Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 70

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Question 1
1 / -0

If $$a=\cos\alpha \cos\beta+\sin \alpha \sin\beta \cos\gamma$$
$$b=\cos\alpha \sin \beta-\sin\alpha \cos\beta \cos\gamma$$
and $$c=\sin \alpha \sin\gamma$$, then $$a^2+b^2+c^2$$ is equal to

A
$$-1$$

B
$$0$$

C
$$1$$

D
none of these

Solution

Given $$a=\cos\alpha \cos\beta+ \sin \alpha \sin\beta \cos\gamma$$
$$b=\cos\alpha \sin \beta-\sin\alpha \cos\beta \cos\gamma$$
and $$c=\sin \alpha \sin\gamma$$

Consider, $$a^{ 2 }+b^{ 2 }+c^{ 2 }=\cos ^{ 2 }{ \alpha } \cos ^{ 2 }{ \beta } +\sin ^{ 2 }{ \alpha } \sin ^{ 2 }{ \beta } \cos ^{ 2 }{ \gamma } +2\cos\alpha \cos\beta \sin\alpha \sin\beta \cos\gamma\\ +\cos ^{ 2 }{ \alpha } \sin ^{ 2 }{ \beta } +\cos ^{ 2 }{ \beta } \sin ^{ 2 }{ \alpha } \cos ^{ 2 }{ \gamma } -2\cos\alpha \cos\beta \sin\alpha \sin\beta cos\gamma +4\sin ^{ 2 }{ \alpha } \sin ^{ 2 }{ \gamma } $$

$$\Rightarrow a^{ 2 }+b^{ 2 }+c^{ 2 }=\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha } \cos ^{ 2 }{ \gamma } +4\sin ^{ 2 }{ \alpha } \sin ^{ 2 }{ \gamma } $$

$$\Rightarrow a^{ 2 }+b^{ 2 }+c^{ 2 }=1+3{ c }^{ 2 }$$
Question 2
1 / -0

The value of expression $$\displaystyle \frac{\sin 30^{\circ}+\tan 45^{\circ}-\sec 60^{\circ}}{\text{cosec}30^{\circ}-\cot 45^{\circ}-\cos 60^{\circ}}$$ =

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$\displaystyle 2+\sqrt{3}$$

Solution

$$\displaystyle \frac{\sin 30^{\circ}+\tan 45^{\circ}-\sec 60^{\circ}}{\text{cosec}30^{\circ}-\cot 45^{\circ}-\cos 60^{\circ}}$$

= $$\displaystyle \frac{\frac{1}{2} + 1 - 2}{2 - 1 - \frac{1}{2}}$$

= $$\displaystyle \frac{1 + 2 - 4}{4 - 2 - 1}$$

= $$\displaystyle \frac{-1}{1}$$

= $$-1$$
Question 3
1 / -0

$$\text{cosec}^2 A \cot^2A-\sec^2A \tan^2A-(\cot^2A-\tan^2A)(\sec^2A+\text{cosec}^2A-1)$$ is equal to

A
$$0$$

B
$$1$$

C
$$\sec^2A$$

D
$$\text{cosec}^2A$$

Solution

$$\text{cosec}^2 A \cot^2A-\sec^2A \tan^2A-(\cot^2A-\tan^2A)(\sec^2A+\text{cosec}^2A-1)$$

$$=\text{cosec}^2A \cot^2A - \sec^2A \tan^2A-\cot^2A\sec^2 A-\cot^2A \text{cosec}^2A +\cot^2A$$
$$+\tan^2A\sec^2A+\tan^2A \text{cosec}^2A-\tan^2A$$

$$=-\cot^{ 2 }A\sec^{ 2 }A+\cot ^{ 2 }{ A } +\tan^{ 2 }A\text{cosec}^{ 2 }A-\tan ^{ 2 }{ A } $$

$$=-\cot^{ 2 }A(\sec^{ 2 }A-1)+\tan^{ 2 }A(\text{cosec}^{ 2 }A-1)$$

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

$$=-1+1$$

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

If $$x=\dfrac {\sin^3p}{\cos^2p}, y=\dfrac {\cos^3p}{\sin^2p}$$ and $$\sin p + \cos p= \dfrac 12$$, then $$x+y$$ is equal to

A
$$\cfrac {75}{18}$$

B
$$\cfrac {44}{9}$$

C
$$\cfrac {79}{18}$$

D
$$\cfrac {48}{9}$$

Solution

Given $$x=\displaystyle \cfrac {\sin^3p}{\cos^2p}, y=\cfrac {\cos^3p}{\sin^2p}$$; and $$\sin p + \cos p=\displaystyle \frac{1}{2}$$

