Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

If $$\sec x=p+\dfrac 1{4p}$$, then $$\sec x+\tan x$$

A
$$2p\ or\ 1/2p$$

B
$$p\ or\ 1/2p$$

C
$$2p\ or\ 1/p$$

D
None of these

Solution

Given

$$ \sec x =p+\dfrac 1{4p}$$

$$ \sec ^2 x =p^2 +\dfrac 1{16p^2} +2(p)\left(\dfrac 1{4p}\right)$$

$$ 1+\tan ^2 x =p^2 +\dfrac 1{16p^2} +\dfrac 12 $$

$$ \tan ^2 x =p^2 +\dfrac 1{16p^2} +\dfrac 12 -1$$

$$ \tan ^2 x =p^2 +\dfrac 1{16p^2} -\dfrac 12$$

$$ \tan ^2 x =p^2 +\dfrac 1{16p^2} -2(p)\left(\dfrac 1{4p}\right)$$

$$ \tan ^2 x =\left(p-\dfrac 1{4p}\right)^2$$

$$ \tan x = \pm \left(p-\dfrac 1{4p}\right)$$

$$ \sec x+\tan x $$

$$ \implies p+\dfrac 1{4p} \pm \left(p-\dfrac 1{4p}\right)$$

$$ \implies 2p (or) \dfrac 1{2p}$$
Question 2
1 / -0

$$\sin 30^{o}- cos60^{o}$$= ?

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$2$$

Solution

$$\sin{ 30 }^{ o }=\dfrac { 1 }{ 2 } $$

$$\cos { 60 }^{ o }=\dfrac { 1 }{ 2 } $$

$$\sin{30 }^{ o }-\cos { 60 }^{ o }=\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 2 } =0$$
Question 3
1 / -0

In the given figure, $$\angle PAB = 60^o $$ and $$ \angle CAB = 30^o $$ and $$PB = 75m.$$ Find $$BC.$$

A
$$22 \sqrt{3}$$

B
$$25 \sqrt{3}$$

C
$$5 \sqrt{3}$$

D
$$20 \sqrt{3}$$
Question 4
1 / -0

Solve $$\sin A(1+\tan A)+\cos A(1+\cot A)$$.

A
$$\sec A+cosec A$$

B
$$\sin A+cos A$$

C
$$\tan A+\cot A$$

D
$$\cos A+\sin A$$

Solution

$$sinA(1+tanA)+cosA(1+cotA) $$

$$ \Rightarrow sinA\left( { 1+\dfrac { { \sin A } }{ { \cos A } } } \right) +cosA\left( { 1+\dfrac { { \cos A } }{ { \sin A } } } \right) $$

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

$$ \Rightarrow \left( { \sin A+\cos A } \right) \left[ { \dfrac { { { { \sin }^{ 2 } }A+{ { \cos }^{ 2 } }A } }{ { \sin A\cdot \cos A } } } \right] \\$$

$$ \Rightarrow \dfrac { { \sin A } }{ { \sin A\cdot \cos A } } +\dfrac { { \cos A } }{ { \sin A\cdot \cos A } } \\$$

$$ \Rightarrow \sec A+ cosec \, A \\ $$
Question 5
1 / -0

If $$\tan A+\cot A=4,\ then\ {\tan}^{4}\ A+{\cot}^{4}\ A$$ is equal to

A
$$110$$

B
$$191$$

C
$$80$$

D
$$194$$

Solution

$$tan A+ cot A=4$$

(squaring both sides)

$$(tan A+ cot A)^{2}= 4^{2}$$

$$\Rightarrow tan^{2}A+cot^{2}A+ 2 (tan A) (cot A)= 16$$

$$\Rightarrow tan^{2}A+ cot^{2}A+2=16$$

$$\Rightarrow tan^{2}A+ cot^{2}A= 16-2$$

$$\Rightarrow tan^{2}A+cot^{2}A=14$$

(on squaring both sides)

$$\Rightarrow (tan^{2}A- cot^{2}A)^{2}= (14)^{2}$$

$$tan^{4}A+ cot^{4}A+2(tan^{2}A)(cot^{2}A)= 196$$

$$tan^{2}A+ cot^{4}A= 196-2$$

$$= 194$$

Option D is correct
Question 6
1 / -0

If $$x\sin \,{45^ \circ }{\cos ^2}{60^ \circ } = \dfrac{{{{\tan }^2}{{60}^ \circ }\cos ec{{30}^ \circ }}}{{\sec {{45}^ \circ }{{\cot }^2}{{30}^ \circ }}}$$ then x =

