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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 7 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 7
Evaluate: $$\cfrac { \sin { \theta } \cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } }{ \cos { \left( { 90 }^{ o }-\theta \right) } } +\cfrac { \cos { \theta } \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } }{ \sin { \left( { 90 }^{ o }-\theta \right) } } +\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ { 63 }^{ o } } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ { 50 }^{ o } } } $$
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$$1$$
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$$2$$
0%
$$4$$
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$$3$$
Explanation
$$\cfrac { \sin { \theta } \cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } }{ \cos { \left( { 90 }^{ o }-\theta \right) } } =\cfrac { \sin { \theta } \cos { \theta } \cos { \theta } }{ \sin { \theta } } ....(i)$$
$$\cfrac { \cos { \theta } \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } }{ \sin { \left( { 90 }^{ o }-\theta \right) } } =\cfrac { \cos { \theta } \sin { \theta } \sin { \theta } }{ \cos { \theta } } ......(ii)\quad $$
$$\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ { 63 }^{ o } } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ { 50 }^{ o } } } $$
$$=\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ \left( { 90 }^{ o }-{ 27 }^{ o } \right) } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ \left( { 90 }^{ o }-{ 40 }^{ o } \right) } } $$
$$=\cfrac { \sin ^{ 2 }{ { 27 }^{ o }\cos ^{ 2 }{ { 27 }^{ o } } } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\sin ^{ 2 }{ { 40 }^{ o } } } =\cfrac { 1 }{ 1 } $$
Using $$(i), (ii), (iii)$$ we get
$$\sin { \theta } \sin { \theta } +\cos { \theta } \cos { \theta } +1$$
$$\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +1=1+1=2$$
If $$\cos (\alpha+\beta)=0$$, then $$\sin (\alpha-\beta)$$, can be reduced to
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$$\cos {\beta}$$
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$$\cos {2\beta}$$
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$$\cos {2\alpha}$$
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$$\sin {2\alpha}$$
Explanation
$$\textbf{Step 1 : Use the triginometric identities for angle transformation}$$
$$\text{Given ,}$$ $$\cos(\alpha +\beta )=0$$
$$\Rightarrow \cos (\alpha +\beta )=\cos90\quad \quad \quad \quad \boldsymbol{[\because \cos 90^\circ=0]}$$
$$\Rightarrow \alpha +\beta =90$$
$$\Rightarrow\alpha =90-\beta$$
$$\therefore\ sin (\alpha -\beta )\\ =\sin(90-2\beta )=\cos2\beta\ \ \ \quad \quad \quad \boldsymbol{[\because \sin(90-\theta )=\cos\theta]} $$
$$\textbf{Hence , option(B) is the correct answer}$$
Evaluate:
$$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sec ^{ 2 }{ { 50 }^{ o } } -\cot ^{ 2 }{ { 40 }^{ o } } } +2\text{cosec} ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { { 32 }^{ o } } -4\tan { { 13 }^{ o } } \tan { { 37 }^{ o } } \tan { { 45 }^{ o } } \tan { { 53 }^{ o } } \tan { { 77 }^{ o } } $$
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$$2$$
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$$-1$$
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$$-2$$
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$$3$$
Explanation