Consider, $$x+y=\displaystyle \cfrac { \sin^{ 3 }p }{ \cos^{ 2 }p } +\cfrac { \cos^{ 3 }p }{ \sin^{ 2 }p } $$

$$\Rightarrow \displaystyle x+y=\cfrac { \sin^{ 5 }p+\cos^{ 5 }p }{ \sin^{ 2 }p\cos^{ 2 }p } $$

$$\displaystyle =\cfrac { \sin^{ 3 }p(1-\cos ^{ 2 }{ p } )+\cos^{ 3 }p(1-\sin ^{ 2 }{ p } ) }{ \sin^{ 2 }p\cos^{ 2}p } $$

$$\displaystyle =\cfrac { \sin^{ 3 }p+\cos^{ 3 }p-\sin ^{ 2 }{ p } \cos ^{ 2 }{ p } (\sin { p } +\cos { p } ) }{ \sin^{ 2 }p\cos^{ 2 }p } $$

$$\displaystyle =\cfrac { { (\sin { p } +\cos { p } ) }^{ 3 }-3\sin { p } \cos { p } (\sin { p } +\cos { p } )-\sin ^{ 2 }{ p } \cos ^{ 2 }{ p } (\sin { p } +\cos { p } ) }{ \sin^{ 2 }p\cos^{ 2 }p } $$ .....(1)

Since $$(\sin p+\ cos p)^2=\sin^{2} p+\cos ^{2}p+2\sin p\cos p$$
$$\Rightarrow \displaystyle \sin p\cos p =-\cfrac{3}{4}$$

So, using in (1),
$$x+y=\displaystyle \dfrac { \dfrac { 1 }{ 8 } +\dfrac { 9 }{ 16 } -\dfrac { 9 }{ 32 } }{ \dfrac { 9 }{ 16 } } =\dfrac { 79 }{ 18 } $$
Question 5
1 / -0

Find the relation obtained by eliminating $$\displaystyle \theta $$ from the equation $$\displaystyle x=a\cos \theta +b\sin \theta $$ and $$\displaystyle y=a\sin \theta -b\cos \theta $$

A
$$\displaystyle x^{2}+y^{2}=a^{2}-b^{2}$$

B
$$\displaystyle x^{2}-y^{2}=a^{2}+b^{2}$$

C
$$\displaystyle x^{2}-y^{2}=a^{2}-b^{2}$$

D
$$\displaystyle x^{2}+y^{2}=a^{2}+b^{2}$$

Solution

Given, $$ x = a \cos \theta + b \sin \theta $$ and $$ y = a \sin \theta - b \cos \theta $$

Squaring them and adding we get
$$ x^2 + y^2 = a^2 \cos ^2 \theta + b^2 \sin ^2 \theta + 2ab\cos \theta \sin \theta + a^2 \sin ^2 \theta + b^2 \cos ^2 \theta - 2ab\cos \theta \sin \theta $$
$$ \Rightarrow x^2 + y^2 = a^2 \cos ^2 \theta + b^2 \sin ^2 \theta + a^2 \sin ^2 \theta + b^2 \cos ^2 \theta $$
$$ \Rightarrow x^2 + y^2 = a^2 (\cos ^2 \theta + \sin ^2 \theta) + b^2 ( \sin ^2 \theta + \cos ^2 \theta ) $$
$$ \Rightarrow x^2 + y^2 = a^2 + b^2 $$
Question 6
1 / -0

Which one of the following when simplified is not equal to one?

A
$$\displaystyle \tan 18^{\circ}\times \tan 36^{\circ}\times \tan 54^{\circ}\times \tan 72^{\circ}$$

B
$$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}71^{\circ}$$

C
$$\displaystyle \frac{2\sin 62^{\circ}}{\cos 28^{\circ}}-\frac{\sec 42^{\circ}}{\text{cosec}48^{\circ}}$$

D
None of these

Solution

(1) $$\displaystyle \tan 18^{\circ}\times \tan 36^{\circ}\times \tan 54^{\circ}\times \tan 72^{\circ}$$
= $$\displaystyle \tan 18^{\circ}\times \tan (90 - 54)^{\circ}\times \tan 54^{\circ}\times \tan (90 - 18)^{\circ}$$
= $$\displaystyle \tan 18^{\circ}\times \cot 54^{\circ}\times \tan 54^{\circ}\times \cot 18^{\circ}$$
= $$1$$

(2) $$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}71^{\circ}$$
= $$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}(90 - 19)^{\circ}$$
= $$\displaystyle \sin ^{2}19^{\circ}+\cos ^{2}19^{\circ}$$
= $$1$$

(3) $$\displaystyle \frac{2\sin 62^{\circ}}{\cos 28^{\circ}}-\frac{\sec 42^{\circ}}{cosec48^{\circ}}$$

= $$\displaystyle \frac{2\sin 62^{\circ}}{\cos (90 - 62)^{\circ}}-\frac{\sec 42^{\circ}}{cosec (90 - 42)^{\circ}}$$

= $$\displaystyle \frac{2\sin 62^{\circ}}{\sin 62^{\circ}}-\frac{\sec 42^{\circ}}{\sec 42^{\circ}}$$

= $$2 - 1$$
= $$1$$

Thus, all are equal to one. Hence, none of these is the answer.
Question 7
1 / -0

If $$\tan { \theta } =\cfrac { p }{ q } $$, then what is $$\cfrac { p\sec { \theta } -qco\sec { \theta } }{ p\sec { \theta } +qco\sec { \theta } } $$ equal to?

