Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 36

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Question 1
1 / -0

If $$0 < A < \pi /2$$ the value of the expression $$\dfrac {tan A}{1-cot A}+\dfrac {cot A}{1-tan A}-sec A cosec A$$ is equal to

A
-1

B
0

C
1

D
sin A + cos A

Solution

$$\displaystyle \frac {\tan A}{1-\cot A}+\frac {\cot A}{1-\tan A}-\sec A cosec A$$

$$\displaystyle =\frac {\sin^2A}{\cos A(\sin A-\cos A)}+\frac {\cos^2A}{\sin A(\cos A-\sin A)}-\sec A cosec A$$

$$\displaystyle =\frac {\sin^3A-\cos^3A}{\sin A\cos A(\sin A-\cos A)}-\sec A cosec A$$

$$=\displaystyle \frac {\sin^2 A+\cos^2 A+\sin A \cos A}{\sin A \cos A}-\sec A cosec A$$ $$[a^3 -b^3=(a-b)(a^2+ab+b^2)]$$

$$=\sec A cosec A+1-\sec A cosec A$$
$$=1$$.
Question 2
1 / -0

For all values of $$\alpha$$ the given expression is equal to:
$$(sin\alpha+cosec \alpha)^2+(cos\alpha+sec\alpha)^2-(tan^2\alpha+cot^2\alpha)$$

A
$$0$$

B
$$2$$

C
$$4$$

D
$$7$$

Solution

$$(\sin\alpha+cosec \alpha)^2+(\cos\alpha+\sec\alpha)^2-(\tan^2\alpha+\cot^2\alpha)$$

$$=\sin^2\alpha+cosec^2\alpha+2+\cos^2\alpha+\sec^2\alpha+2-\tan^2\alpha-\cot^2\alpha$$

$$=4+\sin^2\alpha+\cos^2\alpha+\sec^2\alpha-tan^2\alpha+cosec^2\alpha-\cot^2\alpha=7$$
Question 3
1 / -0

If $$\displaystyle \tan A=1 $$ and $$\displaystyle \tan B=\sqrt{3}$$ evaluate:$$\displaystyle \cos A\cos B-\sin A\sin B$$

A
$$\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}$$

B
$$\displaystyle \frac{\sqrt{2}-\sqrt{6}}{6}$$

C
$$\displaystyle \frac{1-\sqrt{3}}{2\sqrt2}$$

D
$$\displaystyle \frac{\sqrt{2}+\sqrt{6}}{6}$$

Solution

$$\tan A = 1$$ and $$\tan B = \sqrt{3}$$

$$\tan A = \tan 45$$ and $$\tan B = \tan 60$$

$$A = 45^{\circ}$$ and $$B = 60^{\circ}$$

Now,
= $$\cos A \cos B - \sin A \sin B$$

= $$\cos 45^0 \cos 60^0 - \sin 45^0 \sin 60^0$$

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

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

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

= $$\dfrac{1- \sqrt{3}}{2 \sqrt{2}}$$
Question 4
1 / -0

If $$cot\alpha+tan\alpha=m$$ and $$\dfrac {1}{cos\alpha}-cos\alpha=n$$, then

A
$$m(mn^2)^{1/3}-n(nm^2)^{1/3}=1$$

B
$$m(m^2n)^{1/3}-n(nm^2)^{1/3}=1$$

C
$$n(mn^2)^{1/3}-m(nm^2)^{1/3}=1$$

D
$$n(m^2n)^{1/3}-m(nm^2)^{1/3}=1$$

Solution

$$cot\alpha +tan\alpha =m\Rightarrow \cfrac { 1 }{ tan\alpha } +tan\alpha =m$$

$$ \Rightarrow 1+{ tan }^{ 2 }\alpha =mtan\alpha \Rightarrow { sec }^{ 2 }\alpha =mtan\alpha \quad \quad \quad ...(1)$$

$$\cfrac { 1 }{ cos\alpha } -cos\alpha =n\Rightarrow sec\alpha -\cfrac { 1 }{ sec\alpha } =n$$