$$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sec ^{ 2 }{ { 50 }^{ o } } +\cot ^{ 2 }{ { 40 }^{ o } } } +2\csc ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { { 32 }^{ o } } -4\tan { { 13 }^{ o } } \tan { { 37 }^{ o } } \tan { { 45 }^{ o } } \tan { { 53 }^{ o } } \tan { { 77 }^{ o } } $$
$$=\cfrac { \cos ^{ 2 }{ { ({ 90 }^{ o }-70 }^{ o }) } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sec ^{ 2 }{ { ({ 90 }^{ o }-40 }^{ o } } )+\cot ^{ 2 }{ { 40 }^{ o } } }$$
$$+2\csc ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { ({ 90 }^{ o }-58^{ o }) } -4(\tan { { 13 }^{ o } } \tan { { 77 }^{ o } } )(\tan { { 37 }^{ o } } \tan { { 53 }^{ o } } )\tan { { 45 }^{ o } } $$
$$=\cfrac { \sin ^{ 2 }{ { 70 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ co\sec ^{ 2 }{ { 40 }^{ o } } -\cot ^{ 2 }{ { 40 }^{ o } } } +2(\csc ^{ 2 }{ { 58 }^{ o } } -\cot ^{ 2 }{ { 58 }^{ o } }) -4[\tan { ({ 90 }^{ o }-77^{ o }) } \tan { { 77 }^{ o } } ][\tan { ({ 90 }^{ o }-53^{ o }) } \tan { { 53 }^{ o } } ][1]$$
$$=\cfrac{1}{1}+2-4$$
$$=-1$$
Evaluate:
$$\cfrac { \sec ^{ 2 }{ { 54 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } } }{ co\sec ^{ 2 }{ { 57 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { 52 }^{ o } } -\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 } } \tan { { 17 }^{ o } } \tan { { 60 }^{ o } } \tan { { 73 }^{ o } } $$
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$$\cfrac{6}{5}$$
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$$\cfrac{7}{3}$$
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$$\cfrac{9}{2}$$
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$$\cfrac{1}{4}$$
Explanation
$$\cfrac { \sec ^{ 2 }{ { 54 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } } }{ co\sec ^{ 2 }{ { 57 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { 52 }^{ o } } -\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 } } \tan { { 17 }^{ o } } \tan { { 60 }^{ o } } \tan { { 73 }^{ o } } $$
$$\Rightarrow \cfrac { \sec ^{ 2 }{ { (90 }^{ o }- } { 36 }^{ o })-\cot ^{ 2 }{ { 36 }^{ o } } }{ co\sec ^{ 2 }{ { (90 }^{ o }- } { 33 }^{ o })-\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { (90 }^{ o }- } { 38 }^{ o })-\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 } } \tan { \left( { 90 }^{ o }-{ 73 }^{ o } \right) } \tan { { 73 }^{ o } } \tan { { 60 }^{ o } } $$
$$\Rightarrow \cfrac { co\sec ^{ 2 }{ { 36 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } } }{ \sec ^{ 2 }{ { 33 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } co\sec ^{ 2 }{ { 38 }^{ o } } -{ \left( \cfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+\cfrac { 2 }{ \sqrt { 3 } } \cot { { 73 }^{ o } } \tan { { 73 }^{ o } } \times \sqrt { 3 } $$
$$\Rightarrow \cfrac { 1 }{ 1 } +2\sin ^{ 2 }{ { 38 }^{ o } } \times \cfrac { 1 }{ \sin ^{ 2 }{ { 38 }^{ o } } } -\cfrac { 1 }{ 2 } +\cfrac { 2 }{ \sqrt { 3 } } \times \cfrac { 1 }{ \tan { { 73 }^{ o } } } \times \tan { { 73 }^{ o } } \times \sqrt { 3 } $$
$$\quad [\because \quad co\sec ^{ 2 }{ 0 } -\cot ^{ 2 }{ 0 } =1,\sec ^{ 2 }{ 0 } -\tan ^{ 2 }{ 0 } =1]$$
$$\Rightarrow 1+2-\cfrac{1}{2}+2=5-\cfrac{1}{2}$$
$$\Rightarrow \cfrac{9}{2}$$
$$\displaystyle \cos ^{2}5^{\circ}+\cos ^{2}10^{\circ}+\cos ^{2}15^{\circ}+...+\cos ^{2}85^{\circ}=$$
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$$6$$
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$$8$$
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$$9$$
