Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 67

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Question 1
1 / -0

If $$\displaystyle \frac{\sin x}{a}=\frac{\cos x}{b}=\frac{\tan x}{c}=k$$, then
$$bc+\displaystyle \frac{1}{ck}+\frac{ak}{1+bk}$$ is

A
$$k(\displaystyle a+\frac{1}{a})$$

B
$$\dfrac{a}{k}+\dfrac{1}{ak}$$

C
$$\displaystyle \frac{1}{k^{2}}$$

D
$$^{\dfrac{a}{k}}$$

Solution

$$sin x = ak$$
$$cos x = bk$$
$$tan x = ck$$
Since,$$ \displaystyle tan x = \frac{sin x}{cos x} $$,

$$\displaystyle ck = \frac{a}{b} $$ Or, $$\displaystyle bc = \frac{a}{k} $$
$$\displaystyle ck = tan x$$
$$ \Rightarrow \displaystyle \frac{1}{ck} = cot x $$

$$\Rightarrow \displaystyle \frac{ak}{1+bk} = \frac{sin x}{1+cos x} $$

$$\Rightarrow \displaystyle \frac{1}{ck} + \frac{ak}{1+bk} = cot x + \frac{sin x}{1+cos x} $$

$$\displaystyle = \frac{cos x + {cos}^{2} x + {sin}^{2} x}{sin x(1+cos x)} $$

$$\displaystyle = \frac{1+cos x}{sinx(1+cos x)} $$
$$= cosec x $$
Hence, $$bc + \dfrac{1}{ck} + \dfrac{ak}{1+bk} $$ = $$ \dfrac{a}{k} + \dfrac{1}{ak} $$
Question 2
1 / -0

If $$\text{cosec}\,\theta-\sin\theta=a^{3},\ \sec\theta-\cos\theta=b^{3}$$
then $$a^{2}b^{2}(a^{2}+b^{2})=$$

A
$$2$$

B
$$\sin^{2}\theta$$

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

D
$$1$$

Solution

$$\csc { \theta } -\sin { \theta } ={ a }^{ 3 }$$

$$\Rightarrow \dfrac { 1-\sin ^{ 2 }{ \theta } }{ \sin { \theta } } ={ a }^{ 3 }$$

$$\Rightarrow \dfrac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } ={ a }^{ 3 }\quad -(1)$$

$$\sec { \theta } -\cos { \theta } ={ b }^{ 3 }$$

$$\Rightarrow \dfrac { 1-\cos ^{ 2 }{ \theta } }{ \cos { \theta } } ={ b }^{ 3 }$$

$$\Rightarrow \dfrac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } } ={ b }^{ 3 }\quad -(2)$$

$$(1)\div (2)\Rightarrow \dfrac { \cos ^{ 3 }{ \theta } }{ \sin ^{ 3 }{ \theta } } =\dfrac { { a }^{ 3 } }{ { b }^{ 3 } } $$

$$\tan { \theta } =\dfrac { b }{ a } \quad -(3)$$

$$(1)\times (2)\Rightarrow \dfrac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } \times \dfrac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } } ={ a }^{ 3 }{ b }^{ 3 }$$

$$\sin { \theta } .\cos { \theta } ={ a }^{ 3 }{ b }^{ 3 }\quad -(4)$$

$$(3)\times (4)\Rightarrow \sin ^{ 2 }{ \theta } ={ a }^{ 2 }{ b }^{ 4 }$$

$$(4)\div (3)\Rightarrow \cos ^{ 2 }{ \theta } ={ a }^{ 4 }{ b }^{ 2 }$$

$$(3) \Rightarrow 1+\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } =1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } $$

Given, $${ a }^{ 2 }{ b }^{ 2 }\left( { a }^{ 2 }+{ b }^{ 2 } \right) ={ a }^{ 4 }{ b }^{ 2 }\left( 1+\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } \right) $$

$$=\cos ^{ 2 }{ \theta } \times \sec ^{ 2 }{ \theta } =1$$
Question 3
1 / -0

$$\cos\theta +\cos^{2}\theta =1$$ and $$a\sin^{12}\theta +b\sin^{10}\theta +c\sin^{8}\theta +d\sin^{6}\theta =1.$$ Then $$\displaystyle \frac{b+c}{a+d}=$$?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

