Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The value of the expression $$\text{cosec }(75^0+\theta) - \sec (15^0 - \theta) - \tan (55^0 + \theta) + \cot (35^0 - \theta)$$, is

A
$$-1$$

B
$$0$$

C
$$1$$

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

Solution

Given: $$E = \text{cosec }(75^0 + \theta) - \sec (15^0 - \theta) - \tan (55^0 + \theta) + \cot (35^0 - \theta)$$
As, we know that,
$$\text{cosec }\theta = \sec(90^0-\theta)$$ and
$$\tan\theta = \cot(90^0-\theta)$$
Using these, we get
$$E=\sec(90^0-75^0-\theta)-\sec(15^0-\theta)-\cot(90^0-55^0-\theta)+\cot(35^0-\theta)$$
$$\Rightarrow E=\sec(15^0-\theta)-\sec(15^0-\theta)-\cot(35^0-\theta)+\cot(35^0-\theta) = 0$$
Hence, option B is correct.
Question 2
1 / -0

If $$x=a\cos^{3}\theta\sin^{2}\theta, y=a\sin^{3}\theta\cos^{2}\theta$$ and $$\displaystyle \frac{(x^{2}+y^{2})^{p}}{(xy)^{q}}, (p,q\in N)$$ is independent of $$\theta$$ then:

A
$$p + q = 6$$

B
$$4p = 5q$$

C
$$p=q$$

D
$$pq = 16$$

Solution

$$ x = a{\cos }^{3} \theta {\sin }^{2} \theta $$

$$ y = a{\sin }^{3} \theta {\cos }^{2} \theta $$

Then, $$ {({x}^{2} + {y}^{2})}^{p} = {({a}^{2}{\cos }^{6} \theta {\sin }^{4} \theta + {a}^{2}{\sin }^{6} \theta {\cos }^{4} \theta)}^{p} $$
$$= {[{a}^{2}\sin ^4\theta \cos ^4\theta({\cos }^{2} \theta + {\sin }^{2} \theta )]}^{p}$$
$$=(a^2\sin ^4\theta \cos ^4\theta)^p$$

and, $$ {(xy)}^{q} = {({a}^{2}{\cos }^{5} \theta {\sin }^{5} \theta)}^{q} $$

Now,
$$\displaystyle \frac{{({x}^{2} + {y}^{2})}^{p}}{{(xy)}^{q}} $$
$$=\displaystyle \frac{{({a}^{2}{\sin }^{4} \theta {\cos }^{4} \theta)}^{p}}{(a^2\cos ^5\theta \sin ^5\theta)^q} $$
$$=a^{2p-2q}\times (\sin\theta\cos\theta)^{4p-5q}$$

Now, $$\displaystyle \frac{{({x}^{2} + {y}^{2})}^{p}}{{(xy)}^{q}} $$ is independent of $$ \theta $$ if,
$$4p-5q=0$$
$$\therefore 4p=5q$$

Hence, option 'B' is correct.
Question 3
1 / -0

$$\sin \left ( 45^{0}+\theta \right )-\cos\left ( 45^{0}-\theta \right )$$ is equal to

A
$$2 \cos \theta$$

B
$$0$$

C
$$2 \sin \theta$$

D
$$1$$

Solution

$$\sin (45^0+\theta)-\cos (45^0-\theta)$$

$$=sin(45^0+\theta)-sin(90^0-(45^0-\theta))$$

$$=sin (45^0+\theta)-sin( 45^0+\theta)$$

$$=0$$
Question 4
1 / -0

If $$\mathrm{x}= \mathrm{a}\cos^2 \theta \sin\theta$$ and $$\mathrm{y}=\mathrm{a}\sin^{2}\theta\cos\theta$$, then $$\displaystyle \frac{(\mathrm{x}^{2}+\mathrm{y}^{2})^{3}}{\mathrm{x}^{2}\mathrm{y}^{2}}=$$

A
$$\mathrm{a}$$

B
$$\mathrm{a}^{3}$$

C
$$\mathrm{a}^{2}$$

D
$$\mathrm{a}^{5}$$

Solution

Given : $$x=a\cos^{2} \theta \sin \theta$$ and $$y=a\sin^{2} \theta \cos \theta$$

$$\displaystyle \frac { ({ x }^{ 2 }+{ y }^{ 2 })^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { ({ { a }^{ 2 } }\cos ^{ 4 } \theta \sin ^{ 2 }{ \theta } +{ { a }^{ 2 } }\cos ^{ 2 } \theta \sin ^{ 4 }{ \theta } )^{ 3 } }{ { { a }^{ 4 } }\cos ^{ 6 } \theta \sin ^{ 6 }{ \theta } }$$

$$ =\dfrac { {a}^{6}\cos^{6} \theta \sin^{6}{\theta} \left(\cos^{2} \theta +\sin^{2}{\theta} \right)^{3}}{a^{4} \cos^{6} \theta \sin^{6}{\theta}}$$ ($$\because sin^2 \theta +cos^2 \theta =1)$$

$$=a^{2}(1)^{3}=a^{2}$$
Hence, option C is correct.
Question 5
1 / -0

$$\tan^2\alpha=1-p^2$$ then, $$\sec\alpha + \tan^3\alpha \,\text{cosec}\,\alpha=$$

