Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The value of $\displaystyle \cot 15^{\circ}\cot 16^{\circ}\cot 17^{\circ}.....\cot 73^{\circ}\cot 74^{\circ}\cot 75^{\circ}$ is

A
$\displaystyle \frac{1}{2}$

B
$0$

C
$1$

D
$-1$

Solution

$\cot { 15° } \cot { 16° } \cot { 17° } .........\cot { 73° } \cot { 74° } \cot { 75° }$
$=\cot { \left( 90°-75° \right) } \cot { \left( 90°-74° \right) } \cot { \left( 90°-73° \right) } ......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° }$
$=\tan { 75° } \tan { 74° } \tan { 73° } .......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° }$
$=(\tan { 75° }.\cot75^o) (\tan { 74° }.\cot74^o)( \tan { 73° }.\cot73^o) .......\cot45^o$
$=1\times1\times1\times.........\times1$
$=1$
Hence, the answer is $1.$
Question 2
1 / -0

Evaluate : $\displaystyle \frac{3\cos 53^{\circ} \text{cosec}37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$

A
1

B
3

C
6

D
0

Solution

$\displaystyle \frac{3\cos 53^{\circ} cosec 37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$

= $\displaystyle \frac{3\cos 53^{\circ} \text{cosec }(90 - 53)^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}(90 - 29)^{\circ})}-3\tan ^{2}45^{\circ}$

= $\displaystyle \frac{3\cos 53^{\circ} \sec 53^{\circ}}{(\cos ^{2}29^{\circ}+\sin ^{2}29^{\circ})}-3\tan ^{2}45^{\circ}$

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

= $0$
Question 3
1 / -0

Evaluate: $\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$

A
$\displaystyle \sqrt{3}$

B
$2$

C
$1$

D
$\displaystyle \sqrt{2}$

Solution

$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$
$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\tan (90^{\circ}-23^{\circ}).\tan (90^{\circ}-7^{\circ})$ $\displaystyle \left [ \because \tan (90^{\circ}-\theta )=\cot \theta \right ]$
= $\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\cot 23^{\circ}.\cot 7^{\circ}$
$\displaystyle \tan 7^{\circ}.\cot 7^{\circ}\tan 23^{\circ}.\cot 23^{\circ}.\sqrt{3}$ $\displaystyle (\because \tan \theta .\cot \theta =1)$
$\displaystyle =1\times 1\times \sqrt{3}=\sqrt{3}$
Question 4
1 / -0

Let S = $\displaystyle \sin ^{2}30^{\circ}+\sin ^{2}45^{\circ}+\sin ^{2}60^{\circ}$ and
P = $\displaystyle \text{cosec}^{2}45^{\circ}.\sec ^{2}30^{\circ}.\sin ^{3}90^{\circ}.\cos 60^{\circ}$ , then the correct statement is

A
$S < P$

B
$S = P$

C
$S\cdot P = 2$

D
$S + P > 3$

Solution

$S = \displaystyle \sin ^{2}30^{\circ}+\sin ^{2}45^{\circ}+\sin ^{2}60^{\circ}$ and $P = \displaystyle cosec^{2}45^{\circ}.\sec ^{2}30^{\circ}.\sin ^{3}90^{\circ}.\cos 60^{\circ}$

Now, $S = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{3}{4} = \dfrac{1 + 2 + 3}{4} = \dfrac32$

$P = (\sqrt 2)^2 \cdot \left(\dfrac{2}{\sqrt3}\right)^2 \cdot 1^3 \cdot \dfrac{1}{2} = \dfrac{4}{3}$