Solution

$$\cos\theta +\cos^{ 2 }\theta =1$$
$$\Rightarrow \cos\theta +1-\sin^{ 2 }\theta =1$$
$$\Rightarrow \cos\theta = \sin^{ 2 }\theta $$
$$\Rightarrow { \sqrt { 1-\sin^{2}} \theta }={ \sin^{2}\theta}$$

On squaring both sides:-
$$\Rightarrow { \left( \sqrt { 1-\sin^{2}} \theta \right)}^{2}={\left( \sin^{2}\theta \right)}^{2}$$
$$\Rightarrow 1-\sin^{2}\theta = \sin^{4}\theta $$
$$\Rightarrow \sin^{4}\theta + \sin^{2}\theta =1$$

On taking cube on both sides:-
$$\Rightarrow { \left( \sin^{4}\theta +\sin^{2}\theta \right) }^{3}={\left( 1 \right) }^{3}$$
$$\Rightarrow \sin^{12}\theta +3\sin^{10}\theta +3\sin^{8}\theta +\sin^{6}\theta =1$$

On comparing above obtained equation with $$a\sin^{12}\theta +b\sin^{10}\theta +c\sin^{8}\theta +d\sin^{6}\theta =1$$ we get,
$$a=1,b=3,c=3,d=1$$

Hence,
$$\ \dfrac {b+c}{a+d} =\dfrac { 3+3 }{ 1+1 } $$
$$=\dfrac { 6 }{ 2 }$$
$$ =3$$

Hence, option $$B$$ is correct.
Question 4
1 / -0

If $$a \cos^{3}\alpha+3a\cos\alpha.\sin^{2}\alpha=m$$ and $$a \sin^{3}\alpha+3a\cos^{2}\alpha.\sin\alpha=n$$ then, $$(m+n)^{2/3}+(m-n)^{2/3}$$ is equal to:

A
$$2a^{2}$$

B
$$2a^{1/3}$$

C
$$2a^{2/3}$$

D
$$2a^{3}$$

Solution

$$m= a\ {cos}^{ 3 }\alpha + 3 a\ cos\alpha\ { sin }^{ 2 }\alpha$$
$$n = a{ sin }^{ 3 }\alpha + 3a\ {cos}^{2}\alpha\ sin\alpha$$
$$\therefore m+n = a({ cos }^{ 3 }\alpha +{ sin }^{ 3 }\alpha +3cos\alpha { sin }^{ 2 }\alpha +3{ cos }^{ 2 }\alpha sin\alpha )$$
$$m+n = a\left[ { cos }^{ 3 }\alpha +{ sin }^{ 3 }\alpha +3sin\alpha cos\alpha (sin\alpha +cos\alpha ) \right] $$
$$m+n = a{ \left( cos\alpha +sin\alpha \right) }^{ 3 }$$ ...... $$\left[ \because { (a+b })^{3}={ a }^{ 3 }+{ b }^{ 3 }+3ab(a+b) \right]$$
Similarly, $$m-n = a{ \left( cos\alpha -sin\alpha \right) }^{ 3 }$$
Now, $${ (m+n) }^{ 2/3 }+{ (m-n) }^{ 2/3 }={ a }^{ 2/3 }\left[ { \left( cos\alpha +sin\alpha \right) }^{ 3\times2/3 }+{ \left( cos\alpha -sin\alpha \right) }^{ 3\times2/3 } \right]$$
$$= { a }^{ 2/3 }\left[ { \left( cos\alpha +sin\alpha \right) }^{ 2 }+{ \left( cos\alpha -sin\alpha \right) }^{ 2 } \right]$$
$$= { a }^{ 2/3 }\left[ { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha +2sin\alpha cos\alpha +{ cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha -2sin\alpha cos\alpha \right]$$
$$={ a }^{ 2/3 }(1+1)$$ ....... $$(\because \quad { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha =1)$$
$$=2{ a }^{ 2/3 }$$
Question 5
1 / -0

If $$\displaystyle \frac { \sin { \alpha } }{ \sin { \beta } } =\frac { \sqrt { 3 } }{ 2 } $$ and $$\displaystyle \frac { \cos { \alpha } }{ \cos { \beta } } =\frac { \sqrt { 5 } }{ 2 } ,0<\alpha ,\beta <\frac { \pi }{ 2 } $$, then

A
$$\displaystyle \tan { \alpha } =\frac { \sqrt { 3 } }{ \sqrt { 5 } } ,\tan { \beta } =1$$

