Trigonometric Functions Test 12

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Trigonometric Functions Test 12

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Question 1
1 / -0

$$\left( \text{cosec} { \theta } -\sin { \theta } \right) \left( \sec { \theta } -\cos { \theta } \right) \left( \tan { \theta } +\cot { \theta } \right) $$ simplifies to

A
$$0$$

B
$$1$$

C
$$\tan { \theta } $$

D
$$\cot { \theta } $$

Solution

$$\left( \cos\theta -\sin\theta \right) \left( \sec\theta -\cos\theta \right) \left( \tan\theta +\cot\theta \right) $$
$$=\left( \dfrac { 1 }{ \sin\theta } -\sin\theta \right) \left( \dfrac { 1 }{ \cos\theta } -\cos\theta \right) \left( \dfrac { \sin\theta }{ \cos\theta } +\dfrac { \cos\theta }{ \sin\theta } \right) $$
$$= \dfrac { \left( 1-{ \sin }^{ 2 }\theta \right) }{ \sin\theta } \times \dfrac { \left( 1-{ \cos }^{ 2 }\theta \right) }{ \cos\theta } \times \dfrac { \left( { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta \right) }{ \sin\theta \cos\theta } $$
$$= \dfrac { { \cos }^{ 2 }\theta .{ \sin }^{ 2 }\theta .1 }{ { \sin }^{ 2 }\theta .{ \cos }^{ 2 }\theta } $$
$$=1$$
Question 2
1 / -0

If $$\sin \alpha + \cos \alpha = k$$, then $$|\sin \alpha - \cos \alpha |$$ equals.

A
$$\sqrt {2 - k^{2}}$$

B
$$\sqrt {k^{2} - 2}$$

C
$$|k|$$

D
$$\sqrt {2} - k$$

E
$$k - \sqrt {2}$$

Solution

$$\sin\alpha +\cos\alpha =k$$
$${ \left( \sin\alpha +\cos\alpha \right) }^{ 2 }={ k }^{ 2 }$$
$${ \sin }^{ 2 }\alpha +{ \cos }^{ 2 }\alpha +2\sin\alpha \cos\alpha ={ k }^{ 2 }$$
$$2\sin\alpha \cos\alpha ={ k }^{ 2 }-1$$
$${ \left| \sin\alpha -\cos\alpha \right| }^{ 2 }=\left| { \sin }^{ 2 }\alpha +{ \cos }^{ 2 }\alpha -2\sin\alpha \cos\alpha \right| $$
$${ \left| \sin\alpha -\cos\alpha \right| }^{ 2 }=\left| 1-\left( { k }^{ 2 }-1 \right) \right| $$
$${ \left| \sin\alpha -\cos\alpha \right| }^{ 2 }=\left| 2-{ k }^{ 2 } \right| $$
or
$$\left| \sin\alpha -\cos\alpha \right| =\sqrt { 2-{ k }^{ 2 } } $$
Question 3
1 / -0

The expression $$3(\sin x - \cos x)^{4} + 6(\sin x + \cos x)^{2} + 4(\sin^{6}x + \cos^{6}c)$$ is equal to

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

$$3(\sin x - \cos x)^{4} + 6(\sin x + \cos x)^{2} + 4(\sin^{6}x + \cos^{6}c)$$
$$= 3(1 + 4\sin^{2} x\cos^{2} x - 4\sin x\cos x) + 6(1 + 2\sin x \cos x) + 4(1 - 3\sin^{2}x \cos^{2}x) = 3 + 6 + 4 = 13$$.
Question 4
1 / -0

If $$n = \dfrac {\cos \alpha}{\cos \beta}, m = \dfrac {\sin \alpha}{\sin \beta}$$, then $$(m^{2} - n^{2})\sin^{2}\beta$$ is

A
$$1 - n$$

B
$$1 + n$$

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

D
$$1 + n^{2}$$

Solution

Given,

$$(m^2-n^2)\sin ^2\beta $$

$$=\left ( \dfrac{\sin ^2\alpha }{\sin ^2\beta } -\dfrac{\cos ^2\alpha }{\cos ^2\beta }\right )\sin ^2\beta $$

$$=\left ( \dfrac{\cos ^2\beta \sin ^2\alpha-\sin ^2\beta\cos ^2\alpha }{\sin ^2\beta \cos ^2\beta} \right )\sin ^2\beta $$

$$=\sin ^2\alpha-\dfrac{\sin ^2\beta}{\cos ^2\beta }\cos ^2\alpha$$

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

$$=1-n^2$$
Question 5
1 / -0

$$(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)$$ can be written as

A
$$\sin \theta + \cos \theta$$

B
$$\sin^{3} \theta - \cos^{3}\theta$$

C
$$\sin^{3}\theta + \cos^{3}\theta$$

D
$$\sin \theta - \cos \theta$$

Solution

$$\sin \theta + \cos \theta - \sin^{2} \theta \cos \theta - \sin \theta \cos^{2}\theta$$
$$= \cos \theta (1 - \sin^{2} \theta) + \sin \theta (1 - \cos^{2}\theta) = \cos^{3}\theta + \sin^{3}\theta$$.
Question 6
1 / -0

$$\sin ^{ 6 }{ \theta } +\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } -\sin ^{ 4 }{ \theta } \cos ^{ 4 }{ \theta } -\cos ^{ 6 }{ \theta } $$ equals to