$$ \Rightarrow { sec }^{ 2 }\alpha -1=nsec\alpha \Rightarrow { tan }^{ 2 }\alpha =nsec\alpha $$

$$ \Rightarrow { tan }^{ 4 }\alpha ={ n }^{ 2 }{ sec }^{ 2 }\alpha $$

Substituting value from (1)

$${ tan }^{ 4 }\alpha ={ n }^{ 2 }mtan\alpha \Rightarrow { tan }^{ 3 }\alpha ={ n }^{ 2 }m\Rightarrow tan\alpha ={ \left( { n }^{ 2 }m \right) }^{ \cfrac { 1 }{ 3 } }\quad \quad ...(2)$$

Substituting (2) in (1), we get

$${ sec }^{ 2 }\alpha =mtan\alpha =m{ \left( { n }^{ 2 }m \right) }^{ \cfrac { 1 }{ 3 } }\quad \quad \quad ...(3)$$

From (2) and (3)

$${ sec }^{ 2 }\alpha -{ tan }^{ 2 }\alpha =1\Rightarrow m{ \left( { n }^{ 2 }m \right) }^{ \cfrac { 1 }{ 3 } }-{ \left( { n }^{ 2 }m \right) }^{ \cfrac { 2 }{ 3 } }=1$$

$$ \Rightarrow m{ \left( m{ n }^{ 2 } \right) }^{ \cfrac { 1 }{ 3 } }-n{ \left( n{ m }^{ 2 } \right) }^{ \cfrac { 1 }{ 3 } }=1$$
Question 5
1 / -0

If $$\dfrac {2 \sin\alpha}{1+\cos\alpha+\sin\alpha}=y$$, then $$\dfrac {1-\cos\alpha+\sin\alpha}{1+\sin\alpha}$$ is equal to

A
$$\dfrac 1y$$

B
$$y$$

C
$$1-y$$

D
$$1+y$$

Solution

$$\cfrac { 1-\cos\alpha +\sin\alpha }{ 1+\sin\alpha } =\cfrac { 1-\cos\alpha +\sin\alpha }{ 1+\sin\alpha } .\cfrac { 1+\cos\alpha +\sin\alpha }{ 1+\cos\alpha +\sin\alpha } \\ =\cfrac { { \left( 1+\sin\alpha \right) }^{ 2 }-{ \cos }^{ 2 }\alpha }{ \left( 1+\sin\alpha \right) \left( 1+\cos\alpha +\sin\alpha \right) } =\cfrac { 1+{ \sin }^{ 2 }\alpha +2\sin\alpha -\left( 1-{ \sin }^{ 2 }\alpha \right) }{ \left( 1+\sin\alpha \right) \left( 1+\cos\alpha +\sin\alpha \right) } \\ =\cfrac { 2{ \sin }^{ 2 }\alpha +2\sin\alpha }{ \left( 1+\sin\alpha \right) \left( 1+\cos\alpha +\sin\alpha \right) } =\cfrac { 2\sin\alpha \left( \sin\alpha +1 \right) }{ \left( 1+\sin\alpha \right) \left( 1+\cos\alpha +\sin\alpha \right) } \\ =\cfrac { 2\sin\alpha }{ 1+\cos\alpha +\sin\alpha } =y$$
Question 6
1 / -0

If $$\sin x+\sin^2 x=1$$, then the value of $$\cos^{12}x+3 \cos^{10}x+3 \cos^8x+\cos^6x-1$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

Using $$\sin x+{ \sin }^{ 2 }x=1\Rightarrow \sin x=1-{ \sin }^{ 2 }x={ \cos }^{ 2 }x$$
$${ \cos }^{ 12 }x+3{ \cos }^{ 10 }x+3{ \cos }^{ 8 }x+{ \cos }^{ 6 }x-1\\ ={ \sin }^{ 6 }x+3{ \sin }^{ 5 }x+3{ \sin }^{ 4 }x+{ \sin }^{ 3 }x-1\\ ={ \sin }^{ 3 }x\left( { \sin }^{ 3 }x+3{ \sin }^{ 2 }x+3\sin x+1 \right) -1\\ ={ \sin }^{ 3 }x{ \left( \sin x+1 \right) }^{ 3 }-1={ \left( { \sin }^{ 2 }x+\sin x \right) }^{ 3 }-1\\ ={ 1 }^{ 3 }-1=1-1=0$$
Question 7
1 / -0