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$$\displaystyle 8\frac{1}{2}$$
Explanation
$$\displaystyle \left ( \cos ^{2}5^{\circ}+\cos ^{2}85^{\circ} \right )+\left ( \cos ^{2}10+\cos ^{2}80^{\circ} \right )+\cdot \cdot \cdot +\cos ^{2}45^{\circ}+\cdot \cdot \cdot $$
=$$\displaystyle \left ( \cos ^{2} 5^{\circ}+\sin ^{2}5^{\circ}\right )+\left ( \cos ^{2}10^{\circ} \right )+\cdot \cdot \cdot +\cos ^{2}45^{\circ}+\cdot \cdot \cdot \left [ using\cos \left ( 90^{\circ}-\theta \right ) =\sin \theta \right ]$$
=$$\displaystyle 1+1+1+\left ( 8times \right )+\frac{1}{2}=8\frac{1}{2}$$
$$\displaystyle \tan ^{2}B-\sin^{2} B$$ is equal to
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$$\displaystyle \sec ^{2}B$$
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$$\displaystyle 1+\cos ^{2}B$$
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$$\displaystyle \tan ^{2}B\sin ^{2}B$$
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$$\displaystyle \sec ^{2}B-\sin ^{2}B$$
Explanation
$$\displaystyle \tan ^{2}B-\sin^{2} B = \frac {\sin^{2} B }{\cos^{2} B } - \sin^{2} B $$
$$\displaystyle = \frac {\sin^{2} B - \sin^{2} B \cos^{2} B }{\cos^{2} B } $$
$$\displaystyle = \sin^{2} B (\frac {1 - \cos^{2} B }{\cos^{2} B } )$$
$$\displaystyle = \sin^{2} B (\frac { \sin^{2} B }{\cos^{2} B } )$$
$$\displaystyle = \sin^{2} B \tan^{2} B $$
If $$\displaystyle \tan 32^0.\cot (90^0-\theta )=1$$ find $$\theta $$.
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$$\displaystyle 58^{\circ}$$
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$$\displaystyle 122^{\circ}$$
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$$\displaystyle 32^{\circ}$$
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$$\displaystyle 158^{\circ}$$
Explanation
Given, $$\tan 32^0.\cot (90^0-\theta )=1$$
$$\Rightarrow \tan 32^0=\dfrac1{\cot (90^0-\theta )}$$
$$\Rightarrow \tan 32^0=\tan (90^0-\theta )$$
$$\Rightarrow (90^0-\theta )=32^0$$
$$\Rightarrow \theta=90^0-32^0$$
$$=58^0$$
Option A is correct.
The value of $$\displaystyle \csc (65^0+\theta )-\sec (25^0-\theta )-\tan (55^0-\theta )+\cot (35^0+\theta )$$ is
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$$0$$
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$$-1$$
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$$\displaystyle \theta +90$$
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None of these
Explanation
Given, $$\displaystyle \csc (65^{\circ}+\theta )-\sec (25-\theta )-\tan (55-\theta )+\cot (35+\theta )$$
$$= \displaystyle \csc (90^{\circ}-25^{\circ}+\theta )-\sec (25^{\circ}-\theta )-\tan (55^{\circ}-\theta )+\cot (90^{\circ}-55^{\circ}+\theta )$$
$$= \displaystyle \csc (90^{\circ}-(25^{\circ}-\theta) )-\sec (25^{\circ}-\theta )-\tan (55^{\circ}-\theta )+\cot (90^{\circ}-(55^{\circ}-\theta) )$$
$$= \displaystyle \sec (25^{\circ}-\theta)-\sec (25^{\circ}-\theta )-\tan (55^{\circ}-\theta )+\tan (55^{\circ}-\theta )$$
$$=0$$
Find the value of
$$\displaystyle 4\left( { \sin }^{ 4 }{ 30 }^{ o }+{ \cos }^{ 4 }{ 60 }^{ o } \right) -3\left( { \sin }^{ 2 }{ 45 }^{ o }-2{ \cos }^{ 2 }{ 45 }^{ o } \right) $$
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$$1$$
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$$2$$
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$$0$$
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$$3$$
Explanation
$$4(sin^{4}30^{0}+cos^{4}60^{0})-3(sin^{2}45^{0}-2cos^{2}45^{0})$$
