MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 5

$\displaystyle \left( \sin { \theta } +\cos { \theta } \right) \left( 1-\sin { \theta } \cos { \theta } \right)$ can be written as :

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

Explanation

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

$\displaystyle \left( \sin { \theta } +\cos { \theta } \right) \left( { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta -\sin { \theta } .\cos { \theta } \right)$

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

Evaluate

$\displaystyle \left( \sin { A } +\cos { A } \right) \left( \tan { A } +\cot { A } \right)$

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$\displaystyle \sin { A } +\cos { A }$
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$\displaystyle \sec { A } +cosecA$
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$\displaystyle \sin { A }$
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$\displaystyle \cos { A }$

Explanation

The value of

$\displaystyle \left( \sin { A } +\cos { A } \right) \left( \tan { A } +\cot { A } \right)$ is

$\displaystyle =\left( \sin { A } +\cos { A } \right) \left( \frac { \sin { A } }{ \cos { A } } +\frac { \cos { A } }{ \sin { A } } \right)$

$\displaystyle =\left( \sin { A } +\cos { A } \right) \left( \frac { { \sin }^{ 2 }A+{ \cos }^{ 2 }A }{ \cos { A } \sin { A } } \right)$

$=\displaystyle \frac { \sin { A } +\cos { A } }{ \cos { A } \sin { A } }$

$\displaystyle =\frac { \sin { A } }{ \cos { A } \sin { A } } +\frac { \cos { A } }{ \cos { A } \sin { A } }$

$\displaystyle =\frac { 1 }{ \cos { A } } +\frac { 1 }{ \sin { A } }$

$=\sec { A } +cosecA$

$\displaystyle \sin \theta =\cos (2\theta -45^{\circ}),\,\,\, 0 < (2\theta -45^{\circ})< 90^{\circ},$ then

$\displaystyle \tan \theta$

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$-1$
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$0$
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$1$
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$\displaystyle \frac{1}{\sqrt{3}}$

Explanation

Given,

$\displaystyle \sin \theta =\cos (2\theta -45^{\circ})$
since,

$\displaystyle \sin \theta =\cos (90^{\circ} - \theta)$

$\therefore 2\theta -45^{\circ}= 90^{\circ} - \theta$

$\therefore 3\theta= 135^{\circ}$

$\therefore \theta = 45^{\circ}$
So,

$\tan \theta = \tan 45^{\circ} = 1$

$\displaystyle \sqrt { \frac { 1+\sin { A } }{ 1-\sin { A } } }$ is equal to :

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$\displaystyle \frac { \cot{ A } }{ \sin { A } +\cos { A } }$
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$\displaystyle \frac { \cot{ A } }{ \text{cosec}A-1 }$
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$\displaystyle \frac { \cot{ A } }{ \sec { A } -1 }$
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$\displaystyle \frac { \cot{ A } }{ \tan { A } -1 }$

Explanation

The value of

$\displaystyle \sqrt { \frac { 1+\sin { A } }{ 1-\sin { A } } }$ is

$=\sqrt { \dfrac { 1+\sin { A } }{ 1-\sin { A } } } \times \sqrt { \dfrac { 1+\sin { A } }{ 1+\sin { A } } }$

$\displaystyle =\frac { 1+\sin { A } }{ \cos { A } }$

$\displaystyle =\frac { 1+\sin { A } }{ \cos { A } } \times \frac { \cos { A } }{ \cos { A } } =\frac { \cos { A } \left( 1+\sin { A } \right) }{ { \cos }^{ 2 }A }$

$\displaystyle =\frac { \cos { A } \left( 1+\sin { A } \right) }{ \left( 1+\sin { A } \right) \left( 1-\sin { A } \right) } =\frac { \cos { A } }{ 1-\sin { A } }$

$\displaystyle =\frac { \frac { \cos { A } }{ \sin { A } } }{ \frac { 1-\sin { A } }{ \sin { A } } } =\frac { \cot{ A } }{ \text{cosec}A-1 }$

