Trigonometric Functions Test 9

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Trigonometric Functions Test 9

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Weekly Quiz Competition

Question 1
1 / -0

$$9\tan^{2}\theta - 9 \sec^{2}\theta =$$

A
$$1$$

B
$$0$$

C
$$9$$

D
$$-9$$

Solution

$$\displaystyle 9tan^2\theta-9sec^2\theta=-9(sec^2\theta-tan^2\theta)$$
$$\displaystyle =-9(1)$$
$$=-9$$
Option D is correct.
Question 2
1 / -0

If $$\sin( \pi \cos \theta) = \cos (\pi \sin \theta)$$, then which of the following is correct.

A
$$\cos \theta =\dfrac {3}{2\sqrt {2}}$$

B
$$\cos \left (\theta - \dfrac {\pi}{2}\right ) = \dfrac {1}{2\sqrt {2}}$$

C
$$\cos \left (\theta - \dfrac {\pi}{4}\right ) = \dfrac {1}{2\sqrt {2}}$$

D
$$\cos \left (\theta + \dfrac {\pi}{4}\right ) = -\dfrac {1}{2\sqrt {2}}$$

Solution

Given :
$$\sin(\pi \cos \theta)=\cos(\pi \sin \theta)$$
$$\implies \sin(\pi \cos \theta)=\sin\left(\frac{\pi}{2}-\pi \sin \theta\right)$$ ......... $$[\because \cos x=\sin(90^{o}-x)]$$
$$\implies \pi \cos \theta=\dfrac{\pi}{2}- \pi \sin \theta$$
$$\implies \cos \theta=\dfrac{1}{2}-\sin \theta$$ ...... [Divided by $$\pi$$]
$$\implies \cos \theta + \sin \theta =\dfrac{1}{2}$$
$$\implies \dfrac{\sqrt{2}}{\sqrt{2}}(\cos \theta + \sin \theta) =\dfrac{1}{2}$$
$$\implies \sqrt{2} \left(\dfrac{1}{\sqrt{2}}\cos \theta + \dfrac{1}{\sqrt{2}}\sin \theta\right) =\dfrac{1}{2}$$
$$\implies \sqrt{2} \left(\cos\left(\dfrac{\pi}{4}\right)\cos \theta + \sin\left(\dfrac{\pi}{4}\right)\sin \theta\right) =\dfrac{1}{2}$$
$$\implies \sqrt{2}\cos\left(\theta - \dfrac{\pi}{4}\right)=\dfrac{1}{2}$$
$$\implies \cos\left(\theta - \dfrac{\pi}{4}\right)=\dfrac{1}{2\sqrt{2}}$$
Hence, option C is correct.
Question 3
1 / -0

The number of principal solutions of $$\tan 2\theta = 1$$ is

A
One

B
Two

C
Three

D
Four

Solution

Let $$y=\tan 2\theta$$

$$\tan 2\theta = 1 $$

$$\implies 2\theta={\tan^{-1}(1)}$$

$$\implies 2\theta=\dfrac{\pi}{4}+{k\pi},$$ where $$k$$ is a positive integer
Now,

for $$k=0\Rightarrow \theta=\dfrac{\pi}{4}$$ and for $$k=1\Rightarrow \theta=\dfrac{5\pi}{4}$$

$$0\leq \dfrac{\pi}{4} \leq 2\pi$$ and $$0\leq \dfrac{5\pi}{4} \leq 2\pi$$

$$\therefore 2\theta=\dfrac{\pi}{4}, \dfrac{5\pi}{4}$$

$$\implies \theta=\dfrac{\pi}{8},\dfrac{5\pi}{8}$$

Hence, the required principal solutions are $$\dfrac{\pi}{8}$$ and $$\dfrac{5\pi}{8}$$.
Question 4
1 / -0

The equation $$ \sqrt {3} \sin x + \cos x = 4 $$ has

A
Infinitely many solutions

B
No solution

C
Two solutions

D
Only one solution

Solution

We know that,

$$ - \sqrt {a^2 + b^2} \le a \cos \theta + b \sin \theta \le \sqrt {a^2 + b^2}$$

$$ \therefore - \sqrt {3 + 1} \le \sqrt{3} \sin x + \cos x \le \sqrt {3 + 1}$$

$$ \Rightarrow -2 \le \sqrt{3} \sin x + \cos x \le {2} $$

But $$ \sqrt {3} \sin x + \cos x = 4 $$

Hence, given equation has no solution.
Question 5
1 / -0

Principal solutions of the equation $$\sin { 2x } +\cos { 2x } =0$$, where $$\pi< x< 2\pi$$ are