Put the value of $$\sin$$ and $$\cos$$ we get:
=$$\left [ \left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{4} \right ]-3\left [ \left ( \frac{1}{\sqrt{2}} \right )^{2}-2\left ( \frac{1}{\sqrt{2}} \right )^{2} \right ]$$
=$$4\left ( \frac{1}{16}+\frac{1}{16} \right )-3\left ( \frac{1}{2}-2(\frac{1}{2}) \right )$$
=$$\frac{1}{2}+\frac{3}{2}=2$$
If $$\displaystyle \sqrt{2}\cos A=1$$ then the value of $$\displaystyle \tan ^{4}A+\cot ^{4}A$$
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{3}$$
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2
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1
Explanation
Since, $$\displaystyle \sqrt{2}\cos A=1$$
$$\Rightarrow \displaystyle \cos A=\frac {1}{\sqrt{2}}$$
Since, $$\cos \theta = \frac {\text {base}}{\text {hypotenuse}} $$
Here, base = 1 and hypotenuse = $$\sqrt 2$$
using pythagoras
$$ H^{2} = P^{2} + B^{2}$$
$$ (\sqrt 2)^{2} = P^{2} + 1^{2}$$
$$ P= 1$$
$$\therefore \tan A = \frac {\text {perpendicular}}{\text {base}}= 1 $$
and
$$ \cot A = \frac {\text {base}}{\text {perpendicular}}= 1 $$
$$\therefore \displaystyle \tan ^{4}A+\cot ^{4}A = 1^{4} + 1^{4} = 2$$
Option C is correct.
The value of $$\displaystyle \frac{2\sin 67^{\circ}}{\cos 23^{\circ}}-\frac{\cot 40^{\circ}}{\tan 50^{\circ}}$$
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0
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1
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-1
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None
Explanation
$$\displaystyle \frac{2\sin 67^0}{\cos 23^0}-\frac{\cot 40^0}{\tan 50^0}=\displaystyle \frac{2\sin (90^0-23^0)}{\cos 23^0}-\frac{\cot (90^0-50^0)}{\tan50^0}$$
$$=\displaystyle \frac{2\cos 23^0}{\cos 23^0}-\frac{\tan50^0}{\tan 50^0}$$
$$=2-1=1$$
Option B is correct.
Evaluate: $$\displaystyle 3{\cot }^{ 2 }{ 60 }^{ o }+{ \sec }^{ 4 }{ 45 }^{ o }-{ \tan }^{ 2 }{ 60 }^{ o }$$
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
Given that: $$3\cot ^{ 2 }{ { 60^o } } +\sec ^{ 4 }{ { 45 ^o} } -\tan ^{ 2 }{ { 60^\circ } } $$
$$=3\cot ^{ 2 }{ \left( { 90 ^o- } { 30 ^o } \right) } +\sec ^{ 4 }{{ 45 ^\circ } } -\tan ^{ 2 }{ { 60^o } } $$
$$=3\times( \dfrac { 1 }{ \sqrt { 3 } } )^2-(\sqrt { 3 })^2 +{ \left( \sqrt { 2 } \right) }^{ 2 }$$
$$=1 -3+2$$
$$=0$$
The value of $$tan\;1^{\circ}\;tan\;2^{\circ}\;tan\;3^{\circ}....tan\;89^{\circ}$$ is
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$$1$$
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$$0$$
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$$\infty$$
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$$\displaystyle\frac{1}{2}$$
Explanation
$$tan\;1^{\circ}\;tan\;2^{\circ}\;tan\;3^{\circ}....tan\;45^{\circ}.....tan\;87^{\circ}\;tan\;88^{\circ}\;tan\;89^{\circ}$$
$$=tan\;1^{\circ}\;tan\;2^{\circ}\;tan\;3^{\circ}.....tan\;45^{\circ}.....cot\;3^{\circ}\;cot\;2^{\circ}\;cot\;1^{\circ}=1$$
The value of $$\displaystyle \tan { { 5 }^{ o } } .\tan { { 85 }^{ o } } .\tan { { 31 }^{ o } } .\tan { { 5 }9^{ o } } .\tan { { 45 }^{ o } } $$ is :
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$$0$$
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$$2$$
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$$1$$
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$$\displaystyle \frac { 1 }{ 2 } $$
Explanation
The value of $$\tan 5^o. \tan 85^o.\tan 31^o. \tan 59^o.\tan 45^o$$ is
$$=\tan (90-85)^o.\tan 85^o. \tan (90-59)^o. \tan 45^o$$