A
$$\cfrac { p-q }{ p+q } $$

B
$$\cfrac { { q }^{ 2 }-{ p }^{ 2 } }{ { q }^{ 2 }+{ p }^{ 2 } } $$

C
$$\cfrac { { p }^{ 2 }-{ q }^{ 2 } }{ { q }^{ 2 }+{ p }^{ 2 } } $$

D
$$1$$
Question 8
1 / -0

If $$\frac{1 - cos x}{cos x (1 + cos x)} = \frac{sin \alpha}{cos x} - \frac{2}{1 + cos x}$$, then $$\alpha$$ =

A
$$\frac{\pi}{8}$$

B
$$\frac{\pi}{4}$$

C
Both A and B

D
None of the above

Solution
Question 9
1 / -0

If $$\cos P=\dfrac{1}{7}$$ and $$\cos Q=\dfrac{13}{14}$$, P and Q both are acute angle then the value of $$P-Q$$ will be?

A
$$45^o$$

B
$$30^o$$

C
$$75^o$$

D
$$60^o$$

Solution
Question 10
1 / -0

In the figure given
$$\angle ABD=\angle PQD=\angle CDQ=\cfrac { \pi }{ 2 } $$. If $$AB=x.PQ=z$$ and $$CD=y$$, then which one of the following is correct?

A
$$\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 1 }{ z } $$

B
$$\cfrac { 1 }{ x } +\cfrac { 1 }{ z } =\cfrac { 1 }{ y } $$

C
$$\cfrac { 1 }{ z } +\cfrac { 1 }{ y } =\cfrac { 1 }{ x } $$

D
$$\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 2 }{ z } $$

Solution

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Introduction to Trigonometry Test - 70

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Introduction to Trigonometry Test - 70

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SHARING IS CARING

Question 1

$$-1$$

$$0$$

$$1$$

none of these

Solution

Question 2

$$0$$

$$1$$

$$-1$$

$$\displaystyle 2+\sqrt{3}$$

Solution

Question 3

$$0$$

$$1$$

$$\sec^2A$$

$$\text{cosec}^2A$$

Solution

Question 4

$$\cfrac {75}{18}$$

$$\cfrac {44}{9}$$

$$\cfrac {79}{18}$$

$$\cfrac {48}{9}$$

Solution

Question 5

$$\displaystyle x^{2}+y^{2}=a^{2}-b^{2}$$

$$\displaystyle x^{2}-y^{2}=a^{2}+b^{2}$$

$$\displaystyle x^{2}-y^{2}=a^{2}-b^{2}$$

$$\displaystyle x^{2}+y^{2}=a^{2}+b^{2}$$

Solution

Question 6

$$\displaystyle \tan 18^{\circ}\times \tan 36^{\circ}\times \tan 54^{\circ}\times \tan 72^{\circ}$$

$$\displaystyle \sin ^{2}19^{\circ}+\sin ^{2}71^{\circ}$$

$$\displaystyle \frac{2\sin 62^{\circ}}{\cos 28^{\circ}}-\frac{\sec 42^{\circ}}{\text{cosec}48^{\circ}}$$

None of these

Solution

Question 7

$$\cfrac { p-q }{ p+q } $$

$$\cfrac { { q }^{ 2 }-{ p }^{ 2 } }{ { q }^{ 2 }+{ p }^{ 2 } } $$

$$\cfrac { { p }^{ 2 }-{ q }^{ 2 } }{ { q }^{ 2 }+{ p }^{ 2 } } $$

$$1$$

Question 8

$$\frac{\pi}{8}$$

$$\frac{\pi}{4}$$

Both A and B

None of the above

Solution

Question 9

$$45^o$$

$$30^o$$

$$75^o$$

$$60^o$$

Solution

Question 10

$$\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 1 }{ z } $$

$$\cfrac { 1 }{ x } +\cfrac { 1 }{ z } =\cfrac { 1 }{ y } $$

$$\cfrac { 1 }{ z } +\cfrac { 1 }{ y } =\cfrac { 1 }{ x } $$

$$\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 2 }{ z } $$

Solution

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