A
$$2$$

B
$$4$$

C
$$8$$

D
$$1$$
Question 7
1 / -0

If $$x = a \cos \theta \text { and } y = b \sin \theta$$, then $$b ^ { 2 } x ^ { 2 } + a ^ { 2 } y ^ { 2 }$$ is equal to

A
$$a b$$

B
$$a ^ { 2 } b ^ { 2 }$$

C
$$a ^ { 4 } b ^ { 4 }$$

D
none of these

Solution

$$x = a\cos \theta; y = b\sin \theta$$

$$\cos \theta = \dfrac {x}{a}; \sin \theta = \dfrac {y}{b}$$

$$\cos^{2}\theta + \sin^{2}\theta = 1 = \dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{b^{2}}$$

$$x^{2}b^{2} + a^{2}y^{2} = a^{2}b^{2}$$.
Question 8
1 / -0

If $$\cos 9\theta=\sin \dfrac{\pi}{4}$$, then the value of $$\tan 6\theta$$ is

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

B
$$\sqrt {3}$$

C
$$1$$

D
$$0$$

Solution

We have,

$$\cos 9\theta=\sin \dfrac{\pi}{4}$$
$$ \cos 9\theta =\cos \left( {{90}^{o}}-45^0 \right) $$

$$ 9\theta =45^0 $$

$$ 10\theta ={45}^{o} $$

$$ \theta =5^0$$

Value of $$\tan 6\theta $$

$$ =\tan 6\theta $$

$$ =\tan 6\times {{5}^{o}} $$

$$ =\tan {{30}^{o}} $$
$$=\dfrac{1}{\sqrt 3}$$

Hence, this is the answer.
Question 9
1 / -0

If $$\tan { x } =3\cot { x } ,$$ then x =?

A
$${ 45 }^{ \circ }$$

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

C
$${ 30 }^{ \circ }$$

D
$${ 15 }^{ \circ }$$

Solution

$$\tan x = 3 \cot x$$
$$\tan x = 3 \times 1/ \tan x . (\cot x = t/tanx)$$
$$\tan^2 x = 3$$
$$\tan x = \sqrt{3}$$
So the value of x will be $$60^o$$ because the value of $$\tan \, 60$$ is $$\sqrt{3}$$
Question 10
1 / -0

$$\cos 35 ^ { \circ } + \cos 85 ^ { \circ } + \cos 155 ^ { \circ }$$ is

A
0

B
$$\frac { 1 } { \sqrt { 3 } }$$

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

D
$$\cos 275 ^ { \circ }$$

Solution

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Introduction to Trigonometry Test - 64

Result

Introduction to Trigonometry Test - 64

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SHARING IS CARING

Question 1

$$2p\ or\ 1/2p$$

$$p\ or\ 1/2p$$

$$2p\ or\ 1/p$$

None of these

Solution

Question 2

$$0$$

$$1$$

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

$$2$$

Solution

Question 3

$$22 \sqrt{3}$$

$$25 \sqrt{3}$$

$$5 \sqrt{3}$$

$$20 \sqrt{3}$$

Question 4

$$\sec A+cosec A$$

$$\sin A+cos A$$

$$\tan A+\cot A$$

$$\cos A+\sin A$$

Solution

Question 5

$$110$$

$$191$$

$$80$$

$$194$$

Solution

Question 6

$$2$$

$$4$$

$$8$$

$$1$$

Question 7

$$a b$$

$$a ^ { 2 } b ^ { 2 }$$

$$a ^ { 4 } b ^ { 4 }$$

none of these

Solution

Question 8

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

$$\sqrt {3}$$

$$1$$

$$0$$

Solution

Question 9

$${ 45 }^{ \circ }$$

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

$${ 30 }^{ \circ }$$

$${ 15 }^{ \circ }$$

Solution

Question 10

0

$$\frac { 1 } { \sqrt { 3 } }$$

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

$$\cos 275 ^ { \circ }$$

Solution

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