Evalaute

$\displaystyle \frac { \sin { \theta } }{ 1+\cos { \theta } } +\frac { 1+\cos { \theta } }{ \sin { \theta } }$

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$\displaystyle 2\sin { A }$
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$\displaystyle 2cosecA$
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$\displaystyle 2\tan { A }$
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$\displaystyle 2\cos { A }$

Explanation

The value of

$\displaystyle \frac { \sin { \theta } }{ 1+\cos { \theta } } +\frac { 1+\cos { \theta } }{ \sin { \theta } }$ is equal to

$\displaystyle =\frac { { \sin }^{ 2 }\theta { \left( 1+\cos { \theta } \right) }^{ 2 } }{ \sin{ \theta \left( 1+\cos { \theta } \right) } }$

$\displaystyle =\frac { { \sin }^{ 2 }\theta +1+{ \cos }^{ 2 }\theta +2\cos\theta }{ \sin{ \theta \left( 1+\cos { \theta } \right) } }$

$\displaystyle =\frac { 1+\left( { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta \right) +2\cos\theta }{ \sin{ \theta \left( 1+\cos { \theta } \right) } }$

$\displaystyle =\frac { 1+1+2\cos { \theta } }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } =\frac { 2+2\cos { \theta } }{ \sin{ \theta \left( 1+\cos { \theta } \right) } }$

$\displaystyle =\frac { 2\left( 1+\cos { \theta } \right) }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } =\frac { 2 }{ \sin{ \theta } } =2\text{cosec}\theta$

$\displaystyle \cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta }$ , then

$\displaystyle \cos { \theta } +\sin { \theta }$ is equal to :

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$\displaystyle \sqrt { 2 } cosec\theta$
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$\displaystyle \sqrt { 2 } \sin { \theta }$
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$\displaystyle \sqrt { 2 } \tan { \theta }$
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$\displaystyle \sqrt { 2 } \cos { \theta }$

Explanation

Dividing by

$\sin \theta$ we get,

$\cot \theta-1=\sqrt2\Rightarrow \cot\theta=\sqrt2+1$

So,

$\tan \theta=\dfrac{1}{\sqrt2+1}\Rightarrow \tan \theta=\dfrac{\sqrt2-1}{(\sqrt2+1)(\sqrt2-1)}\ or\ \dfrac{-\sqrt2-1}{(\sqrt2+1)(-\sqrt2-1)} \Rightarrow \tan \theta=\sqrt2-1\ or\ -\sqrt2-1\\ \Rightarrow 1+\tan \theta=\pm\sqrt2\Rightarrow \dfrac{\cos \theta+\sin \theta}{\cos \theta}=\pm\sqrt2\Rightarrow \cos \theta+\sin \theta=\pm\sqrt2\cos \theta$

Therefore, Answer is

$\sqrt2\cos \theta$

Evaluate

$\displaystyle \frac { \tan { A } +\sec { A } -1 }{ \tan { A } -\sec { A } +1 }$

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$\displaystyle \sec { A } +\tan { A }$
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$\displaystyle \sin { A }$
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$\displaystyle 1$
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$\displaystyle 0$

Explanation

The value of

$\displaystyle \frac { \tan { A } +\sec { A } -1 }{ \tan { A } -\sec { A } +1 }$ is equal to

$\displaystyle =\frac { \tan { A } +\sec { A } -\left( { \sec }^{ 2 }A-{ \tan }^{ 2 }A \right) }{ \tan { A } -\sec { A } +1 }$

$\displaystyle =\frac { \left( \tan { A } +\sec { A } \right) -\left( \sec { A } +\tan { A } \right) \left( \sec { A } -\tan { A } \right) }{ \tan { A } -\sec { A } +1 }$

$=\displaystyle \frac { \left( \tan { A } +\sec { A } \right) \left( 1-\sec { A } +\tan { A } \right) }{ \tan { A } -\sec { A } +1 }$

$\displaystyle =\tan { A } +\sec { A }$

$\displaystyle =\frac { \sin{ A } }{ \cos { A } } +\frac { 1 }{ \cos { A } } =\frac { \sin{ A }+1 }{ \cos { A } } =\frac { 1+\sin{ A } }{ \cos { A } }$

