Trigonometric Functions Test 11

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Trigonometric Functions Test 11

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Weekly Quiz Competition

Question 1
1 / -0

The principal solution of $$\sec x = \dfrac {2}{\sqrt {3}}$$ are

A
$$\dfrac {\pi}{3}, \dfrac {11\pi}{6}$$

B
$$\dfrac {\pi}{6}, \dfrac {11\pi}{6}$$

C
$$\dfrac {\pi}{4}, \dfrac {11\pi}{4}$$

D
$$\dfrac {\pi}{3}, \dfrac {11\pi}{4}$$

Solution

Given $$\sec x = \dfrac {2}{\sqrt {3}}$$
We know that, $$\sec\: \dfrac{\pi}{6}=\dfrac{2}{\sqrt{3}}$$ also $$\sec \left(2\pi-\dfrac{\pi}{6}\right)=\sec\: \dfrac{11\pi}{6}=\dfrac{2}{\sqrt{3}}$$

Therefore the principal solutions are $$\dfrac{\pi}{6}, \dfrac{11\pi}{6}$$
Question 2
1 / -0

If $$\sin x =$$ $$\cos^2x$$, then $$\cos^2x (1 + \cos^2x)$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
None of these

Solution

Given, $$\sin x=\cos^2x$$
Thus $$\sin x=1-\sin ^2x$$
$$\Rightarrow \sin ^2x+\sin x=1$$ ....(1)
We need to find value of $$\cos^2x(1+\cos^2x)$$
$$ \Rightarrow \sin x(1+\sin x)$$
$$ \Rightarrow \sin x+\sin^2 x$$
From the equation (1), we have
$$\sin x+\sin^2x=1$$
Option B is correct.
Question 3
1 / -0

Find the value of $$\tan { 1 } \tan { 2 } ....\tan { 89 } $$

A
$$-1$$

B
$$1$$

C
$$\sqrt 2$$

D
$$2$$

Solution

$$\tan 1\cdot \tan 2.....\tan 89$$
but $$\tan 89$$ can be written as $$\cot 1[\tan (90-1)=\cot 1]$$ by $$\tan(90-\theta)=\cot(\theta)$$
likewise...,
$$\tan 88=\cot 2$$
$$\tan 87=\cot 3$$
....... up to $$\tan 46=\cot 44$$
then middle one is $$\tan 45=1$$
so it is
$$\tan 1\cdot \tan 2....\tan 44\cdot \tan 45\cdot \cot 44.....\cot 1$$
$$\tan$$ and $$\cot$$ cancel out by$$[\tan(\theta)^{\ast}\cot(\theta)=1]$$
so, remaining is $$1$$
answer$$=1$$.
Question 4
1 / -0

The number of solution of the equation $$\tan x + \sec x = 2\cos x$$ lying in the interval $$[0, 2\pi]$$ is

A
Zero

B
One

C
Two

D
Three

Solution

$$\tan{x}+\sec{ x}=2\cos{x}$$

$$\Rightarrow$$ $$ \dfrac{\sin {x}}{\cos {x}} +\dfrac {1}{\cos{x}}=2 \cos{x}$$

$$\Rightarrow$$ $$\dfrac{(\sin{x}+1)}{\cos{x}} = 2\cos{x}$$

$$\Rightarrow$$ $$1+\sin{x}=2 {(\cos{x})}^{2}$$

$$\Rightarrow$$ $$1+\sin{x}=2(1-{(\sin{x})}^{2})$$

$$\Rightarrow$$ $$1+\sin{x} -2(1+\sin{x})(1-\sin{x})=0$$

$$\Rightarrow$$ $$(1+\sin{x})(1-2(1-\sin{x}))=0$$

$$\Rightarrow$$ $$(1+\sin{x})(2\sin{x}-1)=0$$

$$\Rightarrow$$ $$\sin{x}=1, \sin{x}=\dfrac{1}{2}$$

Since, $$x$$ is between $$[0,2\pi]$$,

The solutions are $$ x=\dfrac{3\pi}{2}, x=\dfrac{\pi}{6}, x=\dfrac{5\pi}{6}$$.

