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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 13 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 13
If $$\cot\theta+\tan\theta=x$$ and $$\sec\theta-\cos\theta=y$$, then
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$$\sin\theta \cos\theta=\displaystyle \cfrac {1}{x}$$
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$$\sin\theta \tan\theta=y$$
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$$(x^2y)^{2/3}=1$$
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$$(x^2y)^{1/3}+(xy^2)^{1/3}=1$$
Explanation
$$\cot \theta +\tan \theta =x\Rightarrow \cfrac { \cos \theta }{ \sin \theta } +\cfrac { \sin \theta }{ \cos \theta } =x$$
$$\cfrac { 1 }{ \sin \theta \cos \theta } =x\Rightarrow \sin \theta \cos \theta =\cfrac { 1 }{ x } $$ ...(1)
$$sec\theta -\cos \theta =y\Rightarrow \cfrac { 1 }{ \cos \theta } -\cos \theta =y$$
$$\Rightarrow \cfrac { 1-{ \cos }^{ 2 }\theta }{ \cos \theta } =y\Rightarrow \cfrac { { \sin }^{ 2 }\theta }{ \cos \theta } =y$$
$$\Rightarrow \sin \theta \tan \theta =y$$ ...(2)
From (1) and (2)
$$x^{ 2 }y=\cfrac { 1 }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } \times \cfrac { { \sin }^{ 2 }\theta }{ \cos \theta } =\cfrac { 1 }{ { \cos }^{ 3 }\theta } ={ sec }^{ 3 }\theta \\ { y }^{ 2 }x=\cfrac { { \sin }^{ 4 }\theta }{ \cos ^{ 2 }\theta } \times \cfrac { 1 }{ \sin \theta \cos \theta } ={ \tan }^{ 3 }\theta \\ { \left( x^{ 2 }y \right) }^{ \frac { 1 }{ 3 } }+{ \left( { y }^{ 2 }x \right) }^{ \frac { 1 }{ 3 } }=\sec\theta +\tan \theta $$
If $$x=\dfrac {\sin^3p}{\cos^2p}, y=\dfrac {\cos^3p}{\sin^2p}$$ and $$\sin p + \cos p= \dfrac 12$$, then $$x+y$$ is equal to
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$$\cfrac {75}{18}$$
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$$\cfrac {44}{9}$$
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$$\cfrac {79}{18}$$
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$$\cfrac {48}{9}$$
Explanation
Given $$x=\displaystyle \cfrac {\sin^3p}{\cos^2p}, y=\cfrac {\cos^3p}{\sin^2p}$$; and $$\sin p + \cos p=\displaystyle \frac{1}{2}$$
Consider, $$x+y=\displaystyle \cfrac { \sin^{ 3 }p }{ \cos^{ 2 }p } +\cfrac { \cos^{ 3 }p }{ \sin^{ 2 }p } $$
$$\Rightarrow \displaystyle x+y=\cfrac { \sin^{ 5 }p+\cos^{ 5 }p }{ \sin^{ 2 }p\cos^{ 2 }p } $$
$$\displaystyle =\cfrac { \sin^{ 3 }p(1-\cos ^{ 2 }{ p } )+\cos^{ 3 }p(1-\sin ^{ 2 }{ p } ) }{ \sin^{ 2 }p\cos^{ 2}p } $$
$$\displaystyle =\cfrac { \sin^{ 3 }p+\cos^{ 3 }p-\sin ^{ 2 }{ p } \cos ^{ 2 }{ p } (\sin { p } +\cos { p } ) }{ \sin^{ 2 }p\cos^{ 2 }p } $$
$$\displaystyle =\cfrac { { (\sin { p } +\cos { p } ) }^{ 3 }-3\sin { p } \cos { p } (\sin { p } +\cos { p } )-\sin ^{ 2 }{ p } \cos ^{ 2 }{ p } (\sin { p } +\cos { p } ) }{ \sin^{ 2 }p\cos^{ 2 }p } $$ .....(1)
Since $$(\sin p+\ cos p)^2=\sin^{2} p+\cos ^{2}p+2\sin p\cos p$$
$$\Rightarrow \displaystyle \sin p\cos p =-\cfrac{3}{4}$$
So, using in (1),
$$x+y=\displaystyle \dfrac { \dfrac { 1 }{ 8 } +\dfrac { 9 }{ 16 } -\dfrac { 9 }{ 32 } }{ \dfrac { 9 }{ 16 } } =\dfrac { 79 }{ 18 } $$
In the figure given
$$\angle ABD=\angle PQD=\angle CDQ=\cfrac { \pi }{ 2 } $$. If $$AB=x.PQ=z$$ and $$CD=y$$, then which one of the following is correct?