$$=\cot 85^o.\tan 85^o. \cot 59^o. \tan 59^o. \tan 45^o$$
$$=\tan 45^o$$
$$=1$$
If $$\displaystyle 5\sin { A } =3$$, then the value of $$\displaystyle { \sec }^{ 2 }A-{ \tan }^{ 2 }A$$ is :
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0
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5
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3
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1
Explanation
Given, $$\displaystyle \sin { A } =\frac { 3 }{ 5 } =\frac { p }{ h } $$
$$\displaystyle \therefore b=\sqrt { 25-9 } =\sqrt { 16 } =4$$
$$\displaystyle \sec { A } =\frac { h }{ b } =\frac { 5 }{ 4 } ,$$
$$\displaystyle \tan { A } =\frac { p }{ b } =\frac { 3 }{ 4 } $$
Therefore, $$\displaystyle { \sec }^{ 2 }A-{ \tan }^{ 2 }A={ \left( \frac { 5 }{ 4 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 4 } \right) }^{ 2 }$$
$$\displaystyle =\frac { 25 }{ 16 } -\frac { 9 }{ 16 } =\frac { 16 }{ 16 } =1$$
The value of $$\displaystyle { \text{cosec} }^{ 2 }\left( { 90 }^{ o }-\theta \right) -{ \tan }^{ 2 }\theta $$ is :
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$$2$$
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$$3$$
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$$0$$
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$$1$$
Explanation
The value of $$\text{cosec} ^2(90-\theta)-\tan ^2\theta$$ is
$$=\displaystyle { \sec }^{ 2 }\theta -{ \tan }^{ 2 }\theta $$
$$=\dfrac { 1 }{ { \cos }^{ 2 }\theta } -\dfrac { { \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =\dfrac { 1-{ \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta }$$
$$ =\dfrac { { \cos }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } $$
$$=1$$
Find the value of $$\displaystyle \cos { \left( { 90 }^{ o }-A \right) } \tan { \left( { 90 }^{ o }-A \right) } \sec { \left( { 90 }^{ o }-A \right) } $$
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$$\displaystyle \cot{ A }$$
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$$\displaystyle \tan { A } $$
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$$\displaystyle \cos { A } $$
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$$\displaystyle \text{cosec }A$$
Explanation
$$\displaystyle \cos { \left( { 90 }^{ o }-A \right) } .\tan { \left( { 90 }^{ o }-A \right) } \sec { \left( { 90 }^{ o }-A \right) } $$
$$=\displaystyle \sin { A } \times \cot{ A }\times \text{cosec} A$$ ..... $$[\because \cos(90^{o} - \theta)=\sin \theta, \tan(90^{o}-\theta)=\cot \theta, \sec(90^{o}-\theta)=\text{cosec} \theta]$$
$$=\sin A \times \dfrac{\cos A}{\sin A}\times \dfrac{1}{\sin A}$$
$$=\dfrac{\cos A}{\sin A}=\cot A$$
Hence, option A is correct.
The value of $$\displaystyle \frac { \sin { { 60 }^{ o } } }{ { \cos }^{ 2 }{ 45 }^{ o } } -\cot{ { 30 }^{ o } }+5\cos { { 90 }^{ o } } $$ is :
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$$0$$
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$$1$$
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$$2$$
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$$\displaystyle \frac { 1 }{ 2 } $$
Explanation
$$\displaystyle \frac { \sin { { 60 }^{ o } } }{ { \cos }^{ 2 }{ 45 }^{ o } } -\cot{ { 30 }^{ o } }+5\cos { { 90 }^{ o } } $$
$$=\displaystyle \frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 1 }{ 2 } } -\sqrt { 3 } +0=\sqrt { 3 } -\sqrt { 3 } =0$$
What is the value of $$\displaystyle { \sin }^{ 2 }{ 35 }^{ o }+{ \sin }^{ 2 }{ 55 }^{ o }$$ ?