A
$$\sin ^{ 2 }{ \theta } -\cos ^{ 3 }{ \theta } $$

B
$$\sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } \quad $$

C
$$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } $$

D
$$\sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } $$
Question 7
1 / -0

$$\left( 1-\sin ^{ 2 }{ \theta } \right) \left( 1+\tan ^{ 2 }{ \theta } \right) $$ is equal to

A
$$1$$

B
$$1.5$$

C
$$2$$

D
$$2.5$$

Solution

$$\left( 1-\sin ^{ 2 }{ \theta } \right) \left( 1+\tan ^{ 2 }{ \theta } \right) =\cos ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } =1\quad \quad $$
Question 8
1 / -0

$$\cfrac { 3-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } $$ is equal to

A
$$3-\cot ^{ 2 }{ \theta } $$

B
$$3+\cot ^{ 2 }{ \theta } $$

C
$$3-\tan ^{ 2 }{ \theta } $$

D
$$3+\tan ^{ 2 }{ \theta } $$

Solution

$$ \dfrac { 3-4{ \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =\dfrac { 3{ \sin }^{ 2 }\theta +3{ \cos }^{ 2 }\theta -4{ \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } $$

$$=\dfrac { 3{ \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =3-{ \tan }^{ 2 }\theta $$
Question 9
1 / -0

If $$\tan { \theta } =\dfrac {3}{4} $$ and $$0<\theta <{ 90 }^{ 0 }$$, then the value of $$\sin { \theta } \cos { \theta } $$ is

A
$$\dfrac {1}{5}$$

B
$$\dfrac {9}{5}$$

C
$$\dfrac {12}{25}$$

D
$$\dfrac {25}{12}$$

Solution

$$\tan { \theta } =\cfrac { 3 }{ 4 } \Rightarrow \cfrac { P }{ B } =\cfrac { 3 }{ 4 } \Rightarrow H=\sqrt { 9+16 } =5\Rightarrow \sin { \theta } \cos { \theta } =\cfrac { 12 }{ 25 } $$
Question 10
1 / -0

If $$\cfrac { \sin { \left( x+y \right) } }{ \sin { \left( x-y \right) } } =\cfrac { a+b }{ a-b } $$, then what is $$\cfrac { \tan { x } }{ \tan { y } } $$ equal to?

A
$$\cfrac { a }{ b } $$

B
$$\cfrac { b }{ a } $$

C
$$\cfrac { a+b }{ a-b } $$

D
$$\cfrac { a-b }{ a+b } $$

Solution

$$\dfrac{sin(x+y)}{sin(x-y)}=\dfrac{a+b}{a-b}\Rightarrow \dfrac{sinxcosy+cosxsiny}{sinxcosy-sinycosx}=\dfrac{a+b}{a-b}$$

$$\Rightarrow (a-b)(sinxcosy+cosxsiny)=(a+b)(sinxcosy-sinycosx)\\ \Rightarrow asinxcosy+acosxsiny-bsinxcosy-bcosxsiny=asinxcosy-asinycosx+bsinxcosy-bsinycosx\\ \Rightarrow sinxcosy(-2b)=sinycosx(-2a)\\ \Rightarrow \dfrac{\dfrac{sinx}{cosx}}{\dfrac{siny}{cosy}}=\dfrac{a}{b}\Rightarrow \dfrac{tanx}{tany}=\dfrac{a}{b}$$

Therefore, Answer is $$A$$

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Trigonometric Functions Test 12

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Trigonometric Functions Test 12

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$\tan { \theta } $$

$$\cot { \theta } $$

Solution

Question 2

$$\sqrt {2 - k^{2}}$$

$$\sqrt {k^{2} - 2}$$

$$|k|$$

$$\sqrt {2} - k$$

$$k - \sqrt {2}$$

Solution

Question 3

$$10$$

$$11$$

$$12$$

$$13$$

Solution

Question 4

$$1 - n$$

$$1 + n$$

$$1 - n^{2}$$

$$1 + n^{2}$$

Solution

Question 5

$$\sin \theta + \cos \theta$$

$$\sin^{3} \theta - \cos^{3}\theta$$

$$\sin^{3}\theta + \cos^{3}\theta$$

$$\sin \theta - \cos \theta$$

Solution

Question 6

$$\sin ^{ 2 }{ \theta } -\cos ^{ 3 }{ \theta } $$

$$\sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } \quad $$

$$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } $$

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

Question 7

$$1$$

$$1.5$$

$$2$$

$$2.5$$

Solution

Question 8

$$3-\cot ^{ 2 }{ \theta } $$

$$3+\cot ^{ 2 }{ \theta } $$

$$3-\tan ^{ 2 }{ \theta } $$

$$3+\tan ^{ 2 }{ \theta } $$

Solution

Question 9

$$\dfrac {1}{5}$$

$$\dfrac {9}{5}$$

$$\dfrac {12}{25}$$

$$\dfrac {25}{12}$$

Solution

Question 10

$$\cfrac { a }{ b } $$

$$\cfrac { b }{ a } $$

$$\cfrac { a+b }{ a-b } $$

$$\cfrac { a-b }{ a+b } $$

Solution

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