If $$\cot\theta+\tan\theta=x$$ and $$\sec\theta-\cos\theta=y$$, then

$$\sin\theta \cos\theta=\displaystyle \cfrac {1}{x}$$

$$(x^2y)^{1/3}+(xy^2)^{1/3}=1$$

MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 13

$\cot\theta+\tan\theta=x$ 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)

$x^{ 2 }y=\cfrac { 1 }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } \times \cfrac { { \sin }^{ 2 }\theta }{ \cos \theta } =\cfrac { 1 }{ { \cos }^{ 3 }\theta } ={ sec }^{ 3 }\theta \\ { y }^{ 2 }x=\cfrac { { \sin }^{ 4 }\theta }{ \cos ^{ 2 }\theta } \times \cfrac { 1 }{ \sin \theta \cos \theta } ={ \tan }^{ 3 }\theta \\ { \left( x^{ 2 }y \right) }^{ \frac { 1 }{ 3 } }+{ \left( { y }^{ 2 }x \right) }^{ \frac { 1 }{ 3 } }=\sec\theta +\tan \theta$

$x=\dfrac {\sin^3p}{\cos^2p}, y=\dfrac {\cos^3p}{\sin^2p}$ 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.

$\frac{1 - cos x}{cos x (1 + cos x)} = \frac{sin \alpha}{cos x} - \frac{2}{1 + cos x}$ , 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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$\dfrac { 1-cos\theta }{ sin\theta }$

$\cos P=\dfrac{1}{7}$ 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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${ cosec }^{ 2 }$

$cos(80^{\circ}+\Theta )=sin(\frac{k}{3})-\Theta )$ 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

$\sec \theta + \tan \theta = x,$ 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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$2 sec A$

$tanA=\sqrt { 2 } -1,\quad then\quad the\quad value\quad of\quad cosec.\quad A.secA\quad is\quad equal\quad to\quad$

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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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$\cfrac {\pi}4+\cfrac {x}2$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 13 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 13 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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