B
$$\displaystyle \tan { \alpha } =1,\tan { \beta } =\frac { \sqrt { 3 } }{ \sqrt { 5 } } $$

C
$$\displaystyle \tan { \alpha } =\frac { \sqrt { 5 } }{ \sqrt { 3 } } ,\tan { \beta } =1$$

D
none of these

Solution

$$\dfrac{\sin\alpha}{\sin\beta}=\dfrac{\sqrt{3}}{2}$$

Or $$\sqrt{\dfrac{1-\cos^{2}\alpha}{1-\cos^{2}\beta}}=\dfrac{\sqrt{3}}{2}$$

$$\dfrac{1-\cos^{2}\alpha}{1-\cos^{2}\beta}=\dfrac{3}{4}$$

Or $$4-4\cos^{2}\alpha=3-3\cos^{2}\beta$$

$$4\cos^{2}\alpha-1=3\cos^{2}\beta$$ ...(i)

And it is given that

$$\cos\alpha=\dfrac{\sqrt{5}}{2}\cos\beta$$

Or $$\cos^{2}\alpha=\dfrac{5}{4}\cos^{2}\beta$$

Substituting in equation i, we get

$$5cos^{2}\beta-1=3cos^{2}\beta$$

Or $$2cos^{2}\beta=1$$

Or $$cos\beta=\dfrac{1}{\sqrt{2}}=sin\beta$$

Therefore
$$tan\beta=1$$

Now
$$cos\beta=sin\beta=\dfrac{1}{\sqrt{2}}$$

Hence
$$cos\alpha=\dfrac{\sqrt{5}}{2\sqrt{2}}$$

And $$sin\alpha=\dfrac{\sqrt{3}}{2\sqrt{2}}$$

Therefore
$$tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{3}}{\sqrt{5}}$$
Question 6
1 / -0

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

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

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

C
$$\displaystyle \pm \dfrac {\sqrt {a^{2}+b^{2}-c^{2}}}{c}$$

D
$$\displaystyle \pm\dfrac {\sqrt {a^{2}+b^{2}+c^{2}}}{c}$$

Solution

$$a\sin {\theta +b\cos {\theta}} =\quad c$$
$$a\cos {\theta -b\sin {\theta}} = \quad k$$
Squaring on both sides and adding gives
$$a^{2} + b^{2} = c^{2} + k^{2}$$
$$k = \pm \sqrt {a^{2} + b^{2} - c^{2}}$$
$$\dfrac {a-b\tan {\theta}}{b+a\tan {\theta}} = \dfrac {a\cos {\theta - b\sin {\theta}}}{a\sin {\theta + b\cos {\theta}}} = \dfrac {k}{c} = \dfrac {\pm \sqrt {a^{2} + b^{2} - c^{2}}}{c}$$
Question 7
1 / -0

lf $$x=\displaystyle \frac{\sin^{3}p}{\cos^{2}p}, y=\displaystyle \frac{\cos^{3}p}{\sin^{2}p}$$ and $$\displaystyle \sin p+\cos p=\frac{1}{2}$$, then $$x+y=$$

A
$$\dfrac {75}{18}$$

B
$$\dfrac{44}9$$

C
$$\dfrac {79}{18}$$

D
$$\dfrac {48}9$$

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

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

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

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

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

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

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

$$=\cfrac { \cfrac { 1 }{ 8 } +\cfrac { 9 }{ 16 } -\cfrac { 9 }{ 128 } }{ \cfrac { 9 }{ 64 } } =\cfrac { 79 }{ 18 } $$
Question 8
1 / -0

If $$\cos x +\cos^{2} x+\cos^{3} x=1, a\sin^{6} x + b \sin^{4} x + c \sin^{2} x + d =\mathrm{0}$$ and $$ a >0$$ then $$ a + b + c + d =$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

$$\cos x +\cos^2 x+\cos ^3 x=1$$
$$\cos x(1+ \cos ^{2}x) = 1 - \cos ^{2}x$$
$$\cos x(2- \sin ^{2}x) = \sin ^{2}x$$
we square both the sides
$$\cos ^{2}x(2-\sin ^{2}x)^{2}= \sin ^{4}x$$
$$(1-\sin ^{2}x)(2-\sin ^{2}x)^{2}= \sin ^{4}x$$
we simplify this to get $$\sin ^{6}x + 4\sin ^{4}x- 8\sin ^{2}x + 4 = 0$$
equating with given equation we get
$$a=1,b=4,c=-8,d=4$$
Question 9
1 / -0

