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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 5 - MCQExams.com
CBSE
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
$$\displaystyle =\left( \sin { A } +\cos { A } \right) \left( \frac { \sin { A } }{ \cos { A } } +\frac { \cos { A } }{ \sin { A } } \right) $$
$$\displaystyle =\left( \sin { A } +\cos { A } \right) \left( \frac { { \sin }^{ 2 }A+{ \cos }^{ 2 }A }{ \cos { A } \sin { A } } \right) $$
$$=\displaystyle \frac { \sin { A } +\cos { A } }{ \cos { A } \sin { A } } $$
$$\displaystyle =\frac { \sin { A } }{ \cos { A } \sin { A } } +\frac { \cos { A } }{ \cos { A } \sin { A } } $$
$$\displaystyle =\frac { 1 }{ \cos { A } } +\frac { 1 }{ \sin { A } }$$
$$ =\sec { A } +cosecA$$
If $$\displaystyle \sin \theta =\cos (2\theta -45^{\circ}),\,\,\, 0 < (2\theta -45^{\circ})< 90^{\circ},$$ 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
$$\displaystyle =\frac { { \sin }^{ 2 }\theta { \left( 1+\cos { \theta } \right) }^{ 2 } }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } $$
$$\displaystyle =\frac { { \sin }^{ 2 }\theta +1+{ \cos }^{ 2 }\theta +2\cos\theta }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } $$
$$\displaystyle =\frac { 1+\left( { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta \right) +2\cos\theta }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } $$
$$\displaystyle =\frac { 1+1+2\cos { \theta } }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } =\frac { 2+2\cos { \theta } }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } $$
$$\displaystyle =\frac { 2\left( 1+\cos { \theta } \right) }{ \sin{ \theta \left( 1+\cos { \theta } \right) } } =\frac { 2 }{ \sin{ \theta } } =2\text{cosec}\theta $$
If $$\displaystyle \cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta } $$, 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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$$0$$
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$$1$$
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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
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0
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1
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2
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3
Explanation
$$\displaystyle \left( 1+\frac { \cos { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } \right) \left( 1+\frac { \sin { A } }{ \cos { A } } +\frac { 1 }{ \cos { A } } \right) $$
$$\displaystyle =\left( \frac { \sin { A } +\cos { A } -1 }{ \sin { A } } \right) \left( \frac { \cos { A } +\sin { A } +1 }{ \cos { A } } \right) $$
$$\displaystyle =\frac { { \left( \sin { A } +\cos { A } \right) }^{ 2 }-1 }{ \sin { A } .\cos { A } } $$
$$\displaystyle =\frac { { \sin }^{ 2 }A+{ \cos }^{ 2 }A+2\sin { A } .\cos { A } -1 }{ \sin { A } .\cos { A } } $$
$$\displaystyle =\frac { { \sin }^{ 2 }A+{ \cos }^{ 2 }A-1 }{ \sin { A } .\cos { A } } =\frac { 2\sin { A } .\cos { A } }{ \sin { A } .\cos { A } } =2$$
If $$\displaystyle \sin { x } +{ \sin }^{ 2 }x=1$$, then $$\displaystyle { \cos }^{ 2 }x+{ \cos }^{ 4 }x$$ is :
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1
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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, $$\displaystyle { \cos }^{ 2 }x+{ \cos }^{ 4 }x=1$$
If $$\displaystyle \frac { { \sin }^{ 2 }\theta }{ 7 } +\frac { { \cos }^{ 2 }\theta }{ 7 } =\frac { x }{ 21 } $$, 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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$$tanA$$
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$$cot$$
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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.