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$$\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 1 }{ z } $$
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$$\cfrac { 1 }{ x } +\cfrac { 1 }{ z } =\cfrac { 1 }{ y } $$
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$$\cfrac { 1 }{ z } +\cfrac { 1 }{ y } =\cfrac { 1 }{ x } $$
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$$\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 2 }{ z } $$
Find the relation obtained by eliminating $$\displaystyle \theta $$ from the equation $$\displaystyle x=a\cos \theta +b\sin \theta $$ and $$\displaystyle y=a\sin \theta -b\cos \theta $$
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$$\displaystyle x^{2}+y^{2}=a^{2}-b^{2}$$
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$$\displaystyle x^{2}-y^{2}=a^{2}+b^{2}$$
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$$\displaystyle x^{2}-y^{2}=a^{2}-b^{2}$$
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$$\displaystyle x^{2}+y^{2}=a^{2}+b^{2}$$
Explanation
Given, $$ x = a \cos \theta + b \sin \theta $$ and $$ y = a \sin \theta - b \cos \theta $$
Squaring them and adding we get
$$ x^2 + y^2 = a^2 \cos ^2 \theta + b^2 \sin ^2 \theta + 2ab\cos \theta \sin \theta + a^2 \sin ^2 \theta + b^2 \cos ^2 \theta - 2ab\cos \theta \sin \theta $$
$$ \Rightarrow x^2 + y^2 = a^2 \cos ^2 \theta + b^2 \sin ^2 \theta + a^2 \sin ^2 \theta + b^2 \cos ^2 \theta $$
$$ \Rightarrow x^2 + y^2 = a^2 (\cos ^2 \theta + \sin ^2 \theta) + b^2 ( \sin ^2 \theta + \cos ^2 \theta ) $$
$$ \Rightarrow x^2 + y^2 = a^2 + b^2 $$
Which one of the following when simplified is not equal to one?
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$$\displaystyle \tan 18^{\circ}\times \tan 36^{\circ}\times \tan 54^{\circ}\times \tan 72^{\circ}$$
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$$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}71^{\circ}$$
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$$\displaystyle \frac{2\sin 62^{\circ}}{\cos 28^{\circ}}-\frac{\sec 42^{\circ}}{\text{cosec}48^{\circ}}$$
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None of these
Explanation
(1) $$\displaystyle \tan 18^{\circ}\times \tan 36^{\circ}\times \tan 54^{\circ}\times \tan 72^{\circ}$$
= $$\displaystyle \tan 18^{\circ}\times \tan (90 - 54)^{\circ}\times \tan 54^{\circ}\times \tan (90 - 18)^{\circ}$$
= $$\displaystyle \tan 18^{\circ}\times \cot 54^{\circ}\times \tan 54^{\circ}\times \cot 18^{\circ}$$
= $$1$$
(2) $$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}71^{\circ}$$
= $$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}(90 - 19)^{\circ}$$
= $$\displaystyle \sin ^{2}19^{\circ}+\cos ^{2}19^{\circ}$$
= $$1$$
(3) $$\displaystyle \frac{2\sin 62^{\circ}}{\cos 28^{\circ}}-\frac{\sec 42^{\circ}}{cosec48^{\circ}}$$
= $$\displaystyle \frac{2\sin 62^{\circ}}{\cos (90 - 62)^{\circ}}-\frac{\sec 42^{\circ}}{cosec (90 - 42)^{\circ}}$$
= $$\displaystyle \frac{2\sin 62^{\circ}}{\sin 62^{\circ}}-\frac{\sec 42^{\circ}}{\sec 42^{\circ}}$$
= $$2 - 1$$
= $$1$$
Thus, all are equal to one. Hence, none of these is the answer.
If $$\frac{1 - cos x}{cos x (1 + cos x)} = \frac{sin \alpha}{cos x} - \frac{2}{1 + cos x}$$, then $$\alpha$$ =
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$$\frac{\pi}{8}$$
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$$\frac{\pi}{4}$$
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Both A and B
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None of the above
$$\dfrac { cot\theta +cosec\theta -1 }{ cot\theta -cosec\theta +1 } $$ is equal to:
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$$\dfrac { 1+cos\theta }{ sin\theta } $$
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$$\dfrac { sin\theta }{ 1+cos\theta } $$
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$$\dfrac { sin\theta }{ 1-cos\theta } $$
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$$\dfrac { 1-cos\theta }{ sin\theta } $$
If $$\cos P=\dfrac{1}{7}$$ and $$\cos Q=\dfrac{13}{14}$$, P and Q both are acute angle then the value of $$P-Q$$ will be?
