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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 10 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 10
$$\csc (90 - \theta) = $$ is equivalent to
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$$\sec \theta$$
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$$\sin \theta$$
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$$\cos \theta$$
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None of these
Explanation
In the figure,
$$cosec (90 - \theta) = \dfrac {r}{x}$$ and $$\sec \theta = \dfrac {r}{x}$$
So, $$cosec (90 - \theta) = \sec \theta$$
Hence, $$\sec \theta$$ is correct.
Solve: $$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec } 70^{\circ} $$
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
Explanation
Given:
$$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec} 70^{\circ}$$
$$= \sec 70^{\circ} \sin (90^{\circ} - 70^{\circ}) + \cos 20^{\circ} \text{cosec} (90^{\circ} - 20^{\circ})$$
$$= \sec 70^{\circ} \cos 70^{\circ} + \cos 20^{\circ} \sec 20^{\circ}$$ $$[\because \sin(90^{\circ}-\theta)=\cos \theta$$ and $$\text{cosec}(90^{\circ}-\theta)=\sec \theta]$$
$$= 1 + 1$$ $$[\because ~\sec \theta.\cos \theta=1]$$
$$= 2$$
So, $$2$$ is correct.
If $$A$$ and $$B$$ are complementary angles, then $$\sin A \times \sec B =$$
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$$1$$
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$$0$$
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$$-1$$
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$$2$$
Explanation
Given that both the angles are complementary i.e., $$A+B=90^{\circ}$$
So, $$A=90^{\circ}-B$$
We have :
$$=\sin A\times \sec B$$
$$=\sin(90^o-B)\times \sec B$$
$$= \cos B\times \sec B$$
[using $$\cos \theta = \sin (90^o - \theta)$$]
$$=\cos B \times \dfrac{1}{\cos B}$$
$$=1$$
$$\tan 5^{\circ} \tan 25^{\circ} \tan 30^{\circ} \tan 65^{\circ} \tan 85^{\circ} =$$
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$$1$$
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$$\dfrac {1}{2}$$
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$$\sqrt {3}$$
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$$\dfrac {1}{\sqrt {3}}$$
Explanation
$$(\tan 5^{\circ} \tan 85^{\circ}) (\tan 25^{\circ} \tan 65^{\circ}) \tan 30^{\circ}$$
$$= [\tan 5^{\circ} \tan (90^{\circ} - 5^{\circ})] [\tan 25^{\circ} (\tan 90^{\circ} - 25^{\circ})] \times \tan 30^{\circ}$$
$$= \tan 5^{\circ} \cot 5^{\circ} \tan 25^{\circ} \cot 25^{\circ} \times \dfrac {1}{\sqrt {3}}$$
$$= 1\times 1\times \dfrac {1}{\sqrt {3}} = \dfrac {1}{\sqrt {3}}$$.
$$\cot (90^{\circ} - \theta) $$ is equivalent to
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$$\cot \theta$$
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$$\cos \theta$$
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$$\tan \theta$$
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$$-\tan \theta$$
Explanation
we know
$$\cos (90 - \theta)= \tan \theta$$
So, $$\tan \theta$$ is correct.
$$\text{cosec } (75^{\circ} + \theta) - \sec (15^{\circ} - \theta) =$$
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$$2\sec \theta$$
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$$2 cosec \theta$$
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$$0$$
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$$1$$
Explanation
Given,
$$\csc { \left( { 75^o } +\theta \right) } -\sec { \left( { 15^o } -\theta \right) } $$
$$=\csc { \left( { 75^o } +\theta \right) } -\sec { \left[ { { 90^o } -\left( 75^o+\theta \right) } \right] } $$
$$=\csc { \left( { 75^o } +\theta \right) } -\csc { \left( { 75^o } +\theta \right) } $$
$$=0$$
$$0$$ will be the answer
If $$\tan 2A = \cot (A - 21^{\circ})$$, where $$2A$$ is an acute angle, then $$\angle A =$$ ____
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$$24^{\circ}$$
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$$27^{\circ}$$
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$$35^{\circ}$$
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$$37^{\circ}$$
Explanation
Given, $$\tan 2A = \cot (A - 21^{\circ})$$
Therefore, $$ \cot (90^{\circ} - 2A) = \cot (A - 21^{\circ})$$
$$\Rightarrow 90^{\circ} - 2A = A - 21^{\circ}$$
$$\Rightarrow 90^{\circ} + 21^{\circ} = 2A + A$$
$$\Rightarrow 111^{\circ} = 3A$$
$$\Rightarrow A = 37^{\circ}$$
So, $$\angle A=37^{\circ}$$ is correct.
$$\sin^{2} 20^{\circ} + \sin^{2} 70^\circ=$$?