The value of

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

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$0$
0%
$1$
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$\displaystyle \tan { \theta }$
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$\displaystyle \cot { \theta }$

Explanation

The value of

$(\text{cosec} \theta -\sin \theta)(\sec \theta -\cos \theta)(\tan \theta +\cot \theta)$ is

$\displaystyle \left( \frac { 1 }{ \sin { \theta } } -\sin { \theta } \right) \left( \frac { 1 }{ \cos { \theta } } -\cos { \theta } \right) \left( \frac { \sin { \theta } }{ \cos { \theta } } +\frac { \cos { \theta } }{ \sin { \theta } } \right)$

$\displaystyle =\left( \frac { 1-{ \sin }^{ 2 }\theta }{ \sin { \theta } } \right) \left( \frac { 1-{ \cos }^{ 2 }\theta }{ \cos { \theta } } \right) \left( \frac { 1 }{ \sin { \theta } .\cos { \theta } } \right)$

$\displaystyle =\frac { { \cos }^{ 2 }\theta \times { \sin }^{ 2 }\theta }{ { \sin }^{ 2 }\theta \times { \cos }^{ 2 }\theta }$

$=1$

The value of

$\displaystyle \left( 1+\cot { A } -\text{cosec } A \right) \left( 1+\tan { A } +\sec{ A } \right)$ is

0%
0
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1
0%
2
0%
3

Explanation

$\displaystyle \left( 1+\frac { \cos { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } \right) \left( 1+\frac { \sin { A } }{ \cos { A } } +\frac { 1 }{ \cos { A } } \right)$

$\displaystyle =\left( \frac { \sin { A } +\cos { A } -1 }{ \sin { A } } \right) \left( \frac { \cos { A } +\sin { A } +1 }{ \cos { A } } \right)$

$\displaystyle =\frac { { \left( \sin { A } +\cos { A } \right) }^{ 2 }-1 }{ \sin { A } .\cos { A } }$

$\displaystyle =\frac { { \sin }^{ 2 }A+{ \cos }^{ 2 }A+2\sin { A } .\cos { A } -1 }{ \sin { A } .\cos { A } }$

$\displaystyle =\frac { { \sin }^{ 2 }A+{ \cos }^{ 2 }A-1 }{ \sin { A } .\cos { A } } =\frac { 2\sin { A } .\cos { A } }{ \sin { A } .\cos { A } } =2$

$\displaystyle \sin { x } +{ \sin }^{ 2 }x=1$ , then

$\displaystyle { \cos }^{ 2 }x+{ \cos }^{ 4 }x$ is :

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1
0%
2
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3
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4

Explanation

Given,

$\displaystyle \sin{ x }+{ \sin }^{ 2 }x=1$

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

$\Rightarrow \displaystyle \sin{ x }={ \cos }^{ 2 }x$
Then

$\displaystyle { \sin }^{ 2 }x={ \cos }^{ 4 }x$

$\Rightarrow \displaystyle 1-{ \cos }^{ 2 }x={ \cos }^{ 4 }x$
Therefore,

$\displaystyle { \cos }^{ 2 }x+{ \cos }^{ 4 }x=1$

$\displaystyle \frac { { \sin }^{ 2 }\theta }{ 7 } +\frac { { \cos }^{ 2 }\theta }{ 7 } =\frac { x }{ 21 }$ , then

$x$ is :

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$1$
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$2$
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$3$
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$4$

Explanation

$\displaystyle \frac { { \sin }^{ 2 }\theta }{ 7 } +\frac { { \cos }^{ 2 }\theta }{ 7 } =\frac { x }{ 21 }\Rightarrow \frac { { \sin }^{ 2 }\theta +\cos^2\theta }{ 7 }=\dfrac{x}{21}\Rightarrow \dfrac17=\dfrac{x}{21}\Rightarrow x=\dfrac{21}{7}=3$