Hence, $SP = \dfrac{3}{2} \cdot \dfrac{4}{3} = 2$
Question 5
1 / -0

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

A
$1$

B
$2$

C
$4$

D
$3$

Solution

$\cfrac { \sin { \theta } \cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } }{ \cos { \left( { 90 }^{ o }-\theta \right) } } =\cfrac { \sin { \theta } \cos { \theta } \cos { \theta } }{ \sin { \theta } } ....(i)$
$\cfrac { \cos { \theta } \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } }{ \sin { \left( { 90 }^{ o }-\theta \right) } } =\cfrac { \cos { \theta } \sin { \theta } \sin { \theta } }{ \cos { \theta } } ......(ii)\quad$
$\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ { 63 }^{ o } } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ { 50 }^{ o } } }$
$=\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ \left( { 90 }^{ o }-{ 27 }^{ o } \right) } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ \left( { 90 }^{ o }-{ 40 }^{ o } \right) } }$
$=\cfrac { \sin ^{ 2 }{ { 27 }^{ o }\cos ^{ 2 }{ { 27 }^{ o } } } }{ \cos ^{ 2 }{ { 40 }^{ o } } +\sin ^{ 2 }{ { 40 }^{ o } } } =\cfrac { 1 }{ 1 }$
Using $(i), (ii), (iii)$ we get
$\sin { \theta } \sin { \theta } +\cos { \theta } \cos { \theta } +1$
$\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +1=1+1=2$
Question 6
1 / -0

If $\cos (\alpha+\beta)=0$ , then $\sin (\alpha-\beta)$ , can be reduced to

A
$\cos {\beta}$

B
$\cos {2\beta}$

C
$\cos {2\alpha}$

D
$\sin {2\alpha}$

Solution

$\textbf{Step 1 : Use the triginometric identities for angle transformation}$
$\text{Given ,}$ $\cos(\alpha +\beta )=0$
$\Rightarrow \cos (\alpha +\beta )=\cos90\quad \quad \quad \quad \boldsymbol{[\because \cos 90^\circ=0]}$
$\Rightarrow \alpha +\beta =90$
$\Rightarrow\alpha =90-\beta$
$$\therefore\ sin (\alpha -\beta )\\ =\sin(90-2\beta )=\cos2\beta\ \ \ \quad \quad \quad \boldsymbol{[\because \sin(90-\theta )=\cos\theta]} $$

$\textbf{Hence , option(B) is the correct answer}$
Question 7
1 / -0

Evaluate: $\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$

A
$0$

B
$1$

C
$2$

D
$3$

Solution

$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$
$=\sin { \left[ { 90 }^{ o }-{ 40 }^{ o }+\theta \right] } -\cos { \left( { 40 }^{ o }-\theta \right) } +(\tan { { 1 }^{ o } } \tan {{89}^{o}})(\tan{{10}^{o}}\tan{{80}^{o}})(\tan {{20}^{o}}\tan {{70}^{o}})$
$=\sin {\left[{90}^{o}-({40}^{o}-\theta) \right]} -\cos {\left({40}^{o}-\theta \right)} +\tan {({90}^{o}} -{ 89 }^{ o })\tan { { 89 }^{ o } } \tan { { (90 }^{ o } } -{ 80 }^{ o })\tan { { 80}^{ o } } \tan { { (90 }^{ o } } -{ 70 }^{ o })\tan { { 70 }^{ o } }$
$=\cos { \left( { 40 }^{ o }-\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\cot { { 89 }^{ o } } \tan { { 89 }^{ o } } \cot { { 80 }^{ o } } \tan { { 80 }^{ o } } \cot { { 70 }^{ o } } \tan { { 70 }^{ o } }$
$=0+1$ $\because \cot {\theta} \tan {\theta}=1$ ]
$=1$
Hence, $\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}=1$
Question 8
1 / -0

The value of $\displaystyle \cos ^{4}\theta +\sin ^{4}\theta +2\cos ^{2}\theta \sin ^{2}\theta$ when $\displaystyle \theta=60^{\circ}$ is

A
1

B
2

C
$\displaystyle \frac{1}{\sqrt{2}}$

D
$\displaystyle 2\sqrt{2}$

Solution

$\displaystyle \cos ^{4}\theta +\sin ^{4}\theta +2\cos ^{2}\theta \sin ^{2}\theta$
= $\displaystyle (\cos ^{2}\theta +\sin ^{2}\theta)^2$
= $\displaystyle (\cos ^{2} 60^0 +\sin ^{2} 60^0)^2$
= $\displaystyle (\frac{1}{4} +\frac{3}{4})^2$
= $\displaystyle 1^2$
= $1$
Question 9
1 / -0