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
$$\sin^{2} 20^{\circ} = \cos^{2} (90^{\circ} - 20^{\circ}) = \cos^{2} 70^\circ$$
Substituting $$\sin^{2} 20^{\circ} = \cos^{2} 70^\circ$$
$$\sin^220^o+\sin^270^o=\cos^{2} 70^{\circ} + \sin^{2} 70^{\circ} = 1$$ $$[\because \sin^2\theta+\cos^2\theta=1]$$
Evaluate: $$\dfrac {\tan 35^{\circ}}{\cot 55^{\circ}} + \dfrac {\cot 78^{\circ}}{\tan 12^{\circ}} = $$
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$$0$$
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$$1$$
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$$2$$
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None of these
Explanation
Given expression is
$$\dfrac { \tan { { 35^o } } }{ \cot { { 55^o } } } +\dfrac { \cot { { 78^o } } }{ \tan { { 12^o } } } $$
We can rewrite the expression as:
$$\dfrac { \tan { { 35^o } } }{ \cot { \left( { 90^o- } { 35^o } \right) } } +\dfrac { \cot { { 78^o } } }{ \tan { \left( { 90^o- } { 78 } \right) } } $$
We know that, $$\cot { \left( 90^o-\theta \right) } =\tan { \theta } $$ and $$\tan { \left( 90^o-\theta \right) } =\cot { \theta } $$
$$\therefore \ $$ The expression becomes,
$$\dfrac { \tan { { 35^o } } }{ \tan { { 35^o } } } +\dfrac { \cot { { 78^o } } }{ \cot { { 78^o } } } $$
$$\Rightarrow 1+1$$
$$\Rightarrow 2$$
Hence, $$\dfrac { \tan { { 35^o } } }{ \cot { { 55^o } } } +\dfrac { \cot { { 78^o } } }{ \tan { { 12^o } } } =2$$
If $$\sin 15^{\circ} = \cos (n\times15^{\circ})$$, then $$n = .....$$
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$$1$$
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$$2$$
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$$5$$
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$$0$$
Explanation
As we know that
$$\cos (90^{\circ}-\theta)=\sin \theta$$
$$\sin 15^{\circ}=\cos (90^{\circ}-15^{\circ})=\cos 75^{\circ}=\cos (5\times 15^{\circ})$$
So $$n=5$$
$$\cot 1^{\circ} \cot 2^{\circ} ...... \cot 89^{\circ} = $$
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$$\dfrac {1}{2}$$
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$$1$$
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$$0$$
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$$-1$$
Explanation
$$\cot 1^{\circ} \cot 2^{\circ} ...... \cot 89^{\circ} $$
$$=(\cot 1^o\cdot \cot 89^o)(\cot 2^o\cdot \cot 88^o)...............(\cot 44^o\cdot \cot 46^o)(\cot 45^o)$$
$$=(\cot 1^o\cdot \tan 1^o)(\cot 2^o\cdot \tan 2^o)...............(\cot 44^o\cdot \tan 44^o)(\cot 45^o)$$, since $$\cot (90^o-\theta)=\tan\theta$$
$$=(1)(1).......(1)(1)=1$$, since $$\tan\theta\cdot \cot\theta=1$$
Without trigonometric table, evaluate $$\dfrac {\cot 63^{\circ}}{\tan 27^{\circ}}$$
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$$0$$
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$$1$$
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$$2$$
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$$-2$$
Explanation
Taking
$$\dfrac {\cot 63^{\circ}}{\tan 27^{\circ}}$$
$$\dfrac {\cot (90-27)^{\circ}}{\tan 27^{\circ}}$$
We know,
$$\cot { \left( 90^0-\theta \right) } =\tan { \theta } $$
$$\dfrac {\tan27^{\circ}}{\tan 27^{\circ}}$$
$$=1$$
Find the value of $$\tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ} $$
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$$\dfrac {1}{16}$$
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$$0$$
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$$1$$
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None of these
$$\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} ... \tan 89^{\circ} = $$
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$$1$$
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$$-1$$
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$$0$$
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None of above
Explanation
$$(\tan 1^{\circ} \tan 89^{\circ}) (\tan 2^{\circ} \tan 88^{\circ}) ..... (\tan 44^{\circ} \tan 46^{\circ}) \tan 45^{\circ}$$, group pair of terms
$$= \left \{\tan 1^{\circ} \tan (90^{\circ} - 1^{\circ})\right \} \left \{\tan 2^{\circ} \tan (90^{\circ} - 2^{\circ})\right \} ...... \left \{\tan 44^{\circ} \tan (90^{\circ} - 44^{\circ})\right \}\tan 45^{\circ}$$
$$= (\tan 1^{\circ} \cot 1^{\circ})(\tan 2^{\circ} \cot 2^{\circ}) ..... (\tan 44^{\circ} \cot 44^{\circ}) \tan 45^{\circ}$$
$$= (1 \times 1 ..... 1)\times 1 = 1$$
Note that $$\tan\theta\times \cot\theta=1$$
Without trigonometric table, evaluate $$\dfrac {\sec 41^o}{\text{cosec } 49^o}$$
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$$0$$
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$$1$$
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$$2$$
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$$-2$$
Explanation
Given
$$\dfrac { \sec { { 41^o } } }{ \csc { { 49^o } } } $$
$$=\dfrac { \sec { \left( { 90^o- } { 49^o } \right) } }{ \csc { { 49^o } } } $$
$$=\dfrac { \csc { { 49^o } } }{ \csc { { 49^o } } } $$ as we know $$\sec { \left( 90^o-\theta \right) } =\csc { \theta } $$
$$=1$$
Hence $$1$$ is the answer
Find the value of $$\cos^2 \theta (1 + \tan^2 \theta) + \sin^2 \theta (1 + \cot^2 \theta)$$.