Therefore, Answer is

$3$

$cot\displaystyle\frac{A}{2}-tan\displaystyle\frac{A}{2}$ will be equal to

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$tanA$
0%
$cot$
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$2cot$
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$2tanA$

Explanation

$\cot { \dfrac { A }{ 2 } } -\tan { \dfrac { A }{ 2 } }$

$\Rightarrow \dfrac { \cos { \dfrac { A }{ 2 } } }{ \sin { \dfrac { A }{ 2 } } } -\dfrac { \sin { \dfrac { A }{ 2 } } }{ \cos { \dfrac { A }{ 2 } } } =\dfrac { \cos ^{ 2 }{ \dfrac { A }{ 2 } } -\sin ^{ 2 }{ \dfrac { A }{ 2 } } }{ \sin { \dfrac { A }{ 2 } } \cos { \dfrac { A }{ 2 } } }$

$\Rightarrow \dfrac { \cos { A } }{ \sin { \dfrac { A }{ 2 } } \cos { \dfrac { A }{ 2 } } } \left[ \because \cos ^{ 2 }{ \dfrac { \theta }{ 2 } } -\sin ^{ 2 }{ \dfrac { \theta }{ 2 } } =\cos { \theta } \right]$

Multiply and divide by

$2,$ we get

$\Rightarrow \dfrac { 2\cos { A } }{ 2\sin { \dfrac { A }{ 2 } } \cos { \dfrac { A }{ 2 } } } =\dfrac { 2\cos { A } }{ \sin { A } } \left[ \because 2\sin { \dfrac { \theta }{ 2 } } \cos { \dfrac { \theta }{ 2 } } =\sin { \theta } \right]$

$=2\cot A$

Hence, the answer is

$2\cot A.$

The value of

$\displaystyle { \sin }^{ 2 }{ 20 }^{ o }+{ \sin }^{ 2 }{ 70 }^{ o }$ is :

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$1$
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$0$
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$2$
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$3$

Explanation

$\displaystyle {\sin}^{ 2 }{ 20 }^{ o }+{\sin}^{ 2 }{ 70 }^{ o }$

$={\sin}^{ 2 }{ 20 }^{ o }+{\sin}^{ 2 }\left( { 90 }^{ o }-{ 20 }^{ o } \right)$

$\displaystyle ={\sin}^{ 2 }{ 20 }^{ o }+{\cos}^{ 2 }{ 20 }^{ o }=1$

$\displaystyle \left[ \because \sin { \left( { 90 }^{ o }-\theta \right) =\cos { \theta } } \right]$

Hence, option A is correct.

$\theta$ is in the first quadrant and

$\cos\theta=\dfrac{3}{5}$ , then the value of

$\displaystyle\frac{5\tan\theta-4 cosec\theta}{5 sec\theta-4\cot\theta}$ will be

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$\displaystyle\frac{5}{16}$
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$\displaystyle\frac{5}{34}$
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$\displaystyle-\frac{5}{34}$
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$\displaystyle-\frac{5}{16}$

Explanation

$\because$ Since

$\theta$ is in the first quadrant, all trigonometrical ratios will be positive.

and

$\cos\theta=\dfrac{3}{5}$

then

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

$=\sqrt{1-\displaystyle\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

and

$\tan\theta=\dfrac{4}{3}$

$\therefore$ Given, Exp. =

$\displaystyle\dfrac{5\tan\theta-4cosec\theta}{5sec\theta-4\cot\theta}$

$=\displaystyle\dfrac{5\times \dfrac{4}{3}-4\times \dfrac{5}{4}}{5\times \displaystyle \dfrac{5}{3}-4\times \dfrac{3}{4}}$

$=\displaystyle\dfrac{5\left(\dfrac{4}{3}-1\right)}{\displaystyle\dfrac{25}{3}-3}$

$=\displaystyle\dfrac{\dfrac{5}{3}}{\dfrac{16}{3}}=\dfrac{5}{16}$

The value of

$\displaystyle { sin }^{ 2 }{ 60 }^{ o }+{ cos }^{ 2 }{ 60 }^{ o }$ is :