If $$\sec C$$ is $$\displaystyle \frac{m}{2\sqrt{2}}$$, then $$m$$ is:

A
1

B
3

C
7

D
5

Solution

Given, $$\triangle ABC$$, A perpendicular from B on AC, let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$

In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$

In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$

Now, $$\sec C = \frac{H}{P} = \frac{12}{8 \sqrt{2}} = \frac{3}{2\sqrt{2}}$$
Question 8
1 / -0

If $$\cos 1^o \cos 2^o \cos 3^o ... \cos {179}^o=x+1$$, then $$x$$ is equal to

A
$$-1$$

B
$$0$$

C
$$1$$

D
none of these

Solution

$$\cos1^{0}.\cos2^{0}...\cos90^{0}...\cos(179^{0})$$
$$=0$$ since $$\cos90^{0}=0$$
Hence
$$x+1=0$$
$$x=-1$$
Question 9
1 / -0

In the following figure, cosec A is $$\displaystyle \frac{5}{m}$$, m is

A
$$1$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

Given, $$\triangle ABC$$, A perpendicular from B on AC, let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$

In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$

In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$

Now, $$cosec A = \frac{H}{P} = \frac{AB}{BD} = \frac{5}{4} $$
$$\therefore m=4$$
Question 10
1 / -0

Find the value of : $$\cot^{2} C\, -\, \displaystyle \frac{1}{\sin^{2} C}$$.

A
- 1

B
2

C
4

D
0

Solution

Given, $$\triangle ABC$$, A perpendicular from B on AC, let it cut AC at D, such that$$\angle BDA = \angle BDC = 90^{\circ}$$
$$AD = 3$$
$$BD = 4$$
$$BC = 12$$

In $$\triangle ABD$$,
$$AB^2 = BD^2 + AD^2$$
$$AB^2 = 3^2 + 4^2$$
$$AB = 5$$

In $$\triangle BCD$$
$$BC^2 = CD^2 + BD^2$$
$$12^2 = CD^2 + 4^2$$
$$CD^2 = 128$$
$$CD = 8 \sqrt{2}$$

Now, $$cot^2 C - \frac{1}{\sin^2 C} = cot^2 C - csc^2 C = (\frac{B}{P})^2 - (\frac{H}{P})^2$$
= $$(\frac{8\sqrt{2}}{4})^2 - (\frac{12}{4})^2$$
= $$ 8 - 9$$
= $$-1$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 36

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Introduction to Trigonometry Test - 36

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Rank

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SHARING IS CARING

Question 1

-1

0

1

sin A + cos A

Solution

Question 2

$$0$$

$$2$$

$$4$$

$$7$$

Solution

Question 3

$$\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}$$

$$\displaystyle \frac{\sqrt{2}-\sqrt{6}}{6}$$

$$\displaystyle \frac{1-\sqrt{3}}{2\sqrt2}$$

$$\displaystyle \frac{\sqrt{2}+\sqrt{6}}{6}$$

Solution

Question 4

$$m(mn^2)^{1/3}-n(nm^2)^{1/3}=1$$

$$m(m^2n)^{1/3}-n(nm^2)^{1/3}=1$$

$$n(mn^2)^{1/3}-m(nm^2)^{1/3}=1$$

$$n(m^2n)^{1/3}-m(nm^2)^{1/3}=1$$

Solution

Question 5

$$\dfrac 1y$$

$$y$$

$$1-y$$

$$1+y$$

Solution

Question 6

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 7

1

3

7

5

Solution

Question 8

$$-1$$

$$0$$

$$1$$

none of these

Solution

Question 9

$$1$$

$$4$$

$$3$$

$$2$$

Solution

Question 10

- 1

2

4

0

Solution

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