Trigonometric Functions Test 25

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Trigonometric Functions Test 25

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\dfrac{\cos^{2}\theta+\tan^{2}\theta\ -1}{\sin^{2}\theta}$$ is

A
$$0$$

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

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

D
$$\dfrac{1}{\sin^{2}\theta}$$

Solution

$$\dfrac{\cos^2 \theta+\tan^2 \theta-1}{\sin^2 \theta}=\dfrac{\cos^2 \theta}{\sin^2 \theta}+\dfrac{\sin^2 \theta}{\sin^2 \theta\cos^2 \theta}-\dfrac{1}{\sin^2 \theta}$$
$$=\cot^2 \theta+\text{sec}^2 \theta-\text{cosec}^{2}\theta$$
$$=\text{sec}^2 \theta-(\text{cosec}^2 \theta-\cot^2 \theta)$$
$$=\text{sec}^2 \theta-1$$
$$=\tan^2 \theta$$
Question 2
1 / -0

If $${ cosec }\theta -cot\theta =p$$, then $$cosec\theta +cot\theta =$$

A
1/p

B
-1/p

C
-p

D
$${ p }^{ 2 }$$

Solution

$$\csc{\theta}-\cot{\theta}=p$$
$$\Rightarrow \left(\csc{\theta}-\cot{\theta}\right)\left(\csc{\theta}+\cot{\theta}\right)=p\left(\csc{\theta}+\cot{\theta}\right)$$
$$\Rightarrow {\csc}^{2}{\theta}-{\cot}^{2}{\theta}=p\left(\csc{\theta}+\cot{\theta}\right)$$
$$\Rightarrow 1=p\left(\csc{\theta}+\cot{\theta}\right)$$
$$\therefore \csc{\theta}+\cot{\theta}=\dfrac{1}{p}$$
Question 3
1 / -0

The value of $$\sin \theta$$ is not equal to

A
$$\pm \frac{1}{2}$$

B
$$\frac{1}{\sqrt 2}$$

C
$$\pm \frac{1}{3}$$

D
$$2$$

Solution
Question 4
1 / -0

The value of $$\sin\theta$$ is not equal to

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

B
$$\dfrac {1}{\sqrt {2}}$$

C
$$\pm\dfrac {1}{3}$$

D
$$2$$

Solution
Question 5
1 / -0

If $$x_{1}$$ and $$x_{2}$$ are two distinct roots of the equation $$a\cos x+b\sin x=c$$, then $$\tan\dfrac {x_{1}+x_{2}}{2}$$ is equal to

A
$$a/b$$

B
$$b/a$$

C
$$c/a$$

D
$$a/c$$

Solution

$$a \cos x + b \sin x =c$$
$$\dfrac{1-\tan^2\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}} +b \dfrac{2\tan \dfrac{x}{2}}{1+\tan^2 \dfrac{x}{2}}=c$$

$$(c+a) \tan^2 \dfrac{x}{2} -2b \tan \dfrac{x}{2} +(c-a) =0$$

Since, $$x_1$$ and $$x_2$$ are roots,
$$\tan \dfrac{x_1}{2}+\tan \dfrac{x_2}{2} =\dfrac{2b}{c+a}$$

and $$\tan \dfrac{x_1}{2}.\tan \dfrac{x_2}{2} =\dfrac{c-a}{c+a}$$

Hence,
$$\tan (\dfrac{x_1+x_2}{2}) = \dfrac{\tan \dfrac{x_1}{2} +\tan \dfrac{x_2}{2}}{1-\tan \dfrac{x_1}{2} \tan \dfrac{x_2}{2}}$$

$$=\dfrac{\dfrac{2b}{c+a}}{1-\dfrac{c-a}{c+a}}=\dfrac{b}{a}$$

Hence, option (b) is correct.
Question 6
1 / -0

The values of x in $$\left(0, \dfrac{\pi}{2}\right)$$ satisfying the equation $$\sin x\cos x=\dfrac{1}{4}$$ are ________.

A
$$\dfrac{\pi}{6}, \dfrac{\pi}{12}$$

B
$$\dfrac{\pi}{12}, \dfrac{5\pi}{12}$$

C
$$\dfrac{\pi}{8}, \dfrac{3\pi}{8}$$

D
$$\dfrac{\pi}{8}, \dfrac{\pi}{4}$$

Solution

We know that, $$\displaystyle \text{sin}(2x) = 2\text{ sin }x\text{ cos }x$$.
So, from $$\displaystyle \text{sin }x \text{ cos }x = \frac{1}{4}$$, we get that $$\text{ sin }(2x) = \frac{1}{2}$$.
So, we have to find the solutions of the equation $$\displaystyle\text{sin} (2x) = \frac{1}{2}$$, where $$\displaystyle x \in (0, \frac{\pi}{2})$$.

