Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 34

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Question 1
1 / -0

If $$x= \sec\phi - \tan\phi $$ and $$y= cosec \phi $$ then:

A
$$\displaystyle x=\frac{y+1}{y-1}$$

B
$$\displaystyle x=\frac{y-2}{x-2}$$

C
$$xy+x+y+1=0$$

D
$$yx^2+x^2-y+1=0$$

Solution

$$x=sec\phi-tan\phi$$ and $$sec^2\phi-tan^2\phi=1$$

$$\dfrac{1}x=\dfrac{1}{(sec\phi-tan\phi)}$$ now multiply and divide by $$sec\phi+tan\phi$$

$$\dfrac{1}x=\dfrac{sec\phi+tan\phi)}{(sec^2\phi-tan^2\phi)}=sec\phi+tan\phi$$

$$x+\dfrac{1}x=2sec\phi$$ and $$x-\dfrac{1}x=-2tan\phi$$

$$y=cosec\phi=sec\phi/tan\phi=\dfrac{(x+\dfrac{1}x)}{-(x-\dfrac{1}x)}=\dfrac{1+x^2}{1-x^2}$$

$$y-yx^2=1+x^2=>x^2+x^2y-y+1=0$$
Question 2
1 / -0

The value of $$\tan 5^{\circ}\tan 10^{\circ}\tan 15^{\circ}\cdots \tan 85^{\circ}$$ is

A
$$0$$

B
Not defined

C
$$1$$

D
$$-1$$

Solution

Using,
$$\tan x=\cot \left( 90^{\circ}-x \right) $$ and $$\tan x.\cot x=1$$

We get ,
$$\tan 5.\tan 10.\tan 15...\tan 80.\tan 85\\ =\tan 5.\tan 10.\tan 15...\tan 45. \cot(90-35)...\cot \left( 90-10 \right) .\cot \left( 90-5 \right) \\ =\tan 5.\cot 5.\tan 10.\cot 10...\tan 45\\ =1$$
Question 3
1 / -0

If $$\displaystyle\tan \alpha +\cot \alpha =p$$, then find $$\displaystyle\tan^{2} \alpha +\cot^{2} \alpha $$

A
$$p$$

B
$$p^{2}$$

C
$$p^{2}-2$$

D
$$2p^{2}$$

Solution

$$\tan { \alpha } +\cot { \alpha } =p\\ \Rightarrow { \left( \tan { \alpha +\cot { \alpha } } \right) }^{ 2 }={ p }^{ 2 }\\ \Rightarrow \tan ^{ 2 }{ \alpha } +\cot ^{ 2 }{ \alpha } +{ 2=p }^{ 2 }\\ \Rightarrow \tan ^{ 2 }{ \alpha } +\cot ^{ 2 }{ \alpha } ={ p }^{ 2 }-2$$
Question 4
1 / -0

$$\cos ^{2} 60^{\circ}+\sin ^{2} 30^{\circ} = \displaystyle \frac{1}{a}$$
Then $$a=$$

A
$$2$$

B
$$3$$

C
$$1$$

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

Solution

$$\cos^2 60^{\circ} + \sin^2 30^{\circ}$$
$$=\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2$$
$$=\cfrac{1}{4} + \cfrac{1}{4}$$
$$=\cfrac{2}{4}$$
$$=\cfrac{1}{2}$$
$$\therefore \dfrac1a = \dfrac12$$
$$\Rightarrow a = 2$$
Question 5
1 / -0

If $$\displaystyle \tan \theta =a-\frac{1}{4a}$$, then $$\sec \theta -\tan \theta $$ equals

A
$$2a$$

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

C
$$\displaystyle -\frac{1}{2a},2a$$

D
$$\displaystyle -2a,\frac{1}{2a}$$

Solution

$$\tan { \theta } =a-\dfrac { 1 }{ 4a } $$

$$Let \,k=\sec { \theta } -\tan { \theta }$$

$$\Rightarrow k=\sqrt { 1+\tan ^{ 2 }{ \theta } } -\tan { \theta } $$ {$$\because \sec^2\theta=1+\tan^2\theta$$}

