Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 27

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Question 1
1 / -0

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

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

B
$$\displaystyle \frac{1}{2p}$$ or $$-2p$$

C
$${-\displaystyle \frac{1}{2p}}$$ or $$2p$$

D
$${-\displaystyle \frac{1}{2p}}$$ or $$-2p$$

Solution

Given that:
$$\tan \theta = p - \displaystyle \frac{1}{4p}$$
Let, $$\sec \theta - \tan \theta = x\dots (1)$$
$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta-\tan \theta}$$
then, $$\sec \theta + \tan \theta = \displaystyle\frac{1}{x}\dots (2)$$
Subtracting $$(1)$$ from $$(2)$$, we get:
$$\displaystyle 2\tan \theta = \frac{1}{x} - x = 2p - \frac{1}{2p}$$

We solve to get $$x=\displaystyle \ -2p$$ or $$\dfrac{1}{2p}$$
$$\implies \sec \theta -\tan \theta =-2p$$ or $$\dfrac{1}{2p}$$
Question 2
1 / -0

If $$a=x\cos^{2}A+y\sin^{2}A$$, then $$(x-a)(y-a)+(x-y)^{2}\sin^{2}A\cos^{2}A$$ is equal to:

A
$$0$$

B
$$1$$

C
$$y+a^{2}$$

D
$$x-a^{2}$$

Solution

Given, $$a=x\cos ^{ 2 }{ A } +y\sin ^{ 2 }{ A } $$
$$\Rightarrow a=x\cos ^{ 2 }{ A } +y(1-\cos ^{ 2 }{ A } )\\ \Rightarrow a-y=(x-y)\cos ^{ 2 }{ A } \\ \Rightarrow \cos ^{ 2 }{ A } =\dfrac { a-y }{ x-y } $$ ......(i)
We know that $$ \cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } =1$$
$$\Rightarrow \sin ^{ 2 }{ A } =1-\cos ^{ 2 }{ A } \\ \Rightarrow \sin ^{ 2 }{ A } =1-\dfrac { a-y }{ x-y } \\ \Rightarrow \sin ^{ 2 }{ A } =\dfrac { x-a }{ x-y } $$ ......(ii)

Given equation is $$(x-a)(y-a)+{ (x-y) }^{ 2 }\sin ^{ 2 }{ A } \cos ^{ 2 }{ A } $$

Substituting (i) and (ii) in the given equation, we get

$$(x-a)(y-a)+{ (x-y) }^{ 2 }\left( \dfrac { x-a }{ x-y } \right) \left( \dfrac { a-y }{ x-y } \right) \\ =(x-a)(y-a)+(x-a)(a-y)\\ =(x-a)(y-a)-(x-a)(y-a)\\ =0$$

Hence, option A is correct.
Question 3
1 / -0

$$\tan A=a\tan B, \displaystyle \sin A=b\sin B\Rightarrow\frac{b^{2}-1}{a^2-1}$$ is equal to

A
$$\sin^{2}A$$

B
$$\sin^{3}A$$

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

D
$$\cos^{3}A$$

Solution

$$\tan A=a\tan B$$ and $$\sin A=b\sin B$$
$$\dfrac{\sin A}{\cos A}=a\dfrac{\sin B}{\cos B}$$

$$\dfrac{b\sin B}{\cos A}=\dfrac{a\sin B}{\cos B}$$...... Given, $$b \sin B= \sin A$$

$$\dfrac{b}{\cos A}=\dfrac{a}{\cos B} \Rightarrow \cos B=\dfrac{a}{b}\cos A$$

$$\sin ^2A=b^2\sin ^2B\Rightarrow 1-\cos ^2A=b^2(1-\cos ^2B)$$

$$\Rightarrow 1-\cos ^2A=b^2\left (1-\dfrac{a^2}{b^2}\cos ^2A\right )$$

$$\Rightarrow1-\cos ^2A=b^2-a^2\cos ^2A$$

$$\Rightarrow \cos ^2A(a^2-1)=b^2-1$$

$$\Rightarrow\dfrac{(b^2-1)}{(a^2-1)}=\cos ^2A$$
Question 4
1 / -0

If $$\tan \theta +\cot \theta =3$$, then $$\tan^{4}\theta+\cot^{4}\theta=$$

A
47

B
162

C
24

D
48

Solution

Given $$\tan \theta +\cot \theta =3$$

$$\Rightarrow (\tan \theta +\cot \theta )^{2} = \tan ^{2}\theta +\cot^{2}\theta +2=9$$

$$\Rightarrow \tan ^{2}\theta +\cot ^{2}\theta =7$$

$$\Rightarrow \tan ^{4}\theta +\cot ^{4} \theta +2=49$$

$$\Rightarrow \tan ^{4}\theta +\cot ^{4}\theta =47$$
Question 5
1 / -0

If $$\sqrt{\sin x}+\cos x=0$$, then $$\sin x=$$

A
$$\displaystyle \frac{\sqrt{5}+1}{2}$$

B
$$\displaystyle \frac{\sqrt{5}+1}{8}$$

C
$$\displaystyle \frac{\sqrt{5}-1}{8}$$

D
$$\displaystyle \frac{\sqrt{5}-1}{2}$$

Solution

$$\sqrt\sin x+\cos x=0$$
$$ \sqrt{\sin x} = -\cos x$$
Squaring both sides:
$$ \sin x = \cos^{2}x$$
$$ = 1 - \sin^{2}x$$
$$\sin^2 x+\sin x-1=0$$