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$1$
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$-1$
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$0$
0%
None of these

Explanation

$\sin ^{ 2 }{ 60° } +\cos ^{ 2 }{ 60° }$

$\because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1$

$\therefore \sin ^{ 2 }{ 60° } +\cos ^{ 2 }{ 60° } =1$

Hence, the answer is

$1.$

$\displaystyle \tan \theta +\frac{1}{\tan \theta }= 2$

then the value of

$\displaystyle \tan ^{^{2}}\theta +\frac{1}{\tan ^{2}\theta }$ will be :

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$4$
0%
$2$
0%
$1$
0%
$8$

Explanation

$\tan { \theta } +\dfrac { 1 }{ \tan { \theta } } =2$

Squaring both sides, we get

$\Rightarrow { \left( \tan { \theta } +\dfrac { 1 }{ \tan { \theta } } \right) }^{ 2 }=4$

$\Rightarrow \tan ^{ 2 }{ \theta } +\dfrac { 1 }{ \tan ^{ 2 }{ \theta } } +2.\tan { \theta } .\dfrac { 1 }{ \tan { \theta } } =4$

$\Rightarrow \tan ^{ 2 }{ \theta } +\dfrac { 1 }{ \tan ^{ 2 }{ \theta } } +2=4$

$\Rightarrow \tan ^{ 2 }{ \theta } +\dfrac { 1 }{ \tan ^{ 2 }{ \theta } } =2$

Hence, the answer is

$2.$

$\displaystyle \sqrt { 3 } \tan { \theta } =3\sin { \theta }$ , then the value of

$\displaystyle { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta$ is :

0%
$\displaystyle \frac { 1 }{ 3 }$
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$\displaystyle \frac { 2 }{ 3 }$
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$\displaystyle \frac { 1 }{ 4 }$
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$\displaystyle \frac { 2 }{ 5 }$

Explanation

$\displaystyle \frac { \sqrt { 3 } \sin { \theta } }{ \cos { \theta } } =3\sin { \theta }$

$\displaystyle \sec { \theta } =\sqrt { 3 }$

$\therefore \displaystyle \cos { \theta } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow { cos }^{ 2 }\theta =\frac { 1 }{ 3 }$

$\displaystyle { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1-{ cos }^{ 2 }\theta -{ cos }^{ 2 }\theta =1-2{ cos }^{ 2 }\theta$ ...

$[\because sin^2\theta = 1-cos^2\theta]$

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

$\displaystyle 3\cot \theta -4 = 0$ , then the value of

$\displaystyle\frac{3\cot \theta +4\cos \theta }{3\sin \theta -2\cos }$ will be

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$\displaystyle \frac{4}{3}$
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$25$
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$1$
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$0$

Explanation

$\displaystyle 3\cot \theta -4=0$

$\displaystyle \cot \theta =\frac{4}{3}$
Now,

$\displaystyle \frac{3\sin \theta +4\cos \theta }{3\sin \theta -2\cos \theta }$

$\displaystyle = \frac{\sin \theta \left [ 3+4\frac{\cos \theta }{\sin \theta } \right ]}{\sin \theta \left [ 3-2\frac{\cos \theta }{\sin \theta } \right ]}$

$\displaystyle = \frac{3+4\cot \theta }{3-2\cot \theta }$

$\displaystyle = \frac{3+4\times \frac{4}{3}}{3-2\times \frac{4}{3}}$

$\displaystyle = \frac{3+\frac{16}{3}}{3-\frac{8}{3}}=\frac{\frac{25}{3}}{\frac{1}{3}}$
Hence,

$\displaystyle = \frac{3\sin \theta +4\cos \theta }{3\sin \theta -2\cos \theta }=25$

$\displaystyle \sin ^{6}+\sin ^{4}\Theta .\cos ^{2}\Theta -\sin ^{2}\Theta .\cos ^{4}\Theta$ is equal to

