Trigonometric Functions Test -2

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Trigonometric Functions Test -2

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Question 1
1 / -0

$$(\text{cosec} \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta)$$ simplifies to

A
$$0$$

B
$$1$$

C
$$\tan \theta$$

D
$$\cot \theta$$

Solution

$$(\text{cosec } \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta)$$

$$=\left ( \dfrac {1}{\sin \theta}-\sin \theta \right )\left ( \dfrac {1}{\cos \theta}-\cos\theta \right )\left ( \dfrac {\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} \right )$$

$$=\dfrac{1-\sin^{2}\theta}{\sin \theta}\times\dfrac{1-\cos^{2} \theta}{\cos \theta}\times\dfrac{\sin^{2} \theta+\cos^{2} \theta}{\sin \theta \cos \theta}$$

$$=\dfrac{\cos^{2}\theta}{\sin \theta}\times\dfrac{\sin^{2} \theta}{\cos \theta}\times\dfrac{1}{\sin \theta \cos \theta}$$

$$=1$$
Question 2
1 / -0

The value of $$ \cos^4\theta-\sin^4 \theta+2 \sin^2 \theta$$ is equal to

A
$$0$$

B
$$\sin^2 \theta$$

C
$$\cos^2\theta$$

D
$$1$$

Solution

$$\cos^4\theta-\sin^4\theta+2\sin^2\theta$$
$$=(\cos^2\theta)^2-\sin^4\theta +2\sin^2\theta$$
$$=(1 - \sin^2\theta)^2 - \sin^4\theta +2\sin^2\theta$$
$$=1 - 2\sin^2\theta + \sin^4\theta-\sin^4\theta +2\sin^2\theta$$
$$=1$$
Hence answer is D
Question 3
1 / -0

Which of the following is/are FALSE $$\forall \theta \in R$$?

A
$$\sin \theta =1 - \cos \theta$$

B
$$\sec \theta - \tan \theta =\dfrac{1}{\sec \theta + \tan \theta}$$

C
$$\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$$

D
$$\sin \dfrac{\pi}{3}=\cos\dfrac{\pi}{6}$$

Solution

Dividing LHS by RHS we get,

in $$A\rightarrow \dfrac{sin\theta}{1-cos\theta}\neq 1$$

in $$B\rightarrow sec^2\theta-tan^2\theta=1$$

in $$C\rightarrow cosec^2\theta -cot^2\theta=1$$

in $$D\rightarrow \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{\sqrt3}{2}}=1$$

Therefore, Answer is $$sin\theta=1-cos\theta$$
Question 4
1 / -0

If $$\tan\theta=\dfrac{3}{4}$$ and $$0< \theta, < 90^0,$$ then the value of $$\sin\theta \cos \theta$$ is

A
$$\dfrac{1}{5}$$

B
$$\dfrac{9}{5}$$

C
$$\dfrac{12}{25}$$

D
$$\dfrac{25}{12}$$

Solution

Given, $$\tan \theta=\dfrac{3}{4}$$

$$\therefore \sec \theta=\sqrt{1+\tan^2\theta}$$

$$=\sqrt{1+\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}$$

or
$$\cos \theta=\dfrac{4}{5}$$

$$\therefore \sin \theta \cos \theta = \tan \theta \cos^2 \theta$$

$$=\dfrac{3}{4}\times \dfrac{16}{25}=\dfrac{12}{25}$$
Question 5
1 / -0

$$\dfrac{3-4 \sin^2 \theta}{\cos^2 \theta}$$ is equal to

A
$$3 - \cot^2 \theta$$

B
$$3 + \cot^2 \theta$$

C
$$3 - \tan^2 \theta$$

D
$$3+ \tan^2 \theta$$

Solution

The given expression is
$$ \dfrac { 3-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =\dfrac { 3\times 1-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =\dfrac { 3\times (\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } )-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =\dfrac { 3\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =3-\dfrac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =3-\tan ^{ 2 }{ \theta } . $$
Question 6
1 / -0

$$(1-\sin^2 \theta)(1+\tan^2 \theta)$$ is equal to

A
$$1$$

B
$$1.5$$

C
$$2$$

D
$$2.5$$

Solution

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

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

A
$$\left ( \dfrac{1-\cos \theta}{1+\cos \theta} \right )$$

B
$$\left ( \dfrac{1+\cos \theta}{1-\cos \theta} \right )$$

C
$$\left ( \dfrac{\cos \theta-1}{\cos \theta+1} \right )$$

D
$$\left ( \dfrac{\cos \theta+1}{\cos \theta-1} \right )$$

Solution

Given, $$ \dfrac { \tan ^{ 2 }{ \theta } }{ { \left( 1+\sec ^{ 2 }{ \theta } \right) }^{ 2 } } =\dfrac { \sec ^{ 2 }{ \theta } -1 }{ { \left( 1+\sec ^{ 2 }{ \theta } \right) }^{ 2 } } $$

$$ =\dfrac { \left( \sec { \theta } +1 \right) \left( \sec { \theta } -1 \right) }{ { \left( 1+\sec ^{ 2 }{ \theta } \right) }^{ 2 } } $$

