Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 26

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Question 1
1 / -0

If $${\rm{sin}}\,{\rm{\theta + cosec\theta = 2,}}\,{\rm{then}}\,\,{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta + cose}}{{\rm{c}}^{\rm{2}}}{\rm{\theta }}\,$$ is equal to:

A
$$1$$

B
$$4$$

C
$$2$$

D
None of these

Solution

We have,
$$\sin \theta +\csc \theta=2$$

On squaring both sides, we get
$$(\sin \theta +\csc \theta)^2=2^2$$

$$\sin^2 \theta +\csc^2 \theta+2\sin \theta\csc \theta=4$$

$$\sin^2 \theta +\csc^2 \theta+2\times 1=4$$

$$\sin^2 \theta +\csc^2 \theta+2=4$$

$$\sin^2 \theta +\csc^2 \theta=2$$

Hence, this is the answer.
Question 2
1 / -0

$$\sin x=\dfrac {\sqrt 3}{2}$$ find $$\sec x$$

A
$$2$$

B
$$\dfrac 12$$

C
$$\sqrt 2$$

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

Solution

$$\sin x=\dfrac {\sqrt 3}{2}\\\sin x=\sin 60^{\circ}\\x=60\\\sec 60=2$$
Question 3
1 / -0

Solve : $$\dfrac { 2tan{ 30 }^{ \circ } }{ 1+{ tan }^{ 2 }{ 30 }^{ \circ } } $$

A
sin $${ 60 }^{ \circ }$$

B
cos $${ 60 }^{ \circ }$$

C
tan $${ 60 }^{ \circ }$$

D
cot $${ 60 }^{ \circ }$$

Solution

We know that,
$$tan(2A) = \dfrac{2tanA}{1+tan^2A}$$
Putting A to be $$30^o$$
We get,
$$tan(60^o) = \dfrac{2tan30^o}{1+tan^230^o}$$

Thus, C is the correct answer.
Question 4
1 / -0

If $$y=cos x$$ then $$x=\frac{\pi}{2}$$, $$y$$ is equal to:

A
$$-1$$

B
$$1$$

C
$$0$$

D
$$\frac{1}{2}$$

Solution

$$y = \cos{x} \quad \left( \text{Given} \right)$$
For $$x = \cfrac{\pi}{2}$$,
$$y = \cos{\left( \cfrac{\pi}{2} \right)} = 0$$
Hence the correct answer is $$0$$.
Question 5
1 / -0

$$\sin x=\dfrac 12\\\tan x=?$$

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

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

C
$$\dfrac{1}{\sqrt{6}}$$

D
$$\dfrac{1}{\sqrt{5}}$$

Solution

$$\sin x=\dfrac 12\\\sin x=\sin 30^{\circ}\\x=30\\\tan 30=\dfrac 1 {\sqrt3}$$
Question 6
1 / -0

sin0$$^0$$ . sin1$$^0$$ ..................... sin90$$^0$$ =

A
0

B
tan0$$^0$$

C
csc90$$^0$$

D
cos90$$^0$$

Solution

Given,

$$\sin 0 \cdot \sin 1.....................\cdot \sin 90$$

Since,

$$\sin 0=0$$

Hence the product is $$0$$
Question 7
1 / -0

The value of $$\tan^{2}60^{o}$$ is:

A
$$3$$

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

C
$$1$$

D
$$\infty$$

Solution

$$\tan^{2}60^{o}=(\surd 3)^{2}=3$$
$$(A)$$ is correct.
Question 8
1 / -0

Given
$$\displaystyle \frac{\sin A}{si{nB}}=\frac{\sqrt{3}}{2}, \displaystyle \frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2},\ 0<\displaystyle A,B<\frac{\pi}{2}$$, then