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$$1$$
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$$3$$
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$$2$$
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$$4$$
Explanation
we know that ,
$$1+\tan ^2\theta=\sec ^2\theta\\1+\cot ^2\theta=\text{cosec} ^2\theta$$
Now, $$\cos^2\theta(1+\tan ^2\theta )+\sin^2\theta(1+\cot ^2\theta)$$
$$=\cos^2\theta \sec ^2\theta+\sin^2\theta \text{cosec} ^2\theta=1+1=2$$
Therefore, Answer is $$2$$
$$\sin 81^{\circ} + \tan 81^{\circ}$$, when expressed in terms of angles between $$0^{\circ}$$ and $$45^{\circ}$$, becomes
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$$\sin 9^{\circ} + \cos 9^{\circ}$$
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$$\cos 9^{\circ} + \tan 9^{\circ}$$
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$$\sin 9^{\circ} + \tan 9^{\circ}$$
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$$\cos 9^{\circ} + \cot 9^{\circ}$$
Explanation
We know that,
$$\sin A = \cos(90˚-A)$$
$$\tan A = \cot(90˚-A)$$
$$\sec A = \text{cosec }(90˚-A)$$
Using these, we get
$$\sin 81^{\circ} = \sin (90^{\circ} - 9^{\circ}) = \cos 9^{\circ}$$
$$\tan 81^{\circ} = \tan (90˚ - 9) = \cot 9^{\circ}$$
$$\therefore \sin 80^{\circ} + \tan 81^{\circ} = \cos 9^{\circ} + \cot 9^{\circ}$$
Find the value of : $$\dfrac {\cos 38^{\circ} \csc 52^{\circ}}{\tan 18^{\circ} \tan 35^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}} =$$
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$$\sqrt {3}$$
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$$\dfrac {1}{3}$$
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$$\dfrac {1}{\sqrt {3}}$$
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$$\dfrac {2}{\sqrt {3}}$$
Explanation
Given:
$$\dfrac { \cos { { 38^o } \csc { { 52^o } } } }{ \tan { { 18^o } } \tan { { 35^o } \tan { { 60^o } } \tan { { 72^o } } \tan { { 55^o } } } } $$
$$=\dfrac { \cos { { 38^o } } \csc { \left( { 90^o- } { 38^o } \right) } }{ \left( \tan { { 18^o\times } } \tan { { 72^o } } \right) \times \left( \tan { { 35^o\times } } \tan { { 55^o } } \right) \times \tan { { 60^o } } } $$
$$=\dfrac { \cos { { 38^o } } \sec { { 38^o } } }{ \tan { { 18^o } } \tan { \left( { 90^o-18^o } \right) \tan { { 35^o } } \tan { \left( { 90^o- } { 35^o } \right) \tan { { 60^o } } } } } $$
( we know that $$\tan { \left( 90^o-\theta \right) } =\cot { \theta } $$ and $$\cot { \theta } =\dfrac { 1 }{ \tan { \theta } } $$and $$\sec { \theta =\dfrac { 1 }{ \cos { \theta } } } $$ and $$\tan { { 60^o=\sqrt { 3 } } } )$$
$$\therefore \dfrac { \cos { { 38^o } \csc { { 52^o } } } }{ \tan { { 18^o } } \tan { { 35^o } \tan { { 60^o } } \tan { { 72^o } } \tan { { 55^o } } } } $$
$$=\dfrac { 1 }{ \left( \tan { { 18^o } } \cot { { 18^o } } \right) \left( \tan { { 35^o } } \cot { { 35^o } } \right) \sqrt { 3 } } $$
$$=\dfrac { 1 }{ \sqrt { 3 } } $$
Evaluate: $$\dfrac {\tan 30^{\circ}}{\cot 60^{\circ}}$$
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$$\dfrac {1}{\sqrt {2}}$$
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$$\dfrac {1}{\sqrt {3}}$$
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$$\sqrt {3}$$
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$$1$$
Explanation
Taking
$$\dfrac {\tan 30^{\circ}}{\cot 60^{\circ}}$$
$$\dfrac {\tan 30^{\circ}}{\cot (90-30)^{\circ}}$$
as we know, $$\cot { \left( 90^0-\theta \right) } =\tan { \theta } $$
$$\dfrac {\tan 30^{\circ}}{\tan30^{\circ}}$$
$$=1$$
Find the name of the person who first produce a table for solving a triangle's length and angles.