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

$$\Rightarrow k=\sqrt { { \left( a+\dfrac { 1 }{ 4a } \right) }^{ 2 } } -a+\dfrac { 1 }{ 4a } =\pm \left( a+\dfrac { 1 }{ 4a } \right) -a+\dfrac { 1 }{ 4a }$$

$$\therefore \quad k=-2a,\dfrac { 1 }{ 2a } $$

Ans: D
Question 6
1 / -0

The expression $$\displaystyle f\left ( \theta \right )= \frac{1}{\tan \theta+\sec \theta+\cot \theta+\text{cosec} \theta}$$ is equivalent to

A
$$\displaystyle \frac{\text{cosec} \theta +\sec \theta+\text{cosec} \theta\sec \theta}{2\text{cosec} \theta\sec\theta}$$

B
$$\displaystyle \frac{\text{cosec} \theta -\sec \theta-\text{cosec} \theta\sec \theta}{2\text{cosec}\theta\sec\theta}$$

C
$$\displaystyle \frac{\text{cosec} \theta +\sec \theta-\text{cosec} \theta\sec \theta}{2\text{cosec}\theta\sec\theta}$$

D
None of these

Solution

$$\displaystyle f(\theta ) = \frac{1}{\tan \theta + \sec \theta + \cot \theta + \text{cosec } \theta }$$

$$\displaystyle= \frac{1}{\frac{\sin \theta }{\cos \theta} +\frac{ 1}{\cos \theta} + \frac{\cos \theta }{\sin \theta} + \frac{1}{\sin \theta} }$$

$$\displaystyle= \frac{\sin \theta . \cos \theta }{\sin^2 \theta + \cos \theta + \cos ^2\theta + \sin \theta } = \frac{\sin \theta . \cos \theta }{\cos \theta + \sin \theta + 1} $$

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

$$\displaystyle= \frac{\cos \theta + \sin \theta - 1}{2} = \frac{\text{cosec } \theta + \sec \theta - \text{cosec } \theta .\sec \theta }{2\text{cosec } \theta .\sec \theta } $$ .....(.dividing by $$\cos \theta\sin\theta$$)
Question 7
1 / -0

Choose the correct option for the following statement.
$$\sin 60^{0} \: \cos 30^{0}+\cos 60^{0} \: \sin 30^{0}= 1$$

A
The given statement is true.

B
The given statement is false.

C
Incomplete information

D
None of these.

Solution

To Prove:$$\sin 60^{0} \: \cos 30^{0}+\cos 60^{0} \: \sin 30^{0}= 1$$

L.H.S.: $$\sin 60^{0} \: \cos 30^{0}+\cos 60^{0} \: \sin 30^{0}$$

= $$\dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2}$$

= $$\dfrac{3}{4} + \dfrac{1}{4}$$

= $$\dfrac{4}{4}$$

= $$1$$

thus, $$L.H.S. = R.H.S.$$
The given statement is true.
Question 8
1 / -0

$$ABC$$ is an isosceles right-angled triangle. Assuming $$AB= BC= x$$, find the value of each of the following trigonometric ratios:
$$\sin 45^{0}$$ is $$\displaystyle \dfrac{1}{\sqrt{m}}$$, the value of $$m$$ is

A
1

B
3

C
2

D
4

Solution

Given: $$\triangle ABC$$, $$AB = BC = x$$, $$\angle ABC = 90^{\circ}$$

By Pythagoras Theorem,
$$AC^2 = AB^2 + BC^2$$
$$AC^2 = x^2 + x^2$$
$$AC^2 = 2x^2$$
$$AC = \sqrt{2} x$$

Since, $$AB = BC$$, then by Isosceles triangle property,
$$\angle ACB = \angle CAB = 45^{\circ}$$