$$ =\dfrac { \sec { \theta } -1 }{ \sec { \theta } +1 } $$

$$ =\dfrac { \dfrac { 1 }{ \cos\theta } -1 }{ \dfrac { 1 }{ \cos\theta } +1 } $$

$$ =\dfrac { 1-\cos\theta }{ 1+\cos\theta }$$
Question 8
1 / -0

$$\sec^2 \theta +\text{cosec}^2 \theta$$ is equal to

A
$$\sec^2 \theta. \cot^2 \theta$$

B
$$\sec^2\theta. \tan^2 \theta$$

C
$$\text{cosec}^2 \theta. \cot^2 \theta$$

D
$$\sec^2 \theta. \text{cosec}^2 \theta$$

Solution

$$\sec ^{ 2 }{ \theta +\text{cosec} ^{ 2 }{ \theta } } \\ =\dfrac { 1 }{ \cos ^{ 2 }{ \theta } } +\dfrac { 1 }{ \sin ^{ 2 }{ \theta } } =\dfrac { \sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta } } }{ \sin ^{ 2 }{ \theta .\cos ^{ 2 }{ \theta } } } \\ =\dfrac { 1 }{ \sin ^{ 2 }{ \theta .\cos ^{ 2 }{ \theta } } } =\text{cosec} ^{ 2 }{ \theta .\sec ^{ 2 }{ \theta } } $$
Question 9
1 / -0

The expression $$\sin^6 \theta+\sin^4 \theta \cos^2 \theta - \sin^2 \theta \cos^4 \theta - \cos^6 \theta$$ can be reduced to

A
$$\sin^2 \theta + \cos^2 \theta$$

B
$$\sin^3 \theta - \cos^3 \theta$$

C
$$ \sin^4 \theta + \cos^4 \theta$$

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

Solution

$$ { \sin }^{ 6 }\theta +{ \sin }^{ 4 }\theta { \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta { \cos }^{ 4 }\theta -{ \cos }^{ 6 }\theta$$
$$={ \sin }^{ 4 }\theta ({ \sin }^{ 2 }\theta +{\cos }^{ 2 }\theta )-{ \cos }^{ 4 }\theta ({ \sin }^{ 2 }\theta+{\cos }^{ 2 }\theta )$$
$$={ \sin }^{ 4 }\theta -{ \cos }^{ 4 }\theta \left[ \because { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right]$$
$$=({ \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta ) ({ \sin }^{ 2 }\theta -{ \cos }^{ 2 }\theta )$$
$$={ \sin }^{ 2 }\theta -{ \cos }^{ 2 }\theta$$
Hence, option $$D$$ is the correct answer.
Question 10
1 / -0

$$\dfrac{\cos A}{1-\tan A} + \dfrac{\sin A}{1 - \cot A}$$is equal to

A
$$ \sin A - \cos A$$

B
$$ \sin A + \cos A$$

C
$$1 -\cos A$$

D
$$1 - \sin A$$

Solution

Simplifying the above equation we get

$$\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-cotA}$$

$$=\dfrac{\cos A}{1-\dfrac{\sin A}{\cos A}}+\dfrac{\sin A}{1-\dfrac{\cos A}{\sin A}}$$

$$=\dfrac{\cos ^2A}{\cos A-\sin A}+\dfrac{\sin ^2A}{\sin A-\cos A}$$

$$=\dfrac{\cos ^2A-\sin ^2A}{\cos A-\sin A}$$

$$=\dfrac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}$$

$$=\cos A+\sin A$$

Hence answer is B

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Trigonometric Functions Test -2

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Trigonometric Functions Test -2

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Question 1

$$0$$

$$1$$

$$\tan \theta$$

$$\cot \theta$$

Solution

Question 2

$$0$$

$$\sin^2 \theta$$

$$\cos^2\theta$$

$$1$$

Solution

Question 3

$$\sin \theta =1 - \cos \theta$$

$$\sec \theta - \tan \theta =\dfrac{1}{\sec \theta + \tan \theta}$$

$$\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$$

$$\sin \dfrac{\pi}{3}=\cos\dfrac{\pi}{6}$$

Solution

Question 4

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

$$\dfrac{9}{5}$$

$$\dfrac{12}{25}$$

$$\dfrac{25}{12}$$

Solution

Question 5

$$3 - \cot^2 \theta$$

$$3 + \cot^2 \theta$$

$$3 - \tan^2 \theta$$

$$3+ \tan^2 \theta$$

Solution

Question 6

$$1$$

$$1.5$$

$$2$$

$$2.5$$

Solution

Question 7

$$\left ( \dfrac{1-\cos \theta}{1+\cos \theta} \right )$$

$$\left ( \dfrac{1+\cos \theta}{1-\cos \theta} \right )$$

$$\left ( \dfrac{\cos \theta-1}{\cos \theta+1} \right )$$

$$\left ( \dfrac{\cos \theta+1}{\cos \theta-1} \right )$$

Solution

Question 8

$$\sec^2 \theta. \cot^2 \theta$$

$$\sec^2\theta. \tan^2 \theta$$

$$\text{cosec}^2 \theta. \cot^2 \theta$$

$$\sec^2 \theta. \text{cosec}^2 \theta$$

Solution

Question 9

$$\sin^2 \theta + \cos^2 \theta$$

$$\sin^3 \theta - \cos^3 \theta$$

$$ \sin^4 \theta + \cos^4 \theta$$

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

Solution

Question 10

$$ \sin A - \cos A$$

$$ \sin A + \cos A$$

$$1 -\cos A$$

$$1 - \sin A$$

Solution

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