Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

If $$ \displaystyle \cos^2 \theta + sec^2 \theta = p$$, then

A
$$ \displaystyle p < 1 $$

B
$$ \displaystyle p = 1 $$

C
$$ \displaystyle 1 < p < 2$$

D
$$ \displaystyle p \geq 2$$

Solution

$$ \displaystyle p = ( \cos^2 \theta + sec^2 \theta )= (cos \theta - sec \theta)^2 + 2 cos \theta sec \theta = (cos\theta - sec \theta )^2 + 2 \geq 2$$.
Question 2
1 / -0

Given that $$ sin \, \alpha = \dfrac{1}{2} , cos \, \beta = \dfrac{1}{2} $$ , then value of $$ ( \alpha + \beta) $$ is

A
$$0^{\circ} $$

B
$$ 30^{\circ} $$

C
$$ 60^{\circ} $$

D
$$ 90^{\circ} $$

Solution

$$ sin \, \alpha = \dfrac{1}{2} $$ [Given]

$$ \Rightarrow \, sin \, \alpha = sin \, 30^{\circ} $$
$$ \Rightarrow \, \alpha = 30^{\circ} $$

Also , $$ cos \, \beta = \dfrac{1}{2} $$

$$ \Rightarrow \, cos \, \beta = cos \, 60^{\circ} $$
$$ \Rightarrow \, \beta = 60^{\circ} $$
$$ \therefore \, \alpha + \beta = 30^{\circ} + 60^{\circ} = 90^{\circ} $$
Hence , verifies the option (d).
Question 3
1 / -0

$$2 \tan 45^{\circ} + \cos 45^{\circ} - \sin 45 ^{\circ} =? $$

A
$$ 0 $$

B
$$ 1 $$

C
$$ 2 $$

D
$$ 3 $$

Solution

We have,
$$\sin45^\circ=\dfrac{1}{\sqrt2}$$
$$\cos45^\circ=\dfrac{1}{\sqrt2}$$
$$\tan45^\circ=1$$

Substitute these values in the given expression to find its value,
$$\begin{aligned}{}2\tan {45^\circ } + \cos {45^\circ } - \sin {45^\circ }&= 2 \times 1 + \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}\\ &= 2\end{aligned}$$
Question 4
1 / -0

Choose the correct alternative answer for the following question.
$$cosec \ 45^\circ= ?$$

A
$$\dfrac12$$

B
$$\sqrt{2}$$

C
$$\dfrac{\sqrt{3}}{2}$$

D
$$\dfrac{2}{\sqrt{3}}$$

Solution

$$\sin 45^{\circ}=\dfrac {1}{\sqrt2}$$

$$cosec \ 45^\circ = \sqrt{2}$$

Hence, the correct answer is $$\sqrt{2}$$.
Question 5
1 / -0

If $$\theta =30^o$$, then $$\dfrac{1-\sin^22\theta}{\cos 2\theta}$$ is

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$-2$$

Solution

$$\dfrac{1-sin^22\theta}{\cos 2\theta}=\dfrac{1-\sin^22\times 30^o}{\cos 2\times 30^o}$$
$$=\dfrac{1-\sin^2 60^o}{\cos 60^o}=\dfrac{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}{\dfrac{1}{2}}$$
$$\dfrac{1-\dfrac{3}{4}}{\dfrac{1}{2}}=\dfrac{\dfrac{4-3}{4}}{\dfrac{1}{2}}\rightarrow \dfrac{1}{4}\times \dfrac{2}{1}=\dfrac{1}{2}$$
Question 6
1 / -0

If $$ \sqrt{3} \cos A = \sin A $$, then the value of $$ \cot A $$ is:

A
$$ \sqrt{3} $$

B
$$ 1 $$

C
$$\dfrac{1}{\sqrt{3}} $$

D
$$ 2$$

Solution
Question 7
1 / -0

$$a \sec \theta+b\tan\theta=1, a^{2}\sec^{2}\theta-b^{2}\tan^{2}\theta=5$$
Then $$ a^{2}(b^{2}+4)=$$

