MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 7

Evaluate:

$\cfrac { \sin { \theta } \cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } }{ \cos { \left( { 90 }^{ o }-\theta \right) } } +\cfrac { \cos { \theta } \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } }{ \sin { \left( { 90 }^{ o }-\theta \right) } } +\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ { 63 }^{ o } } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ { 50 }^{ o } } }$

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

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

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

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

$\cos (\alpha+\beta)=0$ , then

$\sin (\alpha-\beta)$ , can be reduced to

0%
$\cos {\beta}$
0%
$\cos {2\beta}$
0%
$\cos {2\alpha}$
0%
$\sin {2\alpha}$

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:

$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sec ^{ 2 }{ { 50 }^{ o } } -\cot ^{ 2 }{ { 40 }^{ o } } } +2\text{cosec} ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { { 32 }^{ o } } -4\tan { { 13 }^{ o } } \tan { { 37 }^{ o } } \tan { { 45 }^{ o } } \tan { { 53 }^{ o } } \tan { { 77 }^{ o } }$

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

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:

$\cfrac { \sec ^{ 2 }{ { 54 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } } }{ co\sec ^{ 2 }{ { 57 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { 52 }^{ o } } -\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 } } \tan { { 17 }^{ o } } \tan { { 60 }^{ o } } \tan { { 73 }^{ o } }$

0%
$\cfrac{6}{5}$
0%
$\cfrac{7}{3}$
0%
$\cfrac{9}{2}$
0%
$\cfrac{1}{4}$

Explanation

$\Rightarrow \cfrac { \sec ^{ 2 }{ { (90 }^{ o }- } { 36 }^{ o })-\cot ^{ 2 }{ { 36 }^{ o } } }{ co\sec ^{ 2 }{ { (90 }^{ o }- } { 33 }^{ o })-\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { (90 }^{ o }- } { 38 }^{ o })-\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 } } \tan { \left( { 90 }^{ o }-{ 73 }^{ o } \right) } \tan { { 73 }^{ o } } \tan { { 60 }^{ o } }$

$\Rightarrow \cfrac { co\sec ^{ 2 }{ { 36 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } } }{ \sec ^{ 2 }{ { 33 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } } } +2\sin ^{ 2 }{ { 38 }^{ o } } co\sec ^{ 2 }{ { 38 }^{ o } } -{ \left( \cfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+\cfrac { 2 }{ \sqrt { 3 } } \cot { { 73 }^{ o } } \tan { { 73 }^{ o } } \times \sqrt { 3 }$

$\Rightarrow \cfrac { 1 }{ 1 } +2\sin ^{ 2 }{ { 38 }^{ o } } \times \cfrac { 1 }{ \sin ^{ 2 }{ { 38 }^{ o } } } -\cfrac { 1 }{ 2 } +\cfrac { 2 }{ \sqrt { 3 } } \times \cfrac { 1 }{ \tan { { 73 }^{ o } } } \times \tan { { 73 }^{ o } } \times \sqrt { 3 }$

$\quad [\because \quad co\sec ^{ 2 }{ 0 } -\cot ^{ 2 }{ 0 } =1,\sec ^{ 2 }{ 0 } -\tan ^{ 2 }{ 0 } =1]$

$\Rightarrow 1+2-\cfrac{1}{2}+2=5-\cfrac{1}{2}$

$\Rightarrow \cfrac{9}{2}$

$\displaystyle \cos ^{2}5^{\circ}+\cos ^{2}10^{\circ}+\cos ^{2}15^{\circ}+...+\cos ^{2}85^{\circ}=$

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$6$
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$8$
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$9$
0%
$\displaystyle 8\frac{1}{2}$

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

$\displaystyle \tan ^{2}B-\sin^{2} B = \frac {\sin^{2} B }{\cos^{2} B } - \sin^{2} B$

$\displaystyle = \frac {\sin^{2} B - \sin^{2} B \cos^{2} B }{\cos^{2} B }$

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

$\displaystyle = \sin^{2} B (\frac { \sin^{2} B }{\cos^{2} B } )$

$\displaystyle = \sin^{2} B \tan^{2} B$

$\displaystyle \tan 32^0.\cot (90^0-\theta )=1$ 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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$\displaystyle \theta +90$
0%
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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$1$
0%
$2$
0%
$0$
0%
$3$

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:

$\left [ \left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{4} \right ]-3\left [ \left ( \frac{1}{\sqrt{2}} \right )^{2}-2\left ( \frac{1}{\sqrt{2}} \right )^{2} \right ]$

$4\left ( \frac{1}{16}+\frac{1}{16} \right )-3\left ( \frac{1}{2}-2(\frac{1}{2}) \right )$

$\frac{1}{2}+\frac{3}{2}=2$

$\displaystyle \sqrt{2}\cos A=1$ then the value of

$\displaystyle \tan ^{4}A+\cot ^{4}A$

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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{3}$
0%
2
0%
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
0%
1
0%
-1
0%
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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$0$
0%
$1$
0%
$2$
0%
$3$