Now, $$Sin 45^{\circ} = Sin C = \dfrac{P}{H}$$

$$ = \dfrac{AB}{AC}$$

$$ = \dfrac{x}{x\sqrt{2}} = \dfrac{1}{\sqrt{2}}$$
so, the value of $$m$$ is $$=2$$
Question 9
1 / -0

Choose the correct option for the following.
The value of $$\displaystyle \frac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$$ is $$\displaystyle \frac{1-\sqrt{2}}{1+\sqrt{6}}$$, when $$A= 15^{0}$$.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

Given, $$A = 15^{\circ}$$

Thus, $$\displaystyle \dfrac{\cos 3A-2\cos 4A}{\sin 3A+2\sin 4A}$$

Put $$A = 15^{\circ}$$
= $$\displaystyle \dfrac{\cos 45^0 -2\cos 60^0}{\sin 45^0 + 2\sin 60^0}$$

= $$\displaystyle \dfrac{\dfrac{1}{\sqrt{2}} -2 (\dfrac{1}{2})}{\dfrac{1}{\sqrt{2}} + 2(\dfrac{\sqrt{3}}{2})}$$

= $$\displaystyle \dfrac{\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}} + \sqrt{3}}$$

= $$\displaystyle \dfrac{\dfrac{1 - \sqrt{2}}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}} + \sqrt{3}}$$

= $$\displaystyle \dfrac{1 - \sqrt{2}}{1 + \sqrt{6}}$$
Question 10
1 / -0

If $$\tan \angle ADC= 1$$ and $$\sin \angle ABC=4:5$$, then measure of $$CD$$ is:

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

In $$\triangle ABC$$, $$BC = 15$$, $$\sin B = \dfrac {4}{5}$$

Now, $$\sin B = \dfrac {P}{H}$$
$$\dfrac {4}{5} = \dfrac {AC}{AB}$$
Then, let $$AC = 4x$$ and $$AB = 5x$$

Now, $$AB^2 = AC^2 + BC^2$$
$$(5x)^2 = (4x)^2 + 15^2$$
$$25x^2 = 16x^2 + 225$$
$$9x^2 = 225$$
$$x = 5$$

Hence, $$AC= 20$$ and $$AB = 25$$

$$\tan \angle ADC = 1$$
$$\dfrac {AC}{CD} = 1$$
$$AC = CD = 20$$

Now, using Pythagoras theorem,
$$AD^2 = AC^2 + CD^2$$
$$AD^2 = 20^2 + 20^2$$
$$AD = 20 \sqrt{2}$$

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Introduction to Trigonometry Test - 34

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Introduction to Trigonometry Test - 34

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Question 1

$$\displaystyle x=\frac{y+1}{y-1}$$

$$\displaystyle x=\frac{y-2}{x-2}$$

$$xy+x+y+1=0$$

$$yx^2+x^2-y+1=0$$

Solution

Question 2

$$0$$

Not defined

$$1$$

$$-1$$

Solution

Question 3

$$p$$

$$p^{2}$$

$$p^{2}-2$$

$$2p^{2}$$

Solution

Question 4

$$2$$

$$3$$

$$1$$

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

Solution

Question 5

$$2a$$

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

$$\displaystyle -\frac{1}{2a},2a$$

$$\displaystyle -2a,\frac{1}{2a}$$

Solution

Question 6

$$\displaystyle \frac{\text{cosec} \theta +\sec \theta+\text{cosec} \theta\sec \theta}{2\text{cosec} \theta\sec\theta}$$

$$\displaystyle \frac{\text{cosec} \theta -\sec \theta-\text{cosec} \theta\sec \theta}{2\text{cosec}\theta\sec\theta}$$

$$\displaystyle \frac{\text{cosec} \theta +\sec \theta-\text{cosec} \theta\sec \theta}{2\text{cosec}\theta\sec\theta}$$

None of these

Solution

Question 7

The given statement is true.

The given statement is false.

Incomplete information

None of these.

Solution

Question 8

1

3

2

4

Solution

Question 9

True

False

Ambiguous

Data insufficient

Solution

Question 10

$$10$$

$$20$$

$$30$$

$$40$$

Solution

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