Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

If $$\sin A, \cos A$$ and $$\tan A$$ are in $$G.P.,$$ then $$\cot^6 A- \cot^2A$$ is equal to

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$\cfrac {1}{2}$$

Solution

Given,
$$\sin A, \cos A$$ and $$\tan A$$ are in $$G.P.,$$

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

$$\Rightarrow \cos^2A=\sin A\times \dfrac{\sin A}{\cos A}$$

$$\Rightarrow \cos^3 A=\sin^2 A$$

$$\Rightarrow \cos^2A . \cos A = \sin^2 A$$

$$\Rightarrow \cfrac {\cos^2A}{\sin^2A} = \cfrac {1}{\cos A}$$

$$\Rightarrow \cot^2 A= \sec A$$

Squaring on both sides, we get

$$\Rightarrow (\cot^2 A)^2= (\sec A)^2$$

$$\Rightarrow \cot^4 A= \sec^2 A$$

$$\Rightarrow \cot^4A=1+\tan^2A$$

$$\Rightarrow \cot^4A-1 = \tan^2A......(1)$$

Now,
$$\cot^6A - \cot^2A$$

$$ = \cot^2A(\cot^4A-1)$$

$$=\cot^2A \times \tan^2A$$

$$=\dfrac{1}{\tan^2A} \times \tan^2A$$

$$=1$$

Hence, the required value is $$1.$$
Question 2
1 / -0

If $$\displaystyle x=r\sin \theta \cdot \cos \phi,$$ $$y=r\sin \theta \cdot \sin \phi$$ and $$\displaystyle z= r\cos \theta$$, then the value of $$\displaystyle x^{2}+y^{2}+z^{2}$$ is independent of:

A
$$\displaystyle \theta ,\phi$$

B
$$\displaystyle r,\theta$$

C
$$\displaystyle r,\phi$$

D
$$r$$

Solution

Given, $$\displaystyle x=r\sin \theta \cdot \cos \phi,$$ $$y=r\sin \theta \cdot \sin \phi$$ and $$\displaystyle z= r\cos \theta$$

Therefore, $${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$$
$$ ={ \left( r\sin\theta \cos\phi \right) }^{ 2 }+{ \left( r\sin\theta \sin\phi \right) }^{ 2 }+{ \left( r\cos\theta \right) }^{ 2 }$$

$$={ r }^{ 2 }{ \sin }^{ 2 }\theta { \cos }^{ 2 }\phi +{ r }^{ 2 }{ \sin }^{ 2 }\theta { \sin }^{ 2 }\phi +{ r }^{ 2 }