Trigonometric Functions Test -6

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Trigonometric Functions Test -6

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Weekly Quiz Competition

Question 1
1 / -0

sin (180+ϕ) sin(180−ϕ) cosec²ϕ

A
1

B
0

C
- 1

D
none of these

Solution

sin(180 + ϕ)(180 − ϕ) cosec²ϕ = −sinϕ. sinϕ cosec²ϕ = −sin²ϕ cosec²ϕ = −1
Question 2
1 / -0

The solution of equation $$\cos^2 \theta +\sin \theta+1=0$$ lies in the interval

A
$$\displaystyle \left ( - \frac{\pi}{4}, \frac{\pi}{4} \right )$$

B
$$\displaystyle \left ( \frac{\pi}{4}, \frac{3\pi}{4} \right )$$

C
$$\displaystyle \left ( \frac{3\pi}{4}, \frac{5\pi}{4} \right )$$

D
$$\displaystyle \left ( \frac{5\pi}{4}, \frac{7\pi}{4} \right )$$

Solution

$$cos^2 \theta + \sin \theta + 1= 0$$
$$\Rightarrow 1 - \sin ^2 \theta + \sin \theta + 1 = 0$$
$$\Rightarrow \sin ^2 \theta - \sin \theta - 2 = 0$$
$$\Rightarrow (\sin \theta + 1) (\sin \theta - 2) = 0$$
As $$ -1 < \sin \theta < 1$$ so $$\sin \theta + 1 = 0$$
$$\Rightarrow \sin \theta = \sin \displaystyle \frac{3 \pi}{2}$$
$$\Rightarrow \theta = \displaystyle \frac{3\pi}{2} \varepsilon \left( \frac{5 \pi}{4}, \frac{7 \pi}{4} \right )$$
Question 3
1 / -0

In which quadrant does the terminal side of the angle $$250^0$$ lie?

A
Quadrant III

B
Quadrant II

C
Quadrant I

D
Quadrant IV

Solution

Quadrant I: $$0˚$$ to $$90˚$$
Quadrant II: $$90˚$$ to $$180˚$$
Quadrant III: $$180˚$$ to $$270˚$$
Quadrant IV: $$270˚$$ to $$360˚$$

$$180˚ < 250˚ < 270˚$$, so the terminal side of a $$250˚$$ angle lies in quadrant III.
Question 4
1 / -0

In which quadrant does the terminal side of the angle $$330^0$$ lie?

A
Quadrant I

B
Quadrant II

C
Quadrant III

D
Quadrant IV

Solution

Quadrant I: $$0˚$$ to $$90˚$$
Quadrant II: $$90˚$$ to $$180˚$$
Quadrant III: $$1800˚$$ to $$270˚$$
Quadrant IV: $$270˚$$ to $$360˚$$

$$270˚ < 330˚ < 360˚$$, so the terminal side of a $$330˚$$ angle lies in quadrant IV.
Question 5
1 / -0

The minimum value of $$ \displaystyle \sec ^{2}\alpha + \cos ^{2}\alpha $$ is

A
$$1$$

B
$$2$$

C
$$0$$

D
$$-1$$

Solution

Consider the given expression,

$${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha $$

AS it is clear that $${{\cos }^{2}}\alpha $$ and $${{\sec }^{2}}\alpha $$ are positive numbers

So, we can apply the theorem of A.M and G.M

A.M>=G.M

so we have

$$\left( \dfrac{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }{2} \right)=\sqrt{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }$$

Or $$ {{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha =2\sqrt{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }=2.1 $$

$$ {{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha =2 $$

So, minimum value of $${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha $$ is 2.
Question 6
1 / -0

$$\cfrac { \tan { \theta } }{ \sec { \theta } -1 } +\cfrac { \tan { \theta } }{ \sec { \theta } +1 } $$ is equal to

A
$$\text{cosec} { \theta } $$

B
$$2\text{cosec} { \theta } $$

C
$$\sec { \theta } $$

D
$$2\sec { \theta } $$

Solution

$$\cfrac { \tan { \theta } \left( \sec { \theta } +1+\sec { \theta } -1 \right) }{ \sec ^{ 2 }{ \theta } -1 } =\cfrac { 2\tan { \theta } \sec { \theta } }{ \sec ^{ 2 }{ \theta }-1 } =\cfrac { 2\tan { \theta } \sec { \theta } }{ \tan ^{ 2 }{ \theta } } $$
$$\Rightarrow \cfrac { 2\sec { \theta } }{ \tan { \theta } } =2\cfrac { 1 }{ \cos { \theta } .\cfrac { \sin { \theta } }{ \cos { \theta } } } =2\text{cosec} { \theta } $$
Question 7
1 / -0

If $$\sin { \theta } +\cos { \theta } =p$$ and $$\tan { \theta } +\cot { \theta } =q$$, then $$q\left( { p }^{ 2 }-1 \right) =$$