A
$$\cfrac { 7\pi }{ 8 } ,\cfrac { 11\pi }{ 8 } $$

B
$$\cfrac {9 \pi }{ 8 } ,\cfrac { 13\pi }{ 8 } $$

C
$$\cfrac { 11\pi }{ 8 } ,\cfrac { 15\pi }{ 8 } $$

D
$$\cfrac { 15\pi }{ 8 } ,\cfrac { 19\pi }{ 8 } $$

Solution

Given : $$\pi< x< 2\pi$$
$$\implies 2\pi<2x<4\pi$$ ...... $$(i)$$
Also, $$\sin 2x+\cos 2x=0$$ ..... $$(ii)$$ [Given]
$$\implies \sin 2x=-\cos 2x$$
$$\implies \dfrac{\sin 2x}{\cos 2x}=-1$$
$$\implies \tan 2x=-1$$
$$\implies 2x=\tan^{-1}(-1)$$
Now, we know that
$$\tan \dfrac{3\pi}{4}=\tan \dfrac{7\pi}{4}=\tan \dfrac{11\pi}{4}=...... =\tan \dfrac{(4n+3)\pi}{4}=-1$$
So, the possible values for $$2x$$ are $$\dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{4},.....,\dfrac{(4n+3)\pi}{4}$$
But, from $$(i)$$ we get the values for $$2x$$ as $$\dfrac{11\pi}{4}$$ and $$\dfrac{15 \pi}{4}$$.
$$\therefore 2x=\dfrac{11\pi}{4}, \dfrac{15\pi}{4}$$
$$\implies x=\dfrac{11\pi}{8}, \dfrac{15\pi}{8}$$
$$\therefore$$ Principal solutions for $$(ii)$$ are $$\dfrac{11\pi}{8}$$ and $$\dfrac{15\pi}{8}$$.
Question 6
1 / -0

The number of solutions $$\left[ \cos { x } \right] +\left| \sin { x } \right| =1$$ in $$\pi \le x < 3\pi $$, where $$[x]$$ is the greatest integer not exceeding $$x$$ is

A
$$3$$

B
$$4$$

C
$$2$$

D
$$1$$

Solution

We first plotted the graph of $$\sin { x } $$ and $$\cos { x } $$
Then using the properties of integral part and mode function we get the graph of $$\left[ \cos { x } \right] $$ and $$\left| \sin { x } \right| $$
Clearly they collectively give $$1$$ at $$x=\dfrac { 3\pi }{ 2 } ,2\pi ,\dfrac { 5\pi }{ 2 } $$ in the interval $$\left[ \pi ,3\pi \right] $$
So option $$A$$ is correct
Question 7
1 / -0

Principal solutions of the equation $$\sin 2x+\cos2x=0$$, where $$\pi < x < 2\pi$$ are

A
$$7\dfrac{\pi}{8}, 11\dfrac{\pi}{8}$$

B
$$9\dfrac{\pi}{8}, 13\dfrac{\pi}{8}$$

C
$$11\dfrac{\pi}{8}, 15\dfrac{\pi}{8}$$

D
$$15\dfrac{\pi}{8}, 19\dfrac{\pi}{8}$$

Solution

$$\sin { 2x } +\cos { 2x } =0$$
$$\sin { 2x } =-\cos { 2x } $$
$$\cfrac { \sin { 2x } }{ \cos { 2x } } =-1$$
$$\tan { 2x } =-1\longrightarrow 1$$
$$\pi <x<2\pi $$
$$2\pi <2x<4\pi $$
$$\tan { x } $$ is negative in second and fourth quadrant
$$\therefore \tan { \cfrac { 3\pi }{ 4 } }=\tan { \cfrac { 7\pi }{ 4 } } =.....=\tan { \left( 4n-1 \right) } \cfrac { \pi }{ 4 } =-1 $$
$$n=1,2,3,....$$
But $$2\pi <2x<4\pi $$
$$\therefore 2x=\cfrac { 11\pi }{ 4 } $$ or $$2x=\cfrac { 15\pi }{ 4 } $$
$$x=\cfrac { 11\pi }{ 8 } $$ or $$\cfrac { 15\pi }{ 8 } $$
Question 8
1 / -0

$$\displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+\cos\theta \right )=$$

A
$$\tan^{2}\theta-\sec^{2}\theta$$

B
$$\sin^{2}\theta-\cos^{2}\theta$$

C
$$\sec^{2}\theta-\tan^{2}\theta$$

D
$$\cos^{2}\theta-\sin^{2}\theta$$

Solution

We have,
$$\displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+\cos\theta \right )$$

$$\Rightarrow \displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos^2\theta \right )$$ $$\because\ a^2-b^2=(a+b)(a-b)$$

$$\Rightarrow \displaystyle \left ( 1+\cot^{2}\theta \right )\left ( \sin^2\theta \right )$$ $$\because\ \sin^2\theta=1-\cos^2\theta$$

$$\Rightarrow \displaystyle \left ( \csc^{2}\theta \right )\left ( \sin^2\theta \right )$$ $$\because\ 1+\cot^2\theta=\csc^2\theta$$

$$\Rightarrow \displaystyle \dfrac{1}{\sin^2\theta}\times \sin^2\theta$$ $$\because\ \csc^2\theta=\dfrac{1}{\sin^2\theta}$$

$$\Rightarrow 1$$

$$\Rightarrow \displaystyle \sec^2\theta-\tan^2\theta$$ $$\because\ \sec^2\theta-\tan^2\theta=1$$