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

A
$2$

B
$-1$

C
$-2$

D
$3$

Solution

$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sec ^{ 2 }{ { 50 }^{ o } } +\cot ^{ 2 }{ { 40 }^{ o } } } +2\csc ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { { 32 }^{ o } } -4\tan { { 13 }^{ o } } \tan { { 37 }^{ o } } \tan { { 45 }^{ o } } \tan { { 53 }^{ o } } \tan { { 77 }^{ o } }$
$=\cfrac { \cos ^{ 2 }{ { ({ 90 }^{ o }-70 }^{ o }) } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sec ^{ 2 }{ { ({ 90 }^{ o }-40 }^{ o } } )+\cot ^{ 2 }{ { 40 }^{ o } } }$
$+2\csc ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { ({ 90 }^{ o }-58^{ o }) } -4(\tan { { 13 }^{ o } } \tan { { 77 }^{ o } } )(\tan { { 37 }^{ o } } \tan { { 53 }^{ o } } )\tan { { 45 }^{ o } }$
$=\cfrac { \sin ^{ 2 }{ { 70 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ co\sec ^{ 2 }{ { 40 }^{ o } } -\cot ^{ 2 }{ { 40 }^{ o } } } +2(\csc ^{ 2 }{ { 58 }^{ o } } -\cot ^{ 2 }{ { 58 }^{ o } }) -4[\tan { ({ 90 }^{ o }-77^{ o }) } \tan { { 77 }^{ o } } ][\tan { ({ 90 }^{ o }-53^{ o }) } \tan { { 53 }^{ o } } ][1]$
$=\cfrac{1}{1}+2-4$
$=-1$
Question 10
1 / -0

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

A
$\cfrac{6}{5}$

B
$\cfrac{7}{3}$

C
$\cfrac{9}{2}$

D
$\cfrac{1}{4}$

Solution

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

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

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

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

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

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

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

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Introduction to Trigonometry Test - 42

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Introduction to Trigonometry Test - 42

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Question 1

12\displaystyle \frac{1}{2}21​

000

111

−1-1−1

Solution

Question 2

1

3

6

0

Solution

Question 3

3\displaystyle \sqrt{3}3​

222

111

2\displaystyle \sqrt{2}2​

Solution

Question 4

S<PS < PS<P

S=PS = PS=P

S⋅P=2S\cdot P = 2S⋅P=2

S+P>3S + P > 3S+P>3

Solution

Question 5

111

222

444

333

Solution

Question 6

cos⁡β\cos {\beta}cosβ

cos⁡2β\cos {2\beta}cos2β

cos⁡2α\cos {2\alpha}cos2α

sin⁡2α\sin {2\alpha}sin2α

Solution

Question 7

000

111

222

333

Solution

Question 8

1

2

12\displaystyle \frac{1}{\sqrt{2}} 2​1​

22\displaystyle 2\sqrt{2} 22​

Solution

Question 9

222

−1-1−1

−2-2−2

333

Solution

Question 10

65\cfrac{6}{5}56​

73\cfrac{7}{3}37​

92\cfrac{9}{2}29​

14\cfrac{1}{4}41​

Solution

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

$0$

$1$

$-1$

$\displaystyle \sqrt{3}$

$2$

$1$

$\displaystyle \sqrt{2}$

$S < P$

$S = P$

$S\cdot P = 2$

$S + P > 3$

$1$

$2$

$4$

$3$

$\cos {\beta}$

$\cos {2\beta}$

$\cos {2\alpha}$

$\sin {2\alpha}$

$0$

$1$

$2$

$3$

$\displaystyle \frac{1}{\sqrt{2}}$

$\displaystyle 2\sqrt{2}$

$2$

$-1$

$-2$

$3$

$\cfrac{6}{5}$

$\cfrac{7}{3}$

$\cfrac{9}{2}$

$\cfrac{1}{4}$