Using quadratic formula to find the roots of a quadratic equation, we get:

$$\sin x=\dfrac {-1\pm\sqrt{1+4}}{2}$$

$$ \sin x = \dfrac{\sqrt{5} - 1}{2}$$ (The other value is unacceptable as it is less than -1.)
Question 6
1 / -0

If $$x=\tan\theta+\cot\theta, y=\cos\theta-\sin\theta$$, then which of the following is true?

A
$$x=y$$

B
$$\displaystyle \frac{1-y^{2}}{2}=\frac{1}{x}$$

C
$$\displaystyle \frac{y^{2}-1}{2}=\frac{1}{x}$$

D
$$\displaystyle \frac{1+y^{2}}{2}=\frac{1}{x}$$

Solution

Given, $$x=\tan { \theta } +\cot { \theta } $$
$$\Rightarrow x=\dfrac { \sin { \theta } }{ \cos { \theta } } +\dfrac { \cos { \theta } }{ \sin { \theta } } $$
$$\Rightarrow x=\dfrac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \sin { \theta \cos { \theta } } } $$
$$ \Rightarrow x=\dfrac { 1 }{ \sin { \theta \cos { \theta } } } $$
$$ \Rightarrow \sin { \theta \cos { \theta } } =\dfrac { 1 }{ x }$$ ......(i)

Also given, $$y=\cos { \theta } -\sin { \theta } $$

On squaring both sides, we have

$$y^2=(\cos { \theta } -\sin { \theta })^2 \\ \Rightarrow { y }^{ 2 }=\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } -2\sin { \theta \cos { \theta } } \\ \Rightarrow { y }^{ 2 }=1-2\sin { \theta \cos { \theta } } $$

On substituting value of (i), we get

$${ y }^{ 2 }=1-\dfrac { 2 }{ x } \\ \Rightarrow \dfrac { 2 }{ x } =1-{ y }^{ 2 }\\ \Rightarrow \dfrac { 1 }{ x } =\dfrac { 1-{ y }^{ 2 } }{ 2 } $$

So, option B is correct.
Question 7
1 / -0

$$\cos A=a\cos B,\sin A=b\sin B\Rightarrow (b^{2}-a^{2})\sin^{2}B=$$

A
$$1+a^{2}$$

B
$$2+a^{2}$$

C
$$1-a^{2}$$

D
$$2-a^{2}$$

Solution

From question $$a=\dfrac{\cos A}{\cos B}$$ and $$b=\dfrac{\sin A}{\sin B}$$

$$\left( { b }^{ 2 }-{ a }^{ 2 } \right) { \sin }^{ 2 }B=\left( \cfrac { { \sin }^{ 2 }A }{ { \sin }^{ 2 }B } -\cfrac { { \cos }^{ 2 }A }{ { \cos }^{ 2 }B } \right) { \sin }^{ 2 }B$$

$$=\left( \cfrac { { \sin }^{ 2 }A{ \cos }^{ 2 }B{ -\cos }^{ 2 }A{ \sin }^{ 2 }B }{ { { \cos }^{ 2 }B\sin }^{ 2 }B } \right) { \sin }^{ 2 }B$$

$$=\left( \cfrac { \left( 1-{ \cos }^{ 2 }A \right) { \cos }^{ 2 }B{ -\cos }^{ 2 }A\left( 1-{ \cos }^{ 2 }B \right) }{ { { \cos }^{ 2 }B } } \right)$$

$$ =\cfrac { { \cos }^{ 2 }B-{ \cos }^{ 2 }A }{ { \cos }^{ 2 }B } =1-\cfrac { { \cos }^{ 2 }A }{ { \cos }^{ 2 }B } =1-{ a }^{ 2 }$$
Question 8
1 / -0

If $$\displaystyle \cot\theta=3x-\frac{1}{12{x}}$$, then $$\csc \theta-\cot \theta$$ is equal to