If $$\displaystyle a^{2}\cos^{4}\: \theta -b^{2}\sin^{4}\, \theta =0$$, then $$\displaystyle \: \frac{\sin^{8}\, \theta }{a^{3}}+\frac{\cos^{8}\theta }{b^{3}}$$ is

A
$$\displaystyle \frac{1}{a+b}$$

B
$$\displaystyle \frac{1}{(a+b)^{2}}$$

C
$$\displaystyle \frac{1}{(a+b)^{3}}$$

D
$$\dfrac{2}{(a+b)^4}$$

Solution

Given,
$$a^2\cos^4\theta-b^2\sin^4\theta=0$$
then $$\tan^4\theta=\dfrac{a^2}{b^2}$$
$$\tan^2\theta=\dfrac{a}{b}$$
$$\sec^2\theta=\dfrac{a+b}{b}$$
$$\cos^2\theta=\dfrac{b}{a+b}$$
$$\sin^2\theta=\dfrac{a}{a+b}$$
$$\dfrac{\sin^8\theta}{a^3}+\dfrac{\cos^8\theta}{b^3} = \dfrac{a^4}{a^3(a+b)^4}+\dfrac{b^4}{b^3(a+b)^4}$$
$$\dfrac{a}{(a+b)^4}+\dfrac{b}{(a+b)^4}$$
$$=\dfrac{1}{(a+b)^3}$$
Question 10
1 / -0

If $$\sin x+\sin ^{2}x=1$$,then the value of $$\cos ^{12}x+3\cos ^{10}x+3\cos ^{8}x+\cos ^{6}x-2$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

We have $$\sin x+\sin ^{2}x=1$$

$$\Rightarrow \sin x=1-\sin ^{2}x\Rightarrow \sin x=\cos ^{2}x$$

Now $$\cos ^{12}x+3\cos ^{10}x+3\cos ^{8}x+\cos ^{6}x-2=\sin ^{6}x+3\sin ^{5}x+3\sin ^{4}x+\sin ^{3}x-2$$

$$=\left ( \sin ^{2}x \right )^{3}+3 \left ( \sin ^{2}x \right )^{2}(\sin x)+3\left ( \sin ^{2}x \right )\left ( \sin x \right )^{2}+\left ( \sin x \right )^3-2$$

$$=\left ( \sin ^{2}x +\sin x\right )^{3}-2$$.............$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$

$$=\left ( 1 \right )^{3}-2$$

$$=-1$$

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Introduction to Trigonometry Test - 67

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Introduction to Trigonometry Test - 67

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Question 1

$$k(\displaystyle a+\frac{1}{a})$$

$$\dfrac{a}{k}+\dfrac{1}{ak}$$

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

$$^{\dfrac{a}{k}}$$

Solution

Question 2

$$2$$

$$\sin^{2}\theta$$

$$\cos^{2}\theta$$

$$1$$

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$6$$

Solution

Question 4

$$2a^{2}$$

$$2a^{1/3}$$

$$2a^{2/3}$$

$$2a^{3}$$

Solution

Question 5

$$\displaystyle \tan { \alpha } =\frac { \sqrt { 3 } }{ \sqrt { 5 } } ,\tan { \beta } =1$$

$$\displaystyle \tan { \alpha } =1,\tan { \beta } =\frac { \sqrt { 3 } }{ \sqrt { 5 } } $$

$$\displaystyle \tan { \alpha } =\frac { \sqrt { 5 } }{ \sqrt { 3 } } ,\tan { \beta } =1$$

none of these

Solution

Question 6

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

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

$$\displaystyle \pm \dfrac {\sqrt {a^{2}+b^{2}-c^{2}}}{c}$$

$$\displaystyle \pm\dfrac {\sqrt {a^{2}+b^{2}+c^{2}}}{c}$$

Solution

Question 7

$$\dfrac {75}{18}$$

$$\dfrac{44}9$$

$$\dfrac {79}{18}$$

$$\dfrac {48}9$$

Solution

Question 8

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 9

$$\displaystyle \frac{1}{a+b}$$

$$\displaystyle \frac{1}{(a+b)^{2}}$$

$$\displaystyle \frac{1}{(a+b)^{3}}$$

$$\dfrac{2}{(a+b)^4}$$

Solution

Question 10

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

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