A
$$3b^{2}$$

B
$$9b^{2}$$

C
$$b^{2}$$

D
$$4b^{2}$$

Solution

$$a\sec\theta +b\tan\theta =1$$ ............ $$(1)$$
Squaring both the sides we get
$$a^{2} \sec^{2}\theta + b^{2}\tan^{2}\theta + 2ab\sec\theta \tan\theta = 1$$ ..... $$(2)$$
$$a^{2}\sec^{2}\theta - b^{2}\tan^{2}\theta = 5$$ ....... $$(3)$$
Adding equations $$(2)$$ and $$(3)$$ we get,
$$2a^{2}\sec^{2}\theta + 2ab\sec\theta \tan\theta =6$$
$$a^{2}\sec^{2}\theta +ab\sec\theta \tan\theta = 3 $$
$$a\sec\theta \left(a\sec\theta + b\tan\theta \right) = 3$$
$$a\sec\theta = 3$$ .......... (Using eqn $$(1)$$)
$$\implies a=3\cos\theta$$ .......... $$\left(\because \cos\theta = \dfrac{1}{\sec\theta}\right)$$
Substitute $$a$$ in $$(1)$$ we get
$$(3\cos\theta)\sec\theta + b\tan\theta = 1$$
$$3 + b\tan\theta = 1$$
$$btan\theta = -2$$
$$b = -2\cot\theta$$ ............. $$\left(\because \dfrac{1}{\tan \theta}=\cot \theta\right)$$
$$\therefore a^{2}\left(b^{2}+4 \right) = {(3\cos\theta})^2\left[ { (-2\cot\theta ) }^{ 2 }+4 \right]$$
$$ = 9{ cos }^{ 2 }\theta \quad (4{ cot }^{ 2 }\theta +4)$$
$$ = 9{ \cos }^{ 2 }\theta \cdot 4({\cot}^{ 2 }\theta +1)$$
$$ = 9{ \cos }^{ 2 }\theta \cdot 4\text{cosec}^{ 2 }\theta$$
$$ = 36\times \dfrac { \cos^{2}\theta}{\sin^{2}\theta}$$
$$ = 36{\cot}^{2}\theta = 9(-2cot\theta )^{2} = 9{b}^{2}$$
Hence, $$a^2(b^2 + 4)=9b^2$$
Question 8
1 / -0

If $$a \sin^{2}x+b\cos^{2}x=c, b\sin^{2}y+a\cos^{2}y=d$$ and $$a \tan x=b\tan y,$$ then $$\displaystyle \frac{a^{2}}{b^{2}}$$ equals to

A
$$^{\displaystyle \frac{(a-d)(c-a)}{(b-c)(d-b)}}$$

B
$$^{\displaystyle \frac{(b-c)(b-d)}{(a-c)(a-d)}}$$

C
$$\displaystyle \frac{(b-c)(d-b)}{(a-d)(c-a)}$$

D
$$^{\displaystyle \frac{(d-a)(c-a)}{(b-c)(d-b)}}$$

Solution

dividing equation $$asin^2x+bcos^2x=c$$ by $$cos^2x$$
$$a\ tan^2x+b=c\ sec^2x\Rightarrow a\ tan^2x+b=c(1+tan^2x)\Rightarrow tan^2x(c-a)=b-c\Rightarrow tan^2x=\dfrac{b-c}{c-a}$$

Similarly , dividing equation $$bsin^2y+acos^2y=d$$ by $$cos^2y$$
$$b\ tan^2y+a=d\ sec^2y\Rightarrow b\ tan^2y+a=d(1+tan^2y)\Rightarrow tan^2y(d-b)=a-d\Rightarrow tan^2y=\dfrac{a-d}{d-b}$$

Now, $$atanx=btany\Rightarrow a^2tan^2x=b^2tan^2y\Rightarrow \dfrac{a^2}{b^2}=\dfrac{tan^2y}{tan^2x}\Rightarrow \dfrac{a^2}{b^2}=\dfrac{\dfrac{a-d}{d-b}}{\dfrac{b-c}{c-a}}=\dfrac{(a-d)(c-a)}{(b-c)(d-b)}$$

Therefore, Answer is $$\dfrac{(a-d)(c-a)}{(b-c)(d-b)}$$
Question 9
1 / -0

$$\cot^{2}\alpha=1+\mathrm{a}^{2}$$ then $$\text{cosec}\,\alpha+\cot^{3}\alpha\sec\alpha=$$?

A
$$(1+\mathrm{a}^{2})^{\tfrac{2}{3}}$$

B
$$(2+\mathrm{a}^{2})^{\tfrac{2}{3}}$$

C
$$(1+\mathrm{a}^{2})^{\tfrac{3}{2}}$$

D
$$(2+\mathrm{a}^{2})^{\tfrac{3}{2}}$$

Solution

$$\text{cosec}\,\alpha + \cot^{3}\alpha \sec\alpha$$
=$$\dfrac{1}{\sin\alpha}+\dfrac{\cos^2\alpha}{\sin^3\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin^3\alpha}=\dfrac{1}{\sin^3\alpha}$$
$$ = \text{cosec}^{3}\,\alpha$$
$$ = (1+\cot^{2}\alpha)^{\tfrac{3}{2}}$$
$$ = (1 + \mathrm{a^{2}})^{\tfrac{3}{2}}$$
Question 10
1 / -0

$$\displaystyle {x}=\frac{\sin^{3}{p}}{\cos^{2}{p}}, \displaystyle {y}=\frac{\cos^{3}{p}}{\sin^{2}{p}}$$ and $$\sin p$$ $$+$$ $$\cos p $$ $$= \dfrac 12$$ then $${x}+{y}=$$