A
$$6x$$

B
$$-\displaystyle \frac{1}{6x}$$

C
$$\displaystyle \frac{1}{6x}$$ or $$-6x$$

D
$$6x$$ or $$-\displaystyle \frac{1}{6x}$$

Solution

Given: $$\cot \theta=3x-\dfrac {1}{12x}$$

We know $$\csc^{2}\theta - \cot^{2}\theta=1$$

$$\Rightarrow \csc^2 \theta= 1+\cot^2 \theta$$

$$\Rightarrow \csc^2 \theta=1+\bigg( 3x-\dfrac{1}{12x} \bigg)^2$$

$$\Rightarrow \csc^2 \theta=1+9x^2+\dfrac{1}{144x^2}-\dfrac {1}{2}$$

$$\Rightarrow \csc^2 \theta=9x^2+\dfrac{1}{144x^2}+\dfrac {1}{2}$$

$$\Rightarrow \csc^2 \theta=\bigg( 3x+\dfrac{1}{12x} \bigg)^2$$

$$\Rightarrow \csc \theta=\pm \bigg( 3x+\dfrac{1}{12x} \bigg)$$

If $$\csc \theta=3x+\dfrac{1}{12x}$$

Then $$\csc \theta - \cot \theta=3x+\dfrac{1}{12x}-\bigg(3x-\dfrac{1}{12x}\bigg)$$

$$=\dfrac{1}{6x}$$

If $$\csc \theta=-\bigg(3x+\dfrac{1}{12x}\bigg)=-3x-\dfrac{1}{12x}$$

Then $$\csc \theta - \cot \theta=-3x-\dfrac{1}{12x}-\bigg(3x-\dfrac{1}{12x}\bigg)$$

$$=-6x$$

Hence $$\csc \theta-\cot \theta=-6x$$ or $$\dfrac{1}{6x}$$
Question 9
1 / -0

Eiminate $$\theta$$ from $$ x= 1+\tan\theta,y= 2+\cot\theta$$

A
$$xy+1=x+y$$

B
$$xy+2=2x+y$$

C
$$xy+1=2x+y$$

D
$$1=2y+x$$

Solution

We know,
$$ \tan \theta \times \cot \theta = 1 $$
Substituting the corresponding values,
$$(x - 1)(y - 2) = 1$$
$$xy - 2x - y + 2 = 1$$
Or, $$xy + 1 = 2x + y$$
Hence, option 'C' is correct.
Question 10
1 / -0

If $$\cos\theta+\sin\theta=\sqrt{2}\cos\theta$$, then $$\cos\theta-\sin\theta=$$ _____

A
$$\sqrt{2}\sin\theta$$

B
2 $$\sin\theta$$

C
$$-\sqrt{2}\sin\theta$$

D
$$-2\sin\theta$$

Solution

Given,
$$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$$
$$\sin \theta = \sqrt{2} \cos \theta- \cos \theta $$
$$= (\sqrt{2}-1) \cos \theta $$
$$\cos \theta = \dfrac{1}{(\sqrt{2}-1) } \sin \theta = (\sqrt{2}+1)\sin \theta$$

$$\cos \theta - \sin \theta = (\sqrt{2}+1)\sin \theta- \sin \theta $$
$$ = \sqrt{2}\sin \theta $$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 27

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Introduction to Trigonometry Test - 27

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SHARING IS CARING

Question 1

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

$$\displaystyle \frac{1}{2p}$$ or $$-2p$$

$${-\displaystyle \frac{1}{2p}}$$ or $$2p$$

$${-\displaystyle \frac{1}{2p}}$$ or $$-2p$$

Solution

Question 2

$$0$$

$$1$$

$$y+a^{2}$$

$$x-a^{2}$$

Solution

Question 3

$$\sin^{2}A$$

$$\sin^{3}A$$

$$\cos^{2}A$$

$$\cos^{3}A$$

Solution

Question 4

47

162

24

48

Solution

Question 5

$$\displaystyle \frac{\sqrt{5}+1}{2}$$

$$\displaystyle \frac{\sqrt{5}+1}{8}$$

$$\displaystyle \frac{\sqrt{5}-1}{8}$$

$$\displaystyle \frac{\sqrt{5}-1}{2}$$

Solution

Question 6

$$x=y$$

$$\displaystyle \frac{1-y^{2}}{2}=\frac{1}{x}$$

$$\displaystyle \frac{y^{2}-1}{2}=\frac{1}{x}$$

$$\displaystyle \frac{1+y^{2}}{2}=\frac{1}{x}$$

Solution

Question 7

$$1+a^{2}$$

$$2+a^{2}$$

$$1-a^{2}$$

$$2-a^{2}$$

Solution

Question 8

$$6x$$

$$-\displaystyle \frac{1}{6x}$$

$$\displaystyle \frac{1}{6x}$$ or $$-6x$$

$$6x$$ or $$-\displaystyle \frac{1}{6x}$$

Solution

Question 9

$$xy+1=x+y$$

$$xy+2=2x+y$$

$$xy+1=2x+y$$

$$1=2y+x$$

Solution

Question 10

$$\sqrt{2}\sin\theta$$

2 $$\sin\theta$$

$$-\sqrt{2}\sin\theta$$

$$-2\sin\theta$$

Solution

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