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$\displaystyle \sin ^{2}\Theta +\cos ^{2}\Theta$
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$\displaystyle \sin ^{8}\Theta -\cos ^{3}\Theta$
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$\displaystyle \sin ^{4}\Theta +\cos ^{4}\Theta$
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$\displaystyle \sin ^{2}\Theta -\cos ^{2}\Theta$

Explanation

$sin^6\theta+sin^4\theta.cos^2\theta-sin^2\theta.cos^4\theta$

$\Rightarrow sin^4\theta(sin^2\theta+cos^2\theta)-sin^2\theta.cos^4\theta$

$\Rightarrow sin^4\theta- sin^2\theta.cos^4\theta$

$Since\;\;\; sin^2\theta+cos^2\theta=1$

$\displaystyle 3\sin \theta +5\cos \theta =5$ , then find the value of

$\displaystyle \left ( 5\sin \theta -3\cos \theta \right )$ .

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$\displaystyle \left ( 3\sin \theta +5\cos \theta \right )^{2}+\left ( 5\sin \theta +3\cos \theta \right )^{2}$

$\displaystyle =9\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )+25\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$

$\displaystyle =34$

$\displaystyle \therefore \left ( 5\sin \theta -3\cos \theta \right )^{2}=34-25=9$

$\displaystyle \Rightarrow \left ( 5\sin \theta -3\cos \theta \right )=3$

$\displaystyle \frac{\tan x}{\dfrac{\sin ^{3}x}{\cos x}+\sin x\cos x}$

$=1$

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True
0%
False

Explanation

$\displaystyle \frac{\tan x}{\frac{\sin ^{3}x}{\cos x}+\sin x\cos x}=\frac{\tan x\times \cos x}{\sin ^{3}x+\sin x\cos ^{2}x}$

$\displaystyle =\frac{\frac{\sin x}{\cos x}\times \cos x}{\sin x\left ( \sin ^{2}x+\cos ^{2}x \right )}$

$\displaystyle =\frac{\sin x}{\sin x\times 1}=1$

A wheel makes 12 revolutions per hour The radians it turns through in 20 minutes is:

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$\displaystyle 8\pi ^{c}$
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$\displaystyle 16\pi ^{c}$
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$\displaystyle 24\pi ^{c}$
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$\displaystyle 32\pi ^{c}$

Explanation

As the wheel makes

$12$ revolutions per hour,
In

$20$ minutes, it will make

$12 \times \dfrac {20}{60} = 4$ revolutions.

In each revolution, the wheel covers

${360}^{0}$ or

${2\pi }^{c}$
So in

$4$ revolutions, it will cover

$4 \times {2\pi }^{c} = {8\pi }^{c}$

$\displaystyle \sin x+\sin ^{2}x=1$ , then which one of the following is true?

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$\displaystyle \cos x+\cos ^{2}x=1$
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$\displaystyle \cos x-\cos ^{2}x=1$
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$\displaystyle \cos ^{2}x+\cos ^{4}x=1$
0%
$\displaystyle \cos ^{4}x+\cos ^{3}x=1$

Explanation

$\displaystyle \sin x+\sin ^{2}x=1$

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

$\displaystyle \Rightarrow \sin ^{2}x=\cos ^{4}x$

$\displaystyle \Rightarrow 1-\cos ^{2}x=\cos ^{4}x$

$\displaystyle \Rightarrow 1=\cos ^{2}x+\cos ^{4}x$

$\displaystyle \frac{\pi ^{c}}{5}$ in sexagesimal measure is _____

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$\displaystyle 18^{\circ}$
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$\displaystyle 36^{\circ}$
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$\displaystyle 54^{\circ}$
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$\displaystyle 72^{\circ}$

Explanation

$\text{Sexagesimal System}$ , an angle is measured in degrees, minutes and seconds.