A
$$\displaystyle \frac{75}{18}$$

B
$$\displaystyle \frac{44}{9}$$

C
$$\displaystyle \frac{79}{18}$$

D
$$\displaystyle \frac{48}{9}$$

Solution

$$\sin p+\cos p=\cfrac { 1 }{ 2 } \Rightarrow { \left( \sin p+\cos p \right) }^{ 2 }=\cfrac { 1 }{ 4 } \\ 1+2\sin p.\cos p=\cfrac { 1 }{ 4 } \Rightarrow \sin p\cos p=-\cfrac { 3 }{ 8 } $$
Therefore,
$$x+y=\cfrac { { \sin }^{ 3 }p }{ { \cos }^{ 2 }p } +\cfrac { { \cos }^{ 3 }p }{ { \sin }^{ 2 }p } $$

$$ =\cfrac { { \sin }^{ 5 }p+{ \cos }^{ 5 }p }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p } $$

$$=\cfrac { { \sin }^{ 3 }p \, \left( 1-{ \cos }^{ 2 }p \right) +{ \cos }^{ 3 }p \, \left( 1-{ \sin }^{ 2 }p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p } $$

$$ =\cfrac { { \sin }^{ 3 }p-{ \sin }^{ 3 }p \, { \cos }^{ 2 }p+{ \cos }^{ 3 }p-{ \cos }^{ 3 }p \, { \sin }^{ 2 }p }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p } $$

$$ =\cfrac { { \sin }^{ 3 }p+{ \cos }^{ 3 }p-{ \sin }^{ 2 }p \, { \cos }^{ 2 }p \, \left( \sin p+\cos p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p } $$

$$ =\cfrac { { \left( \sin p+\cos p \right) }^{ 3 }-3\sin p \, \cos p\left( \sin p+\cos p \right) -{ \sin }^{ 2 }p \, { \cos }^{ 2 }p\left( \sin p+\cos p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p } $$

$$ =\cfrac { { \left( \cfrac { 1 }{ 2 } \right) }^{ 3 }-3\left( -\cfrac { 3 }{ 8 } \right) \left( \cfrac { 1 }{ 2 } \right) -\left( \cfrac { 9 }{ 64 } \right) \left( \cfrac { 1 }{ 2 } \right) }{ \left( \cfrac { 9 }{ 64 } \right) } $$

$$=\cfrac { \cfrac { 1 }{ 8 } +\cfrac { 9 }{ 16 } -\cfrac { 9 }{ 128 } }{ \cfrac { 9 }{ 64 } } =\cfrac { 79 }{ 18 } $$

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Introduction to Trigonometry Test - 66

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Introduction to Trigonometry Test - 66

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Question 1

$$ \displaystyle p < 1 $$

$$ \displaystyle p = 1 $$

$$ \displaystyle 1 < p < 2$$

$$ \displaystyle p \geq 2$$

Solution

Question 2

$$0^{\circ} $$

$$ 30^{\circ} $$

$$ 60^{\circ} $$

$$ 90^{\circ} $$

Solution

Question 3

$$ 0 $$

$$ 1 $$

$$ 2 $$

$$ 3 $$

Solution

Question 4

$$\dfrac12$$

$$\sqrt{2}$$

$$\dfrac{\sqrt{3}}{2}$$

$$\dfrac{2}{\sqrt{3}}$$

Solution

Question 5

$$1$$

$$-1$$

$$2$$

$$-2$$

Solution

Question 6

$$ \sqrt{3} $$

$$ 1 $$

$$\dfrac{1}{\sqrt{3}} $$

$$ 2$$

Solution

Question 7

$$3b^{2}$$

$$9b^{2}$$

$$b^{2}$$

$$4b^{2}$$

Solution

Question 8

$$^{\displaystyle \frac{(a-d)(c-a)}{(b-c)(d-b)}}$$

$$^{\displaystyle \frac{(b-c)(b-d)}{(a-c)(a-d)}}$$

$$\displaystyle \frac{(b-c)(d-b)}{(a-d)(c-a)}$$

$$^{\displaystyle \frac{(d-a)(c-a)}{(b-c)(d-b)}}$$

Solution

Question 9

$$(1+\mathrm{a}^{2})^{\tfrac{2}{3}}$$

$$(2+\mathrm{a}^{2})^{\tfrac{2}{3}}$$

$$(1+\mathrm{a}^{2})^{\tfrac{3}{2}}$$

$$(2+\mathrm{a}^{2})^{\tfrac{3}{2}}$$

Solution

Question 10

$$\displaystyle \frac{75}{18}$$

$$\displaystyle \frac{44}{9}$$

$$\displaystyle \frac{79}{18}$$

$$\displaystyle \frac{48}{9}$$

Solution

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