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William Rowan Hamilton
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Hipparchus
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Euclid
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Issac Newton
Explanation
$$\text {Hipparchus}$$ gave the first table of chords analogus to modern table of sine values, and used them to solve trigonometric problems
What is the meaning of trigonometry in Greek language?
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Measurement
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Triangle Measure
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Angle Measure
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Degree Measure
Explanation
Trigonometry in the Greek language means triangle measure.
Trigono $$\Rightarrow $$ triangle $$+$$ metron $$\Rightarrow $$ measure
So, option $$B$$ is correct.
If $$A + B = 90^o$$, then ......
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$$\sin A = \sin B$$
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$$\cos A = \cos B$$
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$$\tan A = \tan B$$
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$$\sec A = \text{cosec } B$$
Explanation
$$\sec A = \dfrac {r}{x}$$ and $$\text{cosec } A = \dfrac {r}{x}$$
So, $$\sec A = \text{ cosec } A$$
$$\dfrac {\cos (90 -\theta) \sec (90 - \theta)\tan \theta}{\text{cosec } (90 - \theta)\sin (90 - \theta) \cot (90 - \theta)} + \dfrac {\tan (90 - \theta)}{\cot \theta} = ......$$
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$$1$$
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$$-1$$
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$$2$$
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$$-2$$
Explanation
As we know that,
$$\sin(90^{o}-\theta)=\cos \theta$$,
$$\cos(90^o-\theta)=\sin \theta$$,
$$\text{cosec}(90^o-\theta)=\sec \theta$$,
$$\cot(90^o-\theta)=\tan\theta$$,
$$\sec(90^o - \theta)=\text{cosec} \theta$$, $$\tan(90^o - \theta) = \cot \theta$$
$$\therefore\ \dfrac{\cos(90-\theta) \sec(90-\theta) \tan \theta}{\text{cosec}(90-\theta) \sin(90-\theta) \cot(90-\theta)} + \dfrac{\tan(90-\theta)}{\cot \theta}$$
$$\therefore\ \dfrac{\sin \theta \times \frac{1}{\cos(90-\theta)} \times \tan \theta}{\frac{1}{\sin(90-\theta)} \times \sin(90-\theta) \times \tan \theta} + \dfrac{\cot \theta}{\cot \theta}$$
$$=\dfrac{\sin \theta \times \frac{1}{\sin \theta}}{\frac{1}{\cos \theta} \times \cos \theta} + 1$$
$$=1+1=2$$
Hence, option C is correct.
In the 5th century who created the table of chords with increasing 1 degree?
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Hipparchus
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William Rowan Hamilton
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Euclid
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Ptolemy
Explanation
Ptolemy used length of chords to define his trigonometric functions
The value of $$\cot 1^{\circ} \cot 2^{\circ} .... \cot 89^{\circ}$$ is .....
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$$1$$
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$$0$$
0%
$$-1$$
0%
None of above
Explanation
The given equation can be written as
$$(\cot 1^{\circ} \cot 89^{\circ})(\cot 2^{\circ} \cot 88^{\circ}) ..... (\cot 44^{\circ} \cot 46^{\circ}) \cot 45^{\circ}$$
$$= (\cot 1^{\circ} \cot (90^{\circ} - 1^{\circ})) (\cot 2^{\circ} \cot (90^{\circ} - 2^{\circ})) ..... (\cot 44^{\circ} \cot (90^{\circ} - 44^{\circ}) ) \cot 45^{\circ}$$
$$= \cot 1^{\circ} \tan 1^{\circ} \cdot \cot 2^{\circ} \tan 2^{\circ} .... \cot 44^{\circ} \tan 44^{\circ} \times \cot 45^{\circ}$$
(Since $$\cot (90^{\circ}-x) = \tan x)$$)
$$= 1\times 1 ..... 1\times 1 $$ (Since
$$ \cot x \times \tan x = 1, \cot 45^{\circ}$$)
$$=1$$
Hence, option $$A$$ is correct.
