rightarrow x = \dfrac {2}{4} = \dfrac {1}{2}$$
Question 4
1 / -0

If $$\tan \theta = \dfrac {a}{b}$$, then $$\dfrac {a\sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} = .....$$

A
$$\dfrac {a^{2} + b^{2}}{a^{2} - b^{2}}$$

B
$$\dfrac {a^{2} - b^{2}}{a^{2} + b^{2}}$$

C
$$\dfrac {a^{2}}{a^{2} + b^{2}}$$

D
$$\dfrac {b^{2}}{a^{2} + b^{2}}$$

Solution

Dividing numerator and denominator by $$\cos \theta$$ we get
$$\dfrac {(a \sin \theta - b \cos \theta)}{(a \sin \theta + b\cos \theta)} = \dfrac {a\tan \theta - b}{a\tan \theta + b} = \dfrac {a\times \dfrac {a}{b} - b}{a\times \dfrac {a}{b} + b} = \dfrac {a^{2} - b^{2}}{a^{2} + b^{2}}$$
Question 5
1 / -0

$$\dfrac {\cos (90 -\theta) \sec (90 - \theta)\tan \theta}{\text{cosec } (90 - \theta)\sin (90 - \theta) \cot (90 - \theta)} + \dfrac {\tan (90 - \theta)}{\cot \theta} = ......$$

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$-2$$

Solution

As we know that,
$$\sin(90^{o}-\theta)=\cos \theta$$, $$\cos(90^o-\theta)=\sin \theta$$,
$$\text{cosec}(90^o-\theta)=\sec \theta$$, $$\cot(90^o-\theta)=\tan\theta$$,
$$\sec(90^o - \theta)=\text{cosec} \theta$$, $$\tan(90^o - \theta) = \cot \theta$$

$$\therefore\ \dfrac{\cos(90-\theta) \sec(90-\theta) \tan \theta}{\text{cosec}(90-\theta) \sin(90-\theta) \cot(90-\theta)} + \dfrac{\tan(90-\theta)}{\cot \theta}$$

$$\therefore\ \dfrac{\sin \theta \times \frac{1}{\cos(90-\theta)} \times \tan \theta}{\frac{1}{\sin(90-\theta)} \times \sin(90-\theta) \times \tan \theta} + \dfrac{\cot \theta}{\cot \theta}$$

$$=\dfrac{\sin \theta \times \frac{1}{\sin \theta}}{\frac{1}{\cos \theta} \times \cos \theta} + 1$$

$$=1+1=2$$
Hence, option C is correct.
Question 6
1 / -0

$$\dfrac {1 + \tan^{2}A}{1 + \cot^{2}A} = $$ _____

A
$$\sec^{2}A$$

B
$$-1$$

C
$$\cot^{2}A$$

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

Solution

$$\boxed{\sin^{2}A + \cos^{2}A = 1}$$

$$\dfrac {1 + \tan^{2}A}{1 + \cot^{2}A} \\= \dfrac {1 + \dfrac {\sin^{2}A}{\cos^{2}A}}{1 + \dfrac {\cot^{2}A}{\sin^{2}A}} \\= \dfrac {\dfrac {\sin^{2}A + \cos^{2}A}{\cos^{2}A}}{\dfrac {\sin^{2}A + \cos^{2}A}{\sin^{2}A}}$$
$$=\dfrac {\dfrac {1}{\cos^{2}A}}{\dfrac {1}{\sin^{2}A}} \\= \dfrac {\sin^{2}A}{\cos^{2}A} \\= \tan^{2}A$$
So, $$\tan^{2}A$$ is correct.
Question 7
1 / -0

If $$A + B = 90^o$$, then ......

A
$$\sin A = \sin B$$

B
$$\cos A = \cos B$$

C
$$\tan A = \tan B$$

D
$$\sec A = \text{cosec } B$$

Solution

$$\sec A = \dfrac {r}{x}$$ and $$\text{cosec } A = \dfrac {r}{x}$$

So, $$\sec A = \text{ cosec } A$$
Question 8
1 / -0

The value of $$\cot 1^{\circ} \cot 2^{\circ} .... \cot 89^{\circ}$$ is .....