A
$$\tan A=\sqrt{\frac{3}{5}}$$

B
$$\tan A=\sqrt{\frac{5}{3}}$$

C
$$\tan A=2$$

D
$$\tan B=2$$

Solution

straight away we get $$\sin B = \dfrac{2sinA}{\sqrt{3}} \ and \ \cos B = \dfrac{2cosA}{\sqrt{5}}$$

now $$ \sin^{2}B + \cos^{2}B = 1$$

so we get $$1 = 4(\dfrac{\sin^{2}A}{3} + \dfrac{\cos^{2}A}{5})$$

we divide throughout by $$\cos^{2}A$$ and put $$\sec^{2}A = 1 + \tan^{2}A$$

$$\dfrac{\sec^2A}4 = (\dfrac{\tan^{2}A}{3} + \dfrac{1}{5})$$

$$\dfrac{\tan^{2}A}{3}=\dfrac{1+\tan^2A}4-\dfrac15$$

$$\left (\dfrac13-\dfrac14\right )\tan^2 A=\dfrac14-\dfrac15$$

$$\dfrac{\tan^2A}{12}=\dfrac{1}{20}$$

$$\tan A=\sqrt{\dfrac{3}{5}}$$
Question 9
1 / -0

The maximum value of
$$\cos x\,\left(\displaystyle \dfrac{\cos x}{1-\sin x}+\dfrac{1-\sin x}{\cos x}\right)$$ is

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$\displaystyle f(x)=\cos x(\frac{\cos ^{2}x+1+\sin ^{2}x-2\sin x}{\cos x(1-\sin x)}) $$$$\displaystyle =\cos x(\frac{2(1-\sin x)}{\cos x(1-\sin x)}) =\cos x(\frac{2}{\cos x})=2$$
Question 10
1 / -0

If $$\text{cosec}\displaystyle A=4p+\frac{1}{16p}$$, then the value of $$\text{cosec}A+\cot A$$ is

A
$$8p$$

B
$$\displaystyle \frac{1}{8p}$$

C
$$(-8p)$$ or $$ (\displaystyle \frac{1}{8p})$$

D
$$(8p)$$ or $$(\displaystyle \frac{1}{8p})$$

Solution

$$\text{cosec}A=4p+\dfrac{1}{16p}$$

$$\text{cosec}^{2}A=16p^{2}+\dfrac{1}{256p^{2}}+\dfrac{1}{2}$$
Or
$$1+\cot ^{2}A=16p^{2}+\dfrac{1}{256p^{2}}+\dfrac{1}{2}$$
Or
$$\cot ^{2}A=16p^{2}+\dfrac{1}{256p^{2}}-\dfrac{1}{2}$$
Or
$$\cot ^{2}A=(4p-\dfrac{1}{16p})^{2}$$
Or
$$\cot A=\pm(4p-\dfrac{1}{16p})$$...(i)
Hence
If $$\cot A=(4p-\dfrac{1}{16p})$$
Then

$$\text{cosec}A+\cot A$$

$$=4p+\dfrac{1}{16p}+(4p-\dfrac{1}{16p})$$

$$=8p$$ ...(ii)

If $$\cot A=-(4p-\dfrac{1}{16p})$$

Then

$$\text{cosec}A+\cot A$$

$$=4p+\dfrac{1}{16p}-(4p-\dfrac{1}{16p})$$

$$=\dfrac{2}{16p}$$

$$=\dfrac{1}{8p}$$

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Introduction to Trigonometry Test - 26

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Introduction to Trigonometry Test - 26

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Question 1

$$1$$

$$4$$

$$2$$

None of these

Solution

Question 2

$$2$$

$$\dfrac 12$$

$$\sqrt 2$$

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

Solution

Question 3

sin $${ 60 }^{ \circ }$$

cos $${ 60 }^{ \circ }$$

tan $${ 60 }^{ \circ }$$

cot $${ 60 }^{ \circ }$$

Solution

Question 4

$$-1$$

$$1$$

$$0$$

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

Solution

Question 5

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

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

$$\dfrac{1}{\sqrt{6}}$$

$$\dfrac{1}{\sqrt{5}}$$

Solution

Question 6

0

tan0$$^0$$

csc90$$^0$$

cos90$$^0$$

Solution

Question 7

$$3$$

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

$$1$$

$$\infty$$

Solution

Question 8

$$\tan A=\sqrt{\frac{3}{5}}$$

$$\tan A=\sqrt{\frac{5}{3}}$$

$$\tan A=2$$

$$\tan B=2$$

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 10

$$8p$$

$$\displaystyle \frac{1}{8p}$$

$$(-8p)$$ or $$ (\displaystyle \frac{1}{8p})$$

$$(8p)$$ or $$(\displaystyle \frac{1}{8p})$$

Solution

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