If $$\theta$$ 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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$$0$$
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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 $$1.$$
If $$\displaystyle \tan \theta +\frac{1}{\tan \theta }= 2$$
then the value of $$\displaystyle \tan ^{^{2}}\theta +\frac{1}{\tan ^{2}\theta }$$ will be :
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$$4$$
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$$2$$
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$$1$$
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$$8$$
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 $$2.$$
If $$\displaystyle \sqrt { 3 } \tan { \theta } =3\sin { \theta } $$, then the value of $$\displaystyle { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta $$ is :
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$$\displaystyle \frac { 1 }{ 3 } $$
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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
$$\displaystyle \frac { \sqrt { 3 } \sin { \theta } }{ \cos { \theta } } =3\sin { \theta } $$
$$\displaystyle \sec { \theta } =\sqrt { 3 } $$
$$\therefore \displaystyle \cos { \theta } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow { cos }^{ 2 }\theta =\frac { 1 }{ 3 } $$
$$\displaystyle { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1-{ cos }^{ 2 }\theta -{ cos }^{ 2 }\theta =1-2{ cos }^{ 2 }\theta $$
...$$[\because sin^2\theta = 1-cos^2\theta]$$
$$\displaystyle =1-2\left( \frac { 1 }{ 3 } \right) =1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } $$
If $$\displaystyle 3\cot \theta -4 = 0$$, 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
$$sin^6\theta+sin^4\theta.cos^2\theta-sin^2\theta.cos^4\theta$$
$$\Rightarrow sin^4\theta(sin^2\theta+cos^2\theta)-sin^2\theta.cos^4\theta$$
$$\Rightarrow sin^4\theta- sin^2\theta.cos^4\theta$$ $$Since\;\;\; sin^2\theta+cos^2\theta=1 $$
If $$\displaystyle 3\sin \theta +5\cos \theta =5$$, then find the value of $$\displaystyle \left ( 5\sin \theta -3\cos \theta \right )$$.
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$$4$$
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$$3$$
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$$2$$
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$$1$$
Explanation
$$\displaystyle \left ( 3\sin \theta +5\cos \theta \right )^{2}+\left ( 5\sin \theta +3\cos \theta \right )^{2}$$
$$\displaystyle =9\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )+25\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$$
$$\displaystyle =34$$
$$\displaystyle \therefore \left ( 5\sin \theta -3\cos \theta \right )^{2}=34-25=9$$
$$\displaystyle \Rightarrow \left ( 5\sin \theta -3\cos \theta \right )=3$$
Is $$\displaystyle \frac{\tan x}{\dfrac{\sin ^{3}x}{\cos x}+\sin x\cos x}$$ $$=1$$
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True
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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 $$ 4 \times {2\pi }^{c} = {8\pi }^{c}$$
If $$\displaystyle \sin x+\sin ^{2}x=1$$, 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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$$\displaystyle \cos ^{2}x+\cos ^{4}x=1$$
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$$\displaystyle \cos ^{4}x+\cos ^{3}x=1$$
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
In $$\text{Sexagesimal System}$$, 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
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$$ \displaystyle \frac{\cos A }{1+\sin A} $$
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$$ \displaystyle \frac{\sin A }{1- \cos A} $$
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$$ \displaystyle \frac{\tan A }{1 + \tan A} $$
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$$ \displaystyle \frac{\tan A }{1 + \cos A} $$
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: $$\tan^{2}\theta - \sec^{2}\theta$$
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$$1$$
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$$-1$$
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$$0$$
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$$-2$$
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 $$-1$$
If $$2x = \sec A$$ and $$\dfrac {2}{x} = \tan A$$, then $$x^{2} - \dfrac {1}{x^{2}} = $$
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$$\dfrac {1}{2}$$
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$$2$$
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$$\dfrac {1}{4}$$
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$$\dfrac {1}{16}$$
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 $$\sin^{2}\theta + \cos^{2}\theta$$
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$$1$$
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$$2$$
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$$3$$
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$$0$$
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.
If $$\cot \theta = \dfrac {b}{\sqrt {a^{2} + b^{2}}}$$ and $$0 < \theta < 90^{\circ}$$, then $$\sin \theta =$$
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$$\dfrac {a}{b}$$
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$$\dfrac {a}{\sqrt {a^{2} + b^{2}}}$$
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$$\dfrac {1}{\sqrt {a^{2} + b^{2}}}$$
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$$\dfrac {b}{\sqrt {a^{2} - b^{2}}}$$
Explanation
$$\cos \theta = \dfrac {b}{\sqrt {a^{2} + b^{2}}}$$
By Pythagoras theorem, $$x^{2} + b^{2} =(\sqrt {a^{2} + b^{2}})^{2}$$
$$\therefore x^{2} + b^{2} = a^{2} + b^{2} $$
$$\Rightarrow x^{2} = a^{2} $$
$$\Rightarrow x = a$$
$$\therefore \sin \theta = \dfrac {a}{\sqrt {a^{2} + b^{2}}}$$
$$\cos{\theta}=0.5698$$ $$\therefore \theta=....$$
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$$45^{o}15'$$
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$$55^{o}20'$$
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$$55^{o}16'$$
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$$13^{o}15'$$
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..
0:0:1
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