If $$\cos\theta = \dfrac{14}{4}$$ and $$\sin\theta$$ $$=$$ $$\dfrac{8}{3}$$, what is the value of $$\cot\theta$$?
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$$\dfrac{11}{16}$$
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$$\dfrac{13}{16}$$
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$$\dfrac{16}{21}$$
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$$\dfrac{21}{16}$$
Explanation
$$\cot \theta$$ $$=$$ $$\dfrac{\cos\theta}{\sin\theta}$$ $$=$$
$$\dfrac{\dfrac{14}{4}}{\dfrac{8}{3}}$$
$$=\dfrac{14}{4}\times\dfrac{3}{4}$$
$$\cot\theta = \dfrac{21}{16}$$
So, option $$D$$ is correct.
If $$\cos\theta = \dfrac{2}{21}$$ and $$\sin\theta = \dfrac{6}{7}$$, what is the value of $$\tan\theta$$?
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$$4$$
0%
$$9$$
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$$6$$
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$$7$$
Explanation
$$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$$ $$=$$
$$\dfrac{\dfrac{6}{7}}{\dfrac{2}{21}}$$
$$=\dfrac{6}{7}\times\dfrac{21}{2}$$
$$\tan\theta = 9$$
So, option $$B$$ is correct.
If $$t = 45^{\circ}$$, what is $$\sec (t) \sin (t) - \mathrm{cosec} (t) \cos (t)$$?
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$$-2$$
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$$-1$$
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$$0$$
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$$\dfrac {\pi}{2}$$
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None of the above
Explanation
For $$t=45^0$$, the value of
$$\sec { (t) } \sin { (t) } -\csc { (t) } \cos { (t) }$$ may be determined as follows:
$$\sec { (t) } \sin { (t) } -\csc { (t) } \cos { (t) } =\sec { (45^{ 0 }) } \sin { (45^{ 0 }) } -\csc { (45^{ 0 }) } \cos { (45^{ 0 }) }$$
$$ =\sqrt { 2 } \times \dfrac { 1 }{ \sqrt { 2 } } -\sqrt { 2 } \times \dfrac { 1 }{ \sqrt { 2 } } $$
$$=1-1$$
$$=0$$
If $$\sin \theta + \cos \theta = \sqrt {2}$$, find the value of $$\sin \theta \times \cos \theta$$.
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$$2$$
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$$\sqrt {2} - 1$$
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$$\dfrac {1}{2}$$
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$$2\sec \theta$$
Explanation
$$\sin { \theta } +\cos { \theta } =\sqrt { 2 } $$
Squaring both sides, we get
$$\Rightarrow { \left( \sin { \theta } +\cos { \theta } \right) }^{ 2 }=2$$
$$\Rightarrow \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } .\cos { \theta } =2$$
$$\Rightarrow 1+2\sin { \theta } .\cos { \theta } =2$$
$$\Rightarrow 2\sin { \theta } \cos { \theta } =1$$
$$\Rightarrow \sin { \theta } \cos { \theta } =\dfrac { 1 }{ 2 } .$$
Hence, the answer is $$\dfrac { 1 }{ 2 } .$$
In the following figure $$AB \perp BC$$ and $$\angle ACB = 30^\circ$$, given $$BC = \sqrt{300}\ m$$. The length of $$AB$$ is
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$$10\ m$$
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$$100\ m$$
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$$10 \sqrt 3\ m$$
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$$100 \sqrt 3\ m$$
Explanation
Given that:
From the fig; $$AB\bot BC,\angle ACB=30^\circ, BC=\sqrt300 m$$
To find:
$$AB=?$$
Solution:
In $$\triangle ABC,$$
$$\tan\angle ACB=\cfrac{AB}{BC}$$
or, $$\tan30^\circ=\cfrac{AB}{BC}$$
or, $$AB=\tan30^\circ\times BC$$
or, $$AB=\cfrac{1}{\sqrt3}\times \sqrt{300}m$$
or, $$AB=\cfrac{1}{\sqrt3}\times 10\sqrt{3}m$$
or, $$AB=10m$$
Therefore, A is the correct option.
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