A
$$\dfrac { 1 }{ 2 } $$

B
$$2$$

C
$$1$$

D
$$3$$

Solution

Let $$\left( \sin { \theta } +\cos { \theta } \right) ^{ 2 }=p^{ 2 }$$

$$\Rightarrow 1+\sin { 2\theta } ={ p }^{ 2 }$$

Also $$\tan { \theta } +\dfrac { 1 }{ \tan { \theta } } =q$$

$$\Rightarrow \dfrac { \tan ^{ 2 }{ \theta } +1 }{ 2\tan { \theta } } =\dfrac { q }{ 2 } $$

$$\csc { 2\theta } =\dfrac { q }{ 2 } $$

$$\Rightarrow \dfrac { 1 }{ { p }^{ 2 }-1 } =\dfrac { q }{ 2 }$$

$$ \Rightarrow \left( { p }^{ 2 }-1 \right) q\ =\ 2 $$
Question 8
1 / -0

$$\cfrac { 1 }{ \sec { \theta } -\tan { \theta } } $$ is equal to

A
$$\sec { \theta } -\tan { \theta } $$

B
$$\sec { \theta } \tan { \theta } $$

C
$$\sec { \theta } +\tan { \theta } $$

D
$$\sec { \theta } $$

Solution

$$\cfrac { 1 }{ \sec { \theta } -\tan { \theta } } \times \cfrac { \sec { \theta } +\tan { \theta } }{ \sec { \theta } +\tan { \theta } } =\cfrac { \sec { \theta } -\tan { \theta } }{ \sin ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } } =\sec { \theta } +\tan { \theta } \quad \quad $$
Question 9
1 / -0

If $$\tan\theta +\cot\theta =2$$, then the value of $$\tan^2\theta +\cot^2\theta$$ is __________?

A
$$4$$

B
$$2$$

C
$$\displaystyle\frac{3}{2}$$

D
$$5$$

Solution

Given, $$\tan \theta+\cot \theta=2$$

Squaring both sides, we get

$$(\tan \theta+\cot \theta)^2=2^2$$

$$\Rightarrow \tan^2\theta+\cot^2\theta+2=4$$ .....Since $$\tan\theta\times \cot \theta=\tan \theta\times \dfrac {1}{\tan \theta}=1$$

$$\Rightarrow \tan^2\theta+\cot^2\theta=2$$
Question 10
1 / -0

$$\cos ^{ 4 }{ A } -\sin ^{ 4 }{ A } $$ is equal to

A
$$\sin { 2A } $$

B
$$-\sin { 2A } $$

C
$$\cos { 2A } $$

D
$$-\cos { 2A } $$

Solution

$$\cos ^{ 4 }{ A } -\sin ^{ 4 }{ A } =\left( \cos ^{ 2 }{ A } -\sin ^{ 2 }{ A } \right) \left( \cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } \right) =\cos { 2A } $$
Question 11
1 / -0

If $$\alpha $$ and $$\beta $$ are two different solution lying between $${{ - \pi } \over 2}$$ and $${\pi \over 2}$$ of the equation $$2{\mathop{\rm Tan}\nolimits} \theta + {\mathop{\rm Sec}\nolimits} \theta - 2$$ then $${\mathop{\rm Tan}\nolimits} \alpha + {\mathop{\rm Tan}\nolimits} \beta $$ is

A
$$0$$

B
$$1$$

C
$${4 \over 3}$$

D
$${8 \over 3}$$

Solution

$$2\tan { \theta } +\sec { \theta } -2=0\\ \cfrac { 2\sin { \theta } }{ \cos { \theta } } +\cfrac { 1 }{ \cos { \theta } } -2=0\\ 2\sin { \theta } -2\cos { \theta } +1=0\\ 2(\sin { \theta } -\cos { \theta } )=-1\\ \sin { \theta } -\cos { \theta } =\cfrac { -1 }{ 2 } \\ { (\sin { \theta } -\cos { \theta } ) }^{ 2 }={ (\cfrac { -1 }{ 2 } ) }^{ 2 }\\ { \sin { ^{ 2 }\theta } }+\cos { ^{ 2 }\theta } -2\sin { \theta } \cos { \theta } =\cfrac { 1 }{ 4 } \\ 1-2\sin { \theta } \cos { \theta } =\cfrac { 1 }{ 4 } \\ 1-\sin { 2\theta } =\cfrac { 1 }{ 4 } \\ \Rightarrow \sin { 2\theta } =\cfrac { 3 }{ 4 } \\ \Rightarrow 2\theta =\sin ^{ -1 }{ \cfrac { 3 }{ 4 } } $$
It can be $$ \pi \quad \& \quad -\pi $$
$$ \therefore \tan { \pi } +\tan { (-\pi ) } \\ =0$$

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Trigonometric Functions Test -6

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SHARING IS CARING

Question 1

1

0

- 1

none of these

Solution

Question 2

$$\displaystyle \left ( - \frac{\pi}{4}, \frac{\pi}{4} \right )$$

$$\displaystyle \left ( \frac{\pi}{4}, \frac{3\pi}{4} \right )$$

$$\displaystyle \left ( \frac{3\pi}{4}, \frac{5\pi}{4} \right )$$

$$\displaystyle \left ( \frac{5\pi}{4}, \frac{7\pi}{4} \right )$$

Solution

Question 3

Quadrant III

Quadrant II

Quadrant I

Quadrant IV

Solution

Question 4

Quadrant I

Quadrant II

Quadrant III

Quadrant IV

Solution

Question 5

$$1$$

$$2$$

$$0$$

$$-1$$

Solution

Question 6

$$\text{cosec} { \theta } $$

$$2\text{cosec} { \theta } $$

$$\sec { \theta } $$

$$2\sec { \theta } $$

Solution

Question 7

$$\dfrac { 1 }{ 2 } $$

$$2$$

$$1$$

$$3$$

Solution

Question 8

$$\sec { \theta } -\tan { \theta } $$

$$\sec { \theta } \tan { \theta } $$

$$\sec { \theta } +\tan { \theta } $$

$$\sec { \theta } $$

Solution

Question 9

$$4$$

$$2$$

$$\displaystyle\frac{3}{2}$$

$$5$$

Solution

Question 10

$$\sin { 2A } $$

$$-\sin { 2A } $$

$$\cos { 2A } $$

$$-\cos { 2A } $$

Solution

Question 11

$$0$$

$$1$$

$${4 \over 3}$$

$${8 \over 3}$$

Solution

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