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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$1$
0%
$0$
0%
$\infty$
0%
$\displaystyle\frac{1}{2}$

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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$0$
0%
$2$
0%
$1$
0%
$\displaystyle \frac { 1 }{ 2 }$

Explanation

The value of

$\tan 5^o. \tan 85^o.\tan 31^o. \tan 59^o.\tan 45^o$ is

$=\tan (90-85)^o.\tan 85^o. \tan (90-59)^o. \tan 45^o$

$=\cot 85^o.\tan 85^o. \cot 59^o. \tan 59^o. \tan 45^o$

$=\tan 45^o$

$=1$

$\displaystyle 5\sin { A } =3$ , then the value of

$\displaystyle { \sec }^{ 2 }A-{ \tan }^{ 2 }A$ is :

0%
0
0%
5
0%
3
0%
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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$3$
0%
$0$
0%
$1$

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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$\displaystyle \cot{ A }$
0%
$\displaystyle \tan { A }$
0%
$\displaystyle \cos { A }$
0%
$\displaystyle \text{cosec }A$

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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$0$
0%
$1$
0%
$2$
0%
$\displaystyle \frac { 1 }{ 2 }$

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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$0$
0%
$1$
0%
$\displaystyle \frac { 1 }{ 2 }$
0%
$2$

Explanation

The value of

$\displaystyle { \sin }^{ 2 }{ 35 }^{ o }+{ \sin }^{ 2 }\left( { 90 }^{ o }-{ 35 }^{ o } \right)$

$\displaystyle ={ \sin }^{ 2 }{ 35 }^{ o }+{ \cos }^{ 2 }{ 35 }^{ o }=1$

$\displaystyle \left[ \because { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right]$

$\displaystyle \theta ={ 45 }^{ o }$ , then

$\displaystyle2 \sin { \theta } cos{ \theta }$ is :

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$0$
0%
$1$
0%
$2$
0%
None of these

Explanation

Given,

$\theta=45^o$

Therefore, value of

$\displaystyle2 \sin { \theta } cos{ \theta }$ is

$=\displaystyle 2*\sin \left( { 45 }^{ o } \right) cos {45^0}$

$=2*(1/2)$

$=1$

$\displaystyle \frac { x\text{ cosec }^{ 2 }{ 30 }^{ o }{ \sec }^{ 2 }{ 45 }^{ o } }{ 8{ \cos }^{ 2 }{ 45 }^{ o }{ \sin }^{ 2 }{ 90 }^{ o } } ={ \tan }^{ 2 }{ 60 }^{ o }-{ \tan }^{ 2 }{ 45 }^{ o }$ , then

$x$ is :

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

$\displaystyle \sin { { 40 }^{ o } } .\sec{ { 50 }^{ o } }-\cfrac { \tan { { 40 }^{ o } } }{ \cot { { 50 }^{ o } } } +1$

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

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 :

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

0%
0
0%
2
0%
3
0%
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

$\displaystyle =\frac { \cos { \left( { 90 }^{ o }-{ 20 }^{ o } \right) } }{ \sin { { 20 }^{ o } } } +\frac { \cos { \left( { 90 }^{ o }-{ 31 }^{ o } \right) } }{ \sin { { 31 }^{ o } } } -{ 8\left( \frac { 1 }{ 2 } \right) }^{ 2 }$

$\displaystyle =\frac { \sin { { 20 }^{ o } } }{ \sin { { 20 }^{ o } } } +\frac { \sin { { 31 }^{ o } } }{ \sin { { 31 }^{ o } } } -8\times \frac { 1 }{ 4 }$

$\displaystyle =1+1-2=0$

$\displaystyle \sin \left ( A+B \right ) =\frac{\sqrt{3}}{2}$ and

$\displaystyle \cot \left ( A-B \right )=1$ , then find

$A$

0%
$\displaystyle 27\frac{1^{\circ}}{2}$
0%
$\displaystyle 35\frac{1^{\circ}}{2}$
0%
$\displaystyle 52\frac{1^{\circ}}{2}$
0%
$\displaystyle 55\frac{1^{\circ}}{2}$

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 :

0%
$1$
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 :

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

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 :

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

Explanation

$\displaystyle \sec { { 41 }^{ o } } \sin { { 49 }^{ o }+ } \cos { { 49 }^{ o } } cosec{ 41 }^{ o }$

$\displaystyle =\frac { \sin { { 49 }^{ o } } }{ \cos { { 41 }^{ o } } } +\frac { \cos { { 49 }^{ o } } }{ \sin { { 41 }^{ o } } }$

$\displaystyle =\frac { \sin { { 49 }^{ o } } }{ \cos { \left( { 90 }^{ o }-{ 49 }^{ o } \right) } } +\frac { \cos { { 49 }^{ o } } }{ \sin { \left( { 90 }^{ o }-{ 49 }^{ o } \right) } }$

$\displaystyle =\frac { \sin { { 49 }^{ o } } }{ \sin { { 49 }^{ o } } } +\frac { \cos { { 49 }^{ o } } }{ \cos { { 49 }^{ o } } }$

$\displaystyle =1+1=2$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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