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$$45^o$$
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$$30^o$$
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$$75^o$$
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$$60^o$$
$$\sin { { 48 }^{ 0 } } .\sin { 12^{ 0 } } =$$
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$$\frac { 1+\sqrt { 5 } }{ 8 } $$
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$$\frac { 1-\sqrt { 5 } }{ 8 } $$
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$$\frac { \sqrt { 5 } +1 }{ 8 } $$
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$$\frac { \sqrt { 5 } -1 }{ 8 } $$
$$1+cosec\dfrac { \pi }{ 4 } +cosec\dfrac { \pi }{ 8 } +cosec\dfrac { \pi }{ 16 } =$$
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$$cot\dfrac { \pi }{ 8 } $$
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$$cot\dfrac { \pi }{ 16 } $$
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$$cot\dfrac { \pi }{ 32 } $$
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$$cosec^{ 2 }\dfrac { \pi }{ 16 } $$
$$\frac { \sec { 8A } -1 }{ \sec { 4A } -1 } =$$
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$$0$$
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$$\frac { \tan { 8A } }{ \tan { 2A } } $$
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$$\frac { \cos { 8A } }{ \cos { 2A } } $$
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$$\frac { \sin { 8A } }{ \sin { 2A } } $$
$$\dfrac { tan\theta }{ 1-cot\theta } +\dfrac { cot{\theta}}{ 1-tan\theta } $$ is equal to:
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$$1+sin\theta cost\theta $$
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$$sin\theta cos\theta $$
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$$sec\theta cosec\theta $$
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$$1+sec\theta cosec\theta $$
Suppose $$I_1 = \displaystyle \int_0^{\pi/2} cos(\pi sin^2x)dx; I_2 = \displaystyle \int_0^{\pi/2} cos(2\pi sin^2x)dx \,and \, I_3 = \displaystyle \int_0^{\pi/2} cos(\pi sin x)dx$$ then
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$$I_1 = 0$$
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$$I_2 + I_1 = 0$$
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$$I_1 + I_2 + I_3 = 0$$
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$$I_2 = I_3$$
$$1+cosec\frac { \pi }{ 4 } +cosec\frac { \pi }{ 8 } cosec\frac { \pi }{ 16 } =$$
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$$cot\frac { \pi }{ 8 } $$
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$$cot\frac { \pi }{ 16 } $$
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$$cot\frac { \pi }{ 32 } $$
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$${ cosec }^{ 2 }$$
If $$cos(80^{\circ}+\Theta )=sin(\frac{k}{3})-\Theta )$$ then k=.....
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81
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27
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54
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9
If the median of a triangle ABC passing through A is perpendicular AB then tanA + 2tanB =
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1
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1
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0
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None
An angle is increasing at a constant rate. The rate of increase of tan when the angle is $$ \pi /3 $$ is
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4 times the increase of sine
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8 times the increase of cosine
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8 times the increase of sine
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4 times the increase of cosine
$$\dfrac{cos A}{1+sin A} +\dfrac{cos A}{1-sin A} =$$
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$$ 2 sec A$$
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$$ 2 cos A$$
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$$2$$
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$$0$$
The value of $$\cfrac { cot{ 54 }^{ 0 } }{ cot36^{ 0 } } +\cfrac { tan{ 20 }^{ 0 } }{ cot{ 70 }^{ 0 } } =$$.
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3
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1
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0
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2
In triangle ABC, if $$sinAcosB=\cfrac { 1 }{ 4 } $$ and $$3tanA=tanB$$ then $${ cot }^{ 2 }A$$ is equal to
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2
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3
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4
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5
If $$\sec \theta + \tan \theta = x,$$ then $$\sec \theta = ........... .$$
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$$\frac{x^2 + 1}{x}$$
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$$\frac{x^2 + 1}{2x}$$
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$$\frac{x^2 - 1}{2x}$$
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$$\frac{x^2 - 1}{x}$$
$$\frac { cos\theta }{ sec\theta +tan\theta } +\frac { cos\theta }{ sec\theta -tan\theta } $$=
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1
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$$cos\theta$$
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$$sin\theta$$
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2
$$\dfrac{sin A}{cosec A} + \dfrac{cosA}{sec A}$$ =
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$$1$$
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$$sin^2A$$
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$$ 2 sin^2 A+ cos ^2 A$$
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$$ 2 sec A$$
If $$tanA=\sqrt { 2 } -1,\quad then\quad the\quad value\quad of\quad cosec.\quad A.secA\quad is\quad equal\quad to\quad $$
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$$\dfrac { 1 }{ \sqrt { 2 } } $$
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$$\dfrac { 1 }{ \sqrt { 2 } } $$
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$$2\sqrt { 2 } $$
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$$\dfrac { \sqrt { 3 } }{ 2 } $$
$$tan^{-1} \begin {pmatrix} \cfrac {1+sinx}{cos x}\end {pmatrix}=$$
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$$\cfrac {\pi}4- \cfrac {x}2$$
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$$\cfrac {\pi}4- {x}$$
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$$\cfrac {\pi}4+ {x}$$
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$$\cfrac {\pi}4+\cfrac {x}2$$
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Practice Class 10 Maths Quiz Questions and Answers
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