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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Question 1

000

111

tan⁡θ\tan { \theta } tanθ

cot⁡θ\cot { \theta } cotθ

Solution

Question 2

2−k2\sqrt {2 - k^{2}}2−k2​

k2−2\sqrt {k^{2} - 2}k2−2​

∣k∣|k|∣k∣

2−k\sqrt {2} - k2​−k

k−2k - \sqrt {2}k−2​

Solution

Question 3

101010

111111

121212

131313

Solution

Question 4

1−n1 - n1−n

1+n1 + n1+n

1−n21 - n^{2}1−n2

1+n21 + n^{2}1+n2

Solution

Question 5

sin⁡θ+cos⁡θ\sin \theta + \cos \thetasinθ+cosθ

sin⁡3θ−cos⁡3θ\sin^{3} \theta - \cos^{3}\thetasin3θ−cos3θ

sin⁡3θ+cos⁡3θ\sin^{3}\theta + \cos^{3}\thetasin3θ+cos3θ

sin⁡θ−cos⁡θ\sin \theta - \cos \thetasinθ−cosθ

Solution

Question 6

sin⁡2θ−cos⁡3θ\sin ^{ 2 }{ \theta } -\cos ^{ 3 }{ \theta } sin2θ−cos3θ

sin⁡3θ−cos⁡3θ\sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } \quad sin3θ−cos3θ

sin⁡4θ+cos⁡4θ\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } sin4θ+cos4θ

sin⁡2θ−cos⁡2θ\sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } sin2θ−cos2θ

Question 7

111

1.51.51.5

222

2.52.52.5

Solution

Question 8

3−cot⁡2θ3-\cot ^{ 2 }{ \theta } 3−cot2θ

3+cot⁡2θ3+\cot ^{ 2 }{ \theta } 3+cot2θ

3−tan⁡2θ3-\tan ^{ 2 }{ \theta } 3−tan2θ

3+tan⁡2θ3+\tan ^{ 2 }{ \theta } 3+tan2θ

Solution

Question 9

15\dfrac {1}{5}51​

95\dfrac {9}{5}59​

1225\dfrac {12}{25}2512​

2512\dfrac {25}{12}1225​

Solution

Question 10

ab\cfrac { a }{ b } ba​

ba\cfrac { b }{ a } ab​

a+ba−b\cfrac { a+b }{ a-b } a−ba+b​

a−ba+b\cfrac { a-b }{ a+b } a+ba−b​

Solution

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$0$

$1$

$\tan { \theta }$

$\cot { \theta }$

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

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

$|k|$

$\sqrt {2} - k$

$k - \sqrt {2}$

$10$

$11$

$12$

$13$

$1 - n$

$1 + n$

$1 - n^{2}$

$1 + n^{2}$

$\sin \theta + \cos \theta$

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

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

$\sin \theta - \cos \theta$

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

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

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

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

$1$

$1.5$

$2$

$2.5$

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

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

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

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

$\dfrac {1}{5}$

$\dfrac {9}{5}$

$\dfrac {12}{25}$

$\dfrac {25}{12}$

$\cfrac { a }{ b }$

$\cfrac { b }{ a }$

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

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