From this data, we get that $$\displaystyle 2x = \frac{\pi}{6}$$ or $$\dfrac{5\pi}{6}$$.
That is, $$x = \dfrac{\pi}{12}$$ or $$\dfrac{5\pi}{12}$$.
Question 7
1 / -0

The angles of a triangle are in $$\displaystyle AP$$ and the ratio of the number of degrees in the least to the number of radius in the greatest is $$\displaystyle 60 : \pi$$. The smallest angle is

A
$$\displaystyle 15^0$$

B
$$\displaystyle 30^0$$

C
$$\displaystyle 45^0$$

D
$$\displaystyle 60^0$$

Solution

Let the angles be $$\displaystyle (a - d)^0 , a^0$$ and $$\displaystyle (a + d)^0 $$. Then,
$$\displaystyle a - d + a + a + d = 180 \Rightarrow 3a = 180 \Rightarrow a = 60$$.
So, the angles are $$\displaystyle (60 - d)^0, 60^0$$ and $$\displaystyle(60 + d)^0$$.
$$\displaystyle 180^0 = \pi^c \Rightarrow (60 + d)^0 =\left \{\frac{\pi}{180} \times (60 + d) \right\}^c = \left\{\frac{(60 + d) \pi}{180}\right\}^c$$
$$\therefore$$ $$\displaystyle \frac{(60 - d)}{(60 + d) \frac{\pi}{180}} = \frac{60}{\pi}\Rightarrow \frac{3(60 - d)}{(60 + d)} = 1$$
$$\displaystyle \Rightarrow 180 - 3d = 60 + d \Rightarrow 4d = 120 \Rightarrow d = 30$$.
$$\therefore$$ the smallest angle = $$\displaystyle (60 - 30)^0 = 30^0 $$.
Question 8
1 / -0

$$\sin 15^{\circ} = $$

A
$$\dfrac{\sqrt{3} - 1}{2 \sqrt{2}} $$

B
$$\dfrac{\sqrt{3} + 1}{2 \sqrt{2}} $$

C
$$\dfrac{1- \sqrt{3}}{2 \sqrt{2}} $$

D
$$\dfrac{1+ \sqrt{3}}{ \sqrt{2}} $$

E
$$\dfrac{-(1+ \sqrt{3} )}{2 \sqrt{2}} $$

Solution

$${\textbf{Step -1:}}{\textbf{ Place the values according to the formula to find the value}}{\textbf{.}}$$
$${\text{sin}}\left( {45 - 30} \right) = \sin 45\cos 30 - \cos 45\sin 30$$ $$\left[ {\left( {{\mathbf{\sin A - B = \sin A\cos B - \cos A\sin B}}} \right)} \right]$$
$$\sin 15^\circ = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$$
$$ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$$
$${\textbf{Hence, the value of sin 15}}^\circ {\textbf{ is }}{\mathbf{\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}.}}$$
Question 9
1 / -0

What is $$\cot(\dfrac{A}{2})-\tan(\dfrac{A}{2})$$ equal to ?

A
$$\tan A$$

B
$$\cot A$$

C
$$2\tan A$$

D
$$2\cot A$$

Solution

$$\cot \left(\dfrac{A}{2}\right)-\tan\left(\dfrac{A}{2}\right)$$

$$\Rightarrow$$ $$\dfrac{\cos \dfrac{A}{2}}{\sin\dfrac{A}{2}}-\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$$

$$\Rightarrow$$ $$\dfrac{\cos^2\dfrac{A}{2}-\sin^2\dfrac{A}{2}}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$

$$\Rightarrow$$ $$\dfrac{\cos 2\left(\dfrac{A}{2}\right)}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$

$$\Rightarrow$$ $$\dfrac{\cos A}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$

$$\Rightarrow$$ $$\dfrac{2\cos A}{2\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$

$$\Rightarrow$$ $$\dfrac{2\cos A}{\sin 2\left(\dfrac{A}{2}\right)}$$

$$\Rightarrow$$ $$\dfrac{2\cos A}{\sin A}$$

$$\Rightarrow$$ $$2\cot A$$
Question 10
1 / -0

If $$ x + 1 / x = 2 $$ , the principal value of $$ sin^{-1}x $$ is

A
$$ \pi / 4 $$

B
$$ \pi / 2 $$

C
$$ \pi $$

D
$$ 3\pi / 2 $$

Solution

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Trigonometric Functions Test 25

Result

Trigonometric Functions Test 25

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SHARING IS CARING

Question 1

$$0$$

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

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

$$\dfrac{1}{\sin^{2}\theta}$$

Solution

Question 2

1/p

-1/p

-p

$${ p }^{ 2 }$$

Solution

Question 3

$$\pm \frac{1}{2}$$

$$\frac{1}{\sqrt 2}$$

$$\pm \frac{1}{3}$$

$$2$$

Solution

Question 4

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

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

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

$$2$$

Solution

Question 5

$$a/b$$

$$b/a$$

$$c/a$$

$$a/c$$

Solution

Question 6

$$\dfrac{\pi}{6}, \dfrac{\pi}{12}$$

$$\dfrac{\pi}{12}, \dfrac{5\pi}{12}$$

$$\dfrac{\pi}{8}, \dfrac{3\pi}{8}$$

$$\dfrac{\pi}{8}, \dfrac{\pi}{4}$$

Solution

Question 7

$$\displaystyle 15^0$$

$$\displaystyle 30^0$$

$$\displaystyle 45^0$$

$$\displaystyle 60^0$$

Solution

Question 8

$$\dfrac{\sqrt{3} - 1}{2 \sqrt{2}} $$

$$\dfrac{\sqrt{3} + 1}{2 \sqrt{2}} $$

$$\dfrac{1- \sqrt{3}}{2 \sqrt{2}} $$

$$\dfrac{1+ \sqrt{3}}{ \sqrt{2}} $$

$$\dfrac{-(1+ \sqrt{3} )}{2 \sqrt{2}} $$

Solution

Question 9

$$\tan A$$

$$\cot A$$

$$2\tan A$$

$$2\cot A$$

Solution

Question 10

$$ \pi / 4 $$

$$ \pi / 2 $$

$$ \pi $$

$$ 3\pi / 2 $$

Solution

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