A
$$(2+p^2)^{\tfrac{3}{2}}$$

B
$$(1+p^2)^{\tfrac{3}{2}}$$

C
$$(2-p^2)^{\tfrac{3}{2}}$$

D
$$(1-p^2)^{\tfrac{3}{2}}$$

Solution

$$\displaystyle \sec \alpha + \tan ^{3}\alpha\, \text{cosec}\, \alpha = \frac{1}{\cos \alpha} + \frac{\sin ^{2}\alpha}{\cos ^{3}\alpha} =\frac{\cos^2 \alpha + \sin^2 \alpha}{\cos^3 \alpha} = \frac{1}{\cos^3 \alpha}= \sec ^{3}\alpha$$

$$=(1+\tan ^{2}\alpha)^{\tfrac{3}{2}} = (2-p^{2})^{\tfrac{3}{2}}$$
Question 6
1 / -0

The value of $$sin^{2}30^{\circ}$$ - $$cos^{2}30^{\circ}$$ is:

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

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

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

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

Solution

$$\sin ^{ 2 }{ 30° } -\cos ^{ 2 }{ 30° } $$
$$= { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }$$
$$=\dfrac { 1 }{ 4 } -\dfrac { 3 }{ 4 } $$
$$=-\dfrac { 2 }{ 4 } $$
$$=-\dfrac { 1 }{ 2 } $$
Hence, the answer is $$-\dfrac { 1 }{ 2 }.$$
Question 7
1 / -0

If sin $$A=\dfrac{1}{2}$$ and cos $$B=\dfrac{1}{2}$$, then the value of $$(A+B)$$ is equal to :

A
$$0^{\circ}$$

B
$$60^{\circ}$$

C
$$90^{\circ}$$

D
$$30^{\circ}$$

Solution

$$\sin { A } =\dfrac { 1 }{ 2 } ,\cos { B } =\dfrac { 1 }{ 2 } $$
$$\Rightarrow \sin { A } =\sin { 30° } ,\cos { B } =\cos { 60° } $$
$$\Rightarrow A=30°,B=60°$$
$$\therefore A+B=30°+60°=90°$$
Hence, the answer is $$90°.$$
Question 8
1 / -0

If cot $$\theta =\dfrac{7}{8}$$, then the value of $$tan^{2}\theta$$ equals to :

A
$$\dfrac{8}{7}$$

B
$$\dfrac{49}{64}$$

C
$$\dfrac{64}{49}$$

D
$$\dfrac{7}{8}$$

Solution

$$\cot { \theta } =\dfrac { 7 }{ 8 } $$

$$\Rightarrow \tan { \theta } =\dfrac { 1 }{ \cot { \theta } } \\\ \ \ \ \ \ \ \ \ \ \ \ =\dfrac { 8 }{ 7 } $$

$$\Rightarrow \tan ^{ 2 }{ \theta } =\dfrac { 64 }{ 49 } $$
Hence, the correct option is (C)
Question 9
1 / -0

The value of (sin $$45^{\circ}+cos 45^{\circ}$$) is :

A
1

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

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

D
$$\sqrt{2}$$

Solution

$$\sin { 45° } +\cos { 45° } $$
$$=\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 2 } } $$
$$=\dfrac { 2 }{ \sqrt { 2 } } $$
$$=\sqrt { 2 } $$
Hence, the answer is $$=\sqrt { 2 } .$$
Question 10
1 / -0

In the figure given below, $$\Delta ABC$$ is right angled at B and tan$$A=\dfrac{4}{3}.$$ If $$AC=15$$ cm, then the length of BC is :

A
$$4$$ cm

B
$$3$$ cm

C
$$12$$ cm

D
$$9$$ cm

Solution

$$\tan A=\dfrac43$$
$$\Rightarrow AB=\dfrac34BC$$

Now, in $$\triangle ABC$$

$$AB^2+BC^2=AC^2$$

$$\Bigg(\dfrac34BC\Bigg)^2+BC^2=15^2$$

$$\Rightarrow \dfrac{25}{16}BC^2=15^2$$

$$\Rightarrow BC^2=\dfrac{15^2\times 4^2}{5^2}$$

$$\Rightarrow BC=12$$

Therefore, Answer is $$12\ cm$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 29

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Introduction to Trigonometry Test - 29

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SHARING IS CARING

Question 1

$$-1$$

$$0$$

$$1$$

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

Solution

Question 2

$$p + q = 6$$

$$4p = 5q$$

$$p=q$$

$$pq = 16$$

Solution

Question 3

$$2 \cos \theta$$

$$0$$

$$2 \sin \theta$$

$$1$$

Solution

Question 4

$$\mathrm{a}$$

$$\mathrm{a}^{3}$$

$$\mathrm{a}^{2}$$

$$\mathrm{a}^{5}$$

Solution

Question 5

$$(2+p^2)^{\tfrac{3}{2}}$$

$$(1+p^2)^{\tfrac{3}{2}}$$

$$(2-p^2)^{\tfrac{3}{2}}$$

$$(1-p^2)^{\tfrac{3}{2}}$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$0^{\circ}$$

$$60^{\circ}$$

$$90^{\circ}$$

$$30^{\circ}$$

Solution

Question 8

$$\dfrac{8}{7}$$

$$\dfrac{49}{64}$$

$$\dfrac{64}{49}$$

$$\dfrac{7}{8}$$

Solution

Question 9

1

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

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

$$\sqrt{2}$$

Solution

Question 10

$$4$$ cm

$$3$$ cm

$$12$$ cm

$$9$$ cm

Solution

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