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$$0$$
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$$1$$
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$$\displaystyle \frac { 1 }{ 2 } $$
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$$2$$
Explanation
The value of $$\displaystyle { \sin }^{ 2 }{ 35 }^{ o }+{ \sin }^{ 2 }\left( { 90 }^{ o }-{ 35 }^{ o } \right) $$
$$\displaystyle ={ \sin }^{ 2 }{ 35 }^{ o }+{ \cos }^{ 2 }{ 35 }^{ o }=1$$
$$\displaystyle \left[ \because { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right] $$
If $$\displaystyle \theta ={ 45 }^{ o }$$, then $$\displaystyle2 \sin { \theta } cos{ \theta }$$ is :
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$$0$$
0%
$$1$$
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$$2$$
0%
None of these
Explanation
Given, $$\theta=45^o$$
Therefore, value of
$$\displaystyle2 \sin { \theta } cos{ \theta }$$
is
$$=\displaystyle 2*\sin \left( { 45 }^{ o } \right) cos {45^0}$$
$$=2*(1/2)$$
$$=1$$
If $$\displaystyle \frac { x\text{ cosec }^{ 2 }{ 30 }^{ o }{ \sec }^{ 2 }{ 45 }^{ o } }{ 8{ \cos }^{ 2 }{ 45 }^{ o }{ \sin }^{ 2 }{ 90 }^{ o } } ={ \tan }^{ 2 }{ 60 }^{ o }-{ \tan }^{ 2 }{ 45 }^{ o }$$, then $$x$$ is :
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$$1$$
0%
$$-1$$
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$$2$$
0%
$$0$$
Explanation
$$\displaystyle \frac { { x\left( 2 \right) }^{ 2 }\times { \left( \sqrt { 2 } \right) }^{ 2 } }{ { 8\left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }\times { \sin }^{ 2 }\theta } ={ \left( \sqrt { 3 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 }$$
$$\displaystyle \frac { 8x }{ 8\times \dfrac { 1 }{ 2 } \times 1 } =3-1=2$$
$$\Rightarrow \displaystyle 2x=2$$
$$\Rightarrow \displaystyle x=1$$
Evaluate: $$\displaystyle \sin { { 40 }^{ o } } .\sec{ { 50 }^{ o } }-\cfrac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } } +1$$
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$$0$$
0%
$$1$$
0%
$$-1$$
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$$2$$
Explanation
$$\displaystyle \sin { { 40 }^{ o } } \times \sec { \left( { 90 }^{ o }-{ 40 }^{ o } \right) -\frac { \tan { { 40 }^{ o } } }{ \cot { \left( { 90 }^{ o }-{ 40 }^{ o } \right) } } } +1$$
$$\displaystyle =\sin { { 40 }^{ o } } \times cosec{ 40 }^{ o }-\frac { \tan { { 40 }^{ o } } }{ \tan { { 40 }^{ o } } } +1$$
$$\displaystyle =1-1+1=1$$
The value of $$\displaystyle { \cos }^{ 2 }\left( { 90 }^{ o }-\theta \right) +{ \cos }^{ 2 }\theta $$ is :
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3
0%
0
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2
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1
Explanation
The value of $$\cos^2(90^o-\theta)+\cos^2\theta$$ is
$$=\sin^2\theta+\cos^2\theta$$
$$=1$$
The value of tan $$1^{\circ}$$ tan $$2^{\circ}$$ tan $$3^{\circ}$$$$\times..........\times$$tan $$89^{\circ}$$ is
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$$0$$
0%
$$1$$
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$$2$$
0%
$$\dfrac {1}{2}$$
Explanation