Therefore there are 3 solutions.
Question 5
1 / -0

The equation $$2\tan { x } +5x-2=0$$ has:

A
no solution in $$\left[ 0,\dfrac{\pi}{4} \right] $$

B
at least one real solution in $$\left[ 0,\dfrac{\pi}{4} \right] $$

C
two real solution in $$\left[ 0,\dfrac{\pi}{4} \right] $$

D
None of these

Solution

Given that

$$ 2 \tan x + 5x -2 = 0$$

$$ \tan x = \dfrac{-5}{2}x + 1$$

The functions $$ y = \tan x $$ and $$ y = \dfrac{-5}{2}x + 1$$ can be plotted on the same graph.

The solution to the equation $$ \tan x = \dfrac{-5}{2}x + 1$$ will be at the intersection of the two curves.

The graph is given in the figure above.

From the graph, it is seen that the line intersects the x-axis at
$$ x = \dfrac{2}{5} $$.
For some $$x$$ such that $$ 0 < x <\dfrac{2}{5} $$, the 2 curves meet.

Since $$ 0 <\dfrac{2}{5} < \dfrac{\pi}{4} $$,
Hence, B is true, i.e there is at least one real solution in $$\left[0, \dfrac{\pi}{4}\right]$$.
Question 6
1 / -0

The expression $${ \left( \tan { \theta } +\sec { \theta } \right) }^{ 2 } $$is equal to

A
$$\cfrac { 1+\cos { \theta } }{ 1-\cos { \theta } } $$

B
$$\cfrac { 1+\sin { \theta } }{ 1-\sin { \theta } } $$

C
$$\cfrac { 1-\cos { \theta } }{ 1+\cos { \theta } } $$

D
$$\cfrac { 1-\sin {\theta } }{ 1+\sin { \theta } } $$

Solution

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

If $$\sin { \theta } =\cfrac { 8 }{ 17 } $$ where $${ 0 }^{ o }<\theta <{ 90 }^{ o }$$, then $$\tan { \theta } +\sec { \theta } $$ is

A
$$\dfrac {1}{3}$$

B
$$\dfrac {2}{3}$$

C
$$\dfrac {4}{3}$$

D
$$\dfrac {5}{3}$$

Solution

$$\sin\theta =8/17,\quad \tan\theta =\dfrac { sin\theta }{ \sqrt { 1-{ sin }^{ 2 }\theta } } =8/15$$
$$\sec\theta =\dfrac { 1 }{ \cos\theta } =\dfrac { 1 }{ \sqrt { 1-{
\sin }^{ 2 }\theta } } =\dfrac { 17 }{ 15 } $$
$$\tan\theta +\sec\theta =\dfrac { 8 }{ 15 } +\dfrac { 17 }{ 15 } =\dfrac { 25 }{ 15 } =\dfrac { 5 }{ 3 } $$
Question 8
1 / -0

$$\sec^26 + \text{cosec}^26 $$ is equal to:

A
$$\sec^26. \cot^26$$

B
$$\sec^26 .\tan^26$$

C
$$\text{cosec}^26. \cot^26$$

D
$$\text{cosec}^26. \sec^26$$

Solution

$$ { \sec }^{ 2 }6+ \text{ cosec}^{ 2 }6$$

$$= \dfrac { 1 }{ { \cos }^{ 2 }6 } +\dfrac { 1 }{ { \sin }^{ 2 }6 } $$

$$= \dfrac { { \sin }^{ 2 }6+{ \cos }^{ 2 }6 }{ { \sin }^{ 2 }6{ \cos }^{ 2 }6 } $$

$$={ \sec }^{ 2 }6 \text{ cosec}^{ 2 }6$$
Question 9
1 / -0

$$\sin { \theta } +\cos { \theta } =\sqrt { 2 } $$ and $$\theta$$ is acute, then $$\tan{\theta}$$ is