$\pi = {180}^{0}$

So,

$\dfrac {\pi}{5} = \dfrac {{180}^{0}}{5} = {36}^{0}$

$\displaystyle \frac{1-\sin A}{\cos A}$ is equal to

0%
$\displaystyle \frac{\cos A }{1+\sin A}$
0%
$\displaystyle \frac{\sin A }{1- \cos A}$
0%
$\displaystyle \frac{\tan A }{1 + \tan A}$
0%
$\displaystyle \frac{\tan A }{1 + \cos A}$

Explanation

$\displaystyle \frac{1-\sin A}{\cos A}=\frac{(1-\sin A)(1+\sin A)}{\cos A(1+\sin A)}$

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

$\displaystyle =\frac{\cos A}{1+\sin A}$

Evaluate:

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

0%
$1$
0%
$-1$
0%
$0$
0%
$-2$

Explanation

we know that,

$1+\tan ^2\theta=\sec ^2\theta$

Now,

$\tan ^2\theta-\sec ^2\theta=\tan ^2\theta-(1+\tan ^2\theta)=-1$

Therefore, Answer is

$-1$

$2x = \sec A$ and

$\dfrac {2}{x} = \tan A$ , then

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

0%
$\dfrac {1}{2}$
0%
$2$
0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{16}$

Explanation

Usiong

$\sec^{2}\theta - \tan^{2}\theta = 1$

$\Rightarrow (2x)^{2} - \left (\dfrac {2}{x}\right )^{2} = 1$ , substitute the given values in terms of

$x$

$\Rightarrow 4x^{2} - \dfrac {4}{x^{2}} = 1$

$\Rightarrow 4 \left (x^{2} - \dfrac {1}{x^{2}}\right ) = 1$

$\Rightarrow x^{2} - \dfrac {1}{x^{2}} = \dfrac {1}{4}$

For any acute angle

$\theta$ , find

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

0%
$1$
0%
$2$
0%
$3$
0%
$0$

Explanation

Consider right angled

$\triangle ABC$ , where

$\angle B = 90^{\circ}$ and

$\angle A = \theta$ .

Let

$AB = x, BC = y$ and

$AC = r$

So, by Pythagoras theorem,

$x^{2} + y^{2} = r^{2}$

$\sin^{2} \theta +\cos^{2} \theta = \left (\dfrac {y}{r}\right )^{2} + \left (\dfrac {x}{r}\right )^{2} = \dfrac {y^{2}}{r^{2}} + \dfrac {x^{2}}{r^{2}} = \dfrac {x^{2} + y^{2}}{r^{2}} = \dfrac {r^{2}}{r^{2}} = 1$

So,

$1$ is correct.

$\cot \theta = \dfrac {b}{\sqrt {a^{2} + b^{2}}}$ and

$0 < \theta < 90^{\circ}$ , then

$\sin \theta =$

0%
$\dfrac {a}{b}$
0%
$\dfrac {a}{\sqrt {a^{2} + b^{2}}}$
0%
$\dfrac {1}{\sqrt {a^{2} + b^{2}}}$
0%
$\dfrac {b}{\sqrt {a^{2} - b^{2}}}$

Explanation

$\cos \theta = \dfrac {b}{\sqrt {a^{2} + b^{2}}}$
By Pythagoras theorem,

$x^{2} + b^{2} =(\sqrt {a^{2} + b^{2}})^{2}$

$\therefore x^{2} + b^{2} = a^{2} + b^{2}$

$\Rightarrow x^{2} = a^{2}$

$\Rightarrow x = a$

$\therefore \sin \theta = \dfrac {a}{\sqrt {a^{2} + b^{2}}}$

$\cos{\theta}=0.5698$

$\therefore \theta=....$

0%
$45^{o}15'$
0%
$55^{o}20'$
0%
$55^{o}16'$
0%
$13^{o}15'$

Explanation

$\cos (x) = 0.5698 : \begin{bmatrix} Radians : & x= 0.96453 ... + 2 \pi n, x = 2\pi - 0.96453 ... + 2 \pi n \\ Degrees : & x = 55.2635^{\circ} + 360^{\circ} n , x = 360^{\circ} - 55.2635^{\circ} + 360^{\circ} n \end{bmatrix}$

correct option is B..
1521673_418220_ans_a0135dfefdaa4796af4de57f19854d8c.png

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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