$$ \tan 1^{\circ} \tan 2^{\circ} \tan3^{\circ}\times..........\times \tan 89^{\circ}$$
$$= (\tan 1^{\circ} \tan89^{\circ})( \tan 2^{\circ} \tan 88^{\circ}) \times....\times \tan 45^{\circ}$$
$$= (\tan 1^{\circ} \cot1^{\circ})( \tan 2^{\circ} \cot2^{\circ})\times ...\times \tan45^{\circ}$$ $$[\because \tan \theta=\cot (90^{\circ}-\theta)$$ and $$\tan \theta \times \cot \theta=1 ]$$
$$=1$$
The value of $$\displaystyle \frac { \cos { { 75 }^{ o } } }{ \sin { { 15 }^{ o } } } +\frac { \sin { { 12 }^{ o } } }{ \cos { { 78 }^{ o } } } -\frac { \cos { { 18 }^{ o } } }{ \sin { { 72 }^{ o } } } $$ is :
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0
0%
2
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3
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1
Explanation
we have, $$\displaystyle \cos { { 75 }^{ o } } =\cos { \left( { 90 }^{ o }-{ 15 }^{ o } \right) = } \sin { { 15 }^{ o } } $$
$$\displaystyle \sin { { 12 }^{ o } } =\sin { \left( { 90 }^{ o }-{ 78 }^{ o } \right) } =\cos { { 78 }^{ o } } $$
$$\displaystyle \cos { { 18 }^{ o } } =\cos { \left( { 90 }^{ o }-{ 72 }^{ o } \right) } =\sin { { 72 }^{ o } } $$
$$\displaystyle \therefore \quad \frac { \cos { { 75 }^{ o } } }{ \sin { { 15 }^{ o } } } +\frac { \sin { { 12 }^{ o } } }{ \cos { { 78 }^{ o } } } -\frac { \cos { { 18 }^{ o } } }{ \sin { { 72 }^{ o } } } $$
$$\displaystyle =\frac { \sin { { 15 }^{ o } } }{ \sin { { 15 }^{ o } } } +\frac { \cos { { 78 }^{ o } } }{ \cos { { 78 }^{ o } } } -\frac { \sin { { 72 }^{ o } } }{ \sin { { 72 }^{ o } } } $$
$$\displaystyle =1+1-1=1$$
Hence, option D is correct.
The value of $$\displaystyle \frac { \cos { { 70 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \cos { { 59 }^{ o } } }{ \sin { { 31 }^{ o } } } -8{ \sin }^{ 2 }{ 30 }^{ o }$$ is :
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$$1$$
0%
$$2$$
0%
$$0$$
0%
$$3$$
Explanation
The value of $$\displaystyle \frac { \cos { { 70 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \cos { { 59 }^{ o } } }{ \sin { { 31 }^{ o } } } -8{ \sin }^{ 2 }{ 30 }^{ o }$$ is
$$\displaystyle =\frac { \cos { \left( { 90 }^{ o }-{ 20 }^{ o } \right) } }{ \sin { { 20 }^{ o } } } +\frac { \cos { \left( { 90 }^{ o }-{ 31 }^{ o } \right) } }{ \sin { { 31 }^{ o } } } -{ 8\left( \frac { 1 }{ 2 } \right) }^{ 2 }$$
$$\displaystyle =\frac { \sin { { 20 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \sin { { 31 }^{ o } } }{ \sin { { 31 }^{ o } } } -8\times \frac { 1 }{ 4 } $$
$$\displaystyle =1+1-2=0$$
If $$\displaystyle \sin \left ( A+B \right ) =\frac{\sqrt{3}}{2}$$ and $$\displaystyle \cot \left ( A-B \right )=1$$, then find $$A$$
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$$\displaystyle 27\frac{1^{\circ}}{2}$$
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$$\displaystyle 35\frac{1^{\circ}}{2}$$
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$$\displaystyle 52\frac{1^{\circ}}{2}$$
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$$\displaystyle 55\frac{1^{\circ}}{2}$$
Explanation
Given, $$ \sin(A+B) = \frac {\sqrt{3}}{2} $$
$$ \Rightarrow A +B = 60^o $$ -(1)
And $$ \cot(A-B) = 1 $$