A
$$\dfrac {1}{\sqrt { 3 } }$$

B
$$1$$

C
$$\sqrt { 3 } $$

D
$$\infty$$

Solution

$$\sin { \theta } +\cos { \theta } =\sqrt { 2 } $$
$$\Rightarrow \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } \cos { \theta } =2\quad \Rightarrow 2\sin { \theta } \cos { \theta } =1\quad $$
$$\Rightarrow \sin { 2\theta } =1\Rightarrow \cfrac { 2\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } =1\Rightarrow \tan ^{ 2 }{ \theta }=1 \Rightarrow \tan { \theta =\pm 1\quad } $$
since $$\theta$$ is acute $$\Rightarrow \tan { \theta } =1$$
Question 10
1 / -0

$$\dfrac {\tan \theta}{1 +\tan^{2} \theta} + \dfrac {\cot \theta}{(1 + \cot^{2}\theta)^{2}}$$ is equal to

A
$$2\sin \theta \cdot \cos \theta$$

B
$$\text{cosec} \theta \cdot \sec \theta$$

C
$$\sin \theta \cdot \cos \theta$$

D
$$2\text{cosec} \theta \cdot \sec \theta$$

Solution

$$\dfrac {\tan \theta}{(1 + \tan^{2}\theta)^{2}} + \dfrac {\cot \theta}{(1 + \cot^{2}\theta)^{2}} = \dfrac {\tan \theta}{\sec^{4}\theta} + \dfrac {\cot \theta}{\text{cosec}^{4}\theta}$$

$$= \dfrac {\text{cosec} \theta \sec \theta \left (\dfrac {1}{\sin^{2}\theta} + \dfrac {1}{\cos^{2}\theta}\right )}{\sec^{4}\theta \text{cosec}^{4}\theta} = \dfrac {\sec^{2}\theta \text{cosec}^{2}\theta}{\sec^{3} \theta \text{cosec}^{3}\theta} = \sin \theta \cos \theta$$.

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Trigonometric Functions Test 11

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Trigonometric Functions Test 11

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Question 1

$$\dfrac {\pi}{3}, \dfrac {11\pi}{6}$$

$$\dfrac {\pi}{6}, \dfrac {11\pi}{6}$$

$$\dfrac {\pi}{4}, \dfrac {11\pi}{4}$$

$$\dfrac {\pi}{3}, \dfrac {11\pi}{4}$$

Solution

Question 2

$$0$$

$$1$$

$$2$$

None of these

Solution

Question 3

$$-1$$

$$1$$

$$\sqrt 2$$

$$2$$

Solution

Question 4

Zero

One

Two

Three

Solution

Question 5

no solution in $$\left[ 0,\dfrac{\pi}{4} \right] $$

at least one real solution in $$\left[ 0,\dfrac{\pi}{4} \right] $$

two real solution in $$\left[ 0,\dfrac{\pi}{4} \right] $$

None of these

Solution

Question 6

$$\cfrac { 1+\cos { \theta } }{ 1-\cos { \theta } } $$

$$\cfrac { 1+\sin { \theta } }{ 1-\sin { \theta } } $$

$$\cfrac { 1-\cos { \theta } }{ 1+\cos { \theta } } $$

$$\cfrac { 1-\sin {\theta } }{ 1+\sin { \theta } } $$

Solution

Question 7

$$\dfrac {1}{3}$$

$$\dfrac {2}{3}$$

$$\dfrac {4}{3}$$

$$\dfrac {5}{3}$$

Solution

Question 8

$$\sec^26. \cot^26$$

$$\sec^26 .\tan^26$$

$$\text{cosec}^26. \cot^26$$

$$\text{cosec}^26. \sec^26$$

Solution

Question 9

$$\dfrac {1}{\sqrt { 3 } }$$

$$1$$

$$\sqrt { 3 } $$

$$\infty$$

Solution

Question 10

$$2\sin \theta \cdot \cos \theta$$

$$\text{cosec} \theta \cdot \sec \theta$$

$$\sin \theta \cdot \cos \theta$$

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

Solution

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