Hence, this is the answer.
Question 9
1 / -0

The number of roots of the equation $$x + 2\tan x = \dfrac {\pi}{2}$$ in the interval $$[0, 2\pi]$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
Infinite

Solution

Given, $$x+2 \tan x=\dfrac{\pi}{2}$$
This equation also can be written as:
$$\tan x= \dfrac{\pi}{4}-\dfrac{x}{2}$$
plot the graph b/w $$y=\tan x \text{ and } y= \dfrac{\pi}{4}-\dfrac{x}{2} $$
straight line cuts $$y=\tan x\,,3$$ times in the interval $$[0,2\pi]$$.
$$\therefore$$ no. of roots=$$3$$.
Question 10
1 / -0

If $$0\le x\le 2\pi $$, then the number of solutions of the equation $$\sin ^{ 6 }{ x } +\cos ^{ 6 }{ x } =1$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

E
$$8$$

Solution

Given equation is $$\sin ^{ 8 }{ x } +\cos ^{ 6 }{ x } =1$$
$$\Rightarrow \sin ^{ 8 }{ x } +{ \left( 1-\sin ^{ 2 }{ x } \right) }^{ 3 }=1$$
$$\Rightarrow \sin ^{ 8 }{ x } -\sin ^{ 6 }{ x } +3\sin ^{ 4 }{ x } -3\sin ^{ 2 }{ x } =0$$
$$\Rightarrow \sin ^{ 6 }{ x } \left( \sin ^{ 2 }{ x } -1 \right) +3\sin ^{ 2 }{ x } \left( \sin ^{ 2 }{ x } -1 \right) =0$$
$$\Rightarrow \left( \sin ^{ 6 }{ x } +3\sin ^{ 2 }{ x } \right) \left( \sin ^{ 2 }{ x } -1 \right) =0$$
$$\Rightarrow \sin ^{ 2 }{ x } \left( \sin ^{ 4 }{ x } +3 \right) \left( \sin ^{ 2 }{ x } -1 \right) =0$$
$$\Rightarrow \left( \sin ^{ 4 }{ x } +3 \right) \left[ \sin ^{ 2 }{ x } \left( \sin { x } -1 \right) \left( \sin { x } +1 \right) \right] =0$$
$$\Rightarrow \sin ^{ 2 }{ x } \left( \sin { x } -1 \right) \left( \sin { x } +1 \right) =0$$ $$\left[ \because \sin ^{ 4 }{ x } +3\neq 0 \right] $$
From above, it is clear that it has three roots in $$\left[ 0,2\pi \right] $$.

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Trigonometric Functions Test 9

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SHARING IS CARING

Question 1

$$1$$

$$0$$

$$9$$

$$-9$$

Solution

Question 2

$$\cos \theta =\dfrac {3}{2\sqrt {2}}$$

$$\cos \left (\theta - \dfrac {\pi}{2}\right ) = \dfrac {1}{2\sqrt {2}}$$

$$\cos \left (\theta - \dfrac {\pi}{4}\right ) = \dfrac {1}{2\sqrt {2}}$$

$$\cos \left (\theta + \dfrac {\pi}{4}\right ) = -\dfrac {1}{2\sqrt {2}}$$

Solution

Question 3

One

Two

Three

Four

Solution

Question 4

Infinitely many solutions

No solution

Two solutions

Only one solution

Solution

Question 5

$$\cfrac { 7\pi }{ 8 } ,\cfrac { 11\pi }{ 8 } $$

$$\cfrac {9 \pi }{ 8 } ,\cfrac { 13\pi }{ 8 } $$

$$\cfrac { 11\pi }{ 8 } ,\cfrac { 15\pi }{ 8 } $$

$$\cfrac { 15\pi }{ 8 } ,\cfrac { 19\pi }{ 8 } $$

Solution

Question 6

$$3$$

$$4$$

$$2$$

$$1$$

Solution

Question 7

$$7\dfrac{\pi}{8}, 11\dfrac{\pi}{8}$$

$$9\dfrac{\pi}{8}, 13\dfrac{\pi}{8}$$

$$11\dfrac{\pi}{8}, 15\dfrac{\pi}{8}$$

$$15\dfrac{\pi}{8}, 19\dfrac{\pi}{8}$$

Solution

Question 8

$$\tan^{2}\theta-\sec^{2}\theta$$

$$\sin^{2}\theta-\cos^{2}\theta$$

$$\sec^{2}\theta-\tan^{2}\theta$$

$$\cos^{2}\theta-\sin^{2}\theta$$

Solution

Question 9

$$1$$

$$2$$

$$3$$

Infinite

Solution

Question 10

$$2$$

$$3$$

$$4$$

$$5$$

$$8$$

Solution

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