$$ \Rightarrow A-B = 45^o $$---- (2)
On ddding equations (1) and (2), we get
$$ \Rightarrow 2A = 105^0 $$
$$ \Rightarrow A = 52\dfrac{1}{2}^0 $$
The value of $$\displaystyle \frac { \sin { { 70 }^{ o } } }{ \cos { { 20 }^{ o } } } +\frac { \text{cosec }{ 20 }^{ o } }{ \sec { { 70 }^{ o } } } -2\cos { { 70 }^{ o } } \text{cosec }{ 20 }^{ o }$$ is :
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$$1$$
0%
$$2$$
0%
$$0$$
0%
$$3$$
Explanation
The value of $$\displaystyle \frac { \sin { { 70 }^{ o } } }{ \cos { { 20 }^{ o } } } +\frac { cosec{ 20 }^{ o } }{ \sec { { 70 }^{ o } } } -2\cos { { 70 }^{ o } } cosec{ 20 }^{ o }$$ is
$$\displaystyle =\frac { \sin { { 70 }^{ o } } }{ \cos { \left( { 90 }^{ o }-{ 70 }^{ o } \right) } } +\frac { cosec{ 20 }^{ o } }{ \sec { \left( { 90 }^{ o }-{ 20 }^{ o } \right) } } -\frac { 2\cos { { 70 }^{ o } } }{ \sin { { 20 }^{ o } } } $$
$$\displaystyle =\frac { \sin { { 70 }^{ o } } }{ \sin { { 70 }^{ o } } } +\frac { cosec{ 20 }^{ o } }{ cosec{ 20 }^{ o } } -\frac { 2\cos { { 70 }^{ o } } }{ \sin { \left( { 90 }^{ o }-{ 70 }^{ o } \right) } } $$
$$\displaystyle \left[ \because \quad \cos { \left( { 90 }^{ o }-\theta \right) } \sin { \theta } and\quad \sec { \left( { 90 }^{ o }-\theta \right) =cosec\theta } \right] $$
$$\displaystyle =1+1-\frac { 2\cos { { 70 }^{ o } } }{ \cos { { 70 }^{ o } } } =2-2=0$$
The value of $$\displaystyle \frac { 2\cos { { 67 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } } $$ is :
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Explanation
$$\displaystyle \frac { 2\cos { { 67 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } }$$
$$\displaystyle =\frac { 2\cos { \left( { 90 }^{ o }-{ 23 }^{ o } \right) } }{ \sin { { 23 }^{ o } } } -\frac { \tan { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } }{ \cot { { 50 }^{ o } } } $$
$$\displaystyle =\frac { 2\sin { { 23 }^{ o } } }{ \sin { { 23 }^{ o } } } -\frac { \cot { { 50 }^{ o } } }{ \cot { { 50 }^{ o } } }$$
$$\displaystyle \left[ \because \quad \cos { \left( { 90 }^{ o }-\theta \right) } =\sin { \theta } ,\tan { \left( { 90 }^{ o }-\theta \right) =\cot { \theta } } \right] $$
$$\displaystyle =2-1=1$$
The value of $$\displaystyle \sec { { 41 }^{ o } } \sin { { 49 }^{ o }+ } \cos { { 49 }^{ o } } \text{cosec }{ 41 }^{ o }$$ is :
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Explanation
$$\displaystyle \sec { { 41 }^{ o } } \sin { { 49 }^{ o }+ } \cos { { 49 }^{ o } } cosec{ 41 }^{ o }$$
$$\displaystyle =\frac { \sin { { 49 }^{ o } } }{ \cos { { 41 }^{ o } } } +\frac { \cos { { 49 }^{ o } } }{ \sin { { 41 }^{ o } } } $$
$$\displaystyle =\frac { \sin { { 49 }^{ o } } }{ \cos { \left( { 90 }^{ o }-{ 49 }^{ o } \right) } } +\frac { \cos { { 49 }^{ o } } }{ \sin { \left( { 90 }^{ o }-{ 49 }^{ o } \right) } } $$
$$\displaystyle =\frac { \sin { { 49 }^{ o } } }{ \sin { { 49 }^{ o } } } +\frac { \cos { { 49 }^{ o } } }{ \cos { { 49 }^{ o } } } $$
$$\displaystyle =1+1=2$$
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Practice Class 10 Maths Quiz Questions and Answers
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