A
$$1$$

B
$$0$$

C
$$-1$$

D
None of above

Solution

The given equation can be written as
$$(\cot 1^{\circ} \cot 89^{\circ})(\cot 2^{\circ} \cot 88^{\circ}) ..... (\cot 44^{\circ} \cot 46^{\circ}) \cot 45^{\circ}$$

$$= (\cot 1^{\circ} \cot (90^{\circ} - 1^{\circ})) (\cot 2^{\circ} \cot (90^{\circ} - 2^{\circ})) ..... (\cot 44^{\circ} \cot (90^{\circ} - 44^{\circ}) ) \cot 45^{\circ}$$

$$= \cot 1^{\circ} \tan 1^{\circ} \cdot \cot 2^{\circ} \tan 2^{\circ} .... \cot 44^{\circ} \tan 44^{\circ} \times \cot 45^{\circ}$$ (Since $$\cot (90^{\circ}-x) = \tan x)$$)

$$= 1\times 1 ..... 1\times 1 $$ (Since $$ \cot x \times \tan x = 1, \cot 45^{\circ}$$)

$$=1$$

Hence, option $$A$$ is correct.
Question 9
1 / -0

If $$\cot\theta = \dfrac{6}{8}$$ and $$\theta$$ is acute, what is the value of $$\text{cosec}\,\theta$$?

A
$$\dfrac{6}{10}$$

B
$$\dfrac{8}{10}$$

C
$$\dfrac{10}{8}$$

D
$$\dfrac{5}{10}$$

Solution

First use the identity: $$1 + \cot^{2}\theta = \text{cosec}\,^{2}\theta$$
and substitute the value of $$\cot\theta = \dfrac{6}{8}$$

Therefore, $$1+ \left(\dfrac{6}{8}\right)^{2} = \text{cosec}\,^{2}\theta$$

$$1 + \sqrt{\dfrac{36}{64}} = \text{cosec}\,^{2}\theta$$

$$\text{cosec}\,\theta = \sqrt{\dfrac{100}{64}}$$

Squaring on both the sides, we get
$$\text{cosec}\,\theta = \dfrac{10}{8}$$
Question 10
1 / -0

If $$\tan\theta$$ = $$\dfrac{6}{8}$$ and $$\theta$$ is acute, what is the value of $$\cos\theta$$?

A
$$\dfrac{6}{10}$$

B
$$\dfrac{8}{10}$$

C
$$\dfrac{10}{10}$$

D
$$\dfrac{5}{10}$$

Solution

First use the identity: $$1 + \tan^{2}\theta = \sec^{2}\theta$$
and substitute the value of $$\tan\theta = \dfrac{6}{8}$$

Therefore, $$1+ \left(\dfrac{6}{8}\right)^{2}= \sec^{2}\theta$$

$$1 + \sqrt{\dfrac{36}{64}} = \sec^{2}\theta$$

$$\sec \theta = \sqrt{\dfrac{100}{64}}$$

Taking square roots on both the sides, we get
$$\sec\theta = \dfrac{10}{8}$$

Now we know the value of $$\sec \theta$$, to find $$\cos \theta$$ we just take the reciprocal.

Therefore, $$\cos \theta = \dfrac{8}{10}$$

So, option $$B$$ is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 58

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Introduction to Trigonometry Test - 58

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SHARING IS CARING

Question 1

$$1$$

$$-1$$

$$0$$

$$-\sqrt{2}$$

Solution

Question 2

$$\cos\theta$$

$$\sec\theta$$

$$\text{cosec}\,\theta$$

$$\tan\theta$$

Solution

Question 3

$$2$$

$$-2$$

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

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

Solution

Question 4

$$\dfrac {a^{2} + b^{2}}{a^{2} - b^{2}}$$

$$\dfrac {a^{2} - b^{2}}{a^{2} + b^{2}}$$

$$\dfrac {a^{2}}{a^{2} + b^{2}}$$

$$\dfrac {b^{2}}{a^{2} + b^{2}}$$

Solution

Question 5

$$1$$

$$-1$$

$$2$$

$$-2$$

Solution

Question 6

$$\sec^{2}A$$

$$-1$$

$$\cot^{2}A$$

$$\tan^{2}A$$

Solution

Question 7

$$\sin A = \sin B$$

$$\cos A = \cos B$$

$$\tan A = \tan B$$

$$\sec A = \text{cosec } B$$

Solution

Question 8

$$1$$

$$0$$

$$-1$$

None of above

Solution

Question 9

$$\dfrac{6}{10}$$

$$\dfrac{8}{10}$$

$$\dfrac{10}{8}$$

$$\dfrac{5}{10}$$

Solution

Question 10

$$\dfrac{6}{10}$$

$$\dfrac{8}{10}$$

$$\dfrac{10}{10}$$

$$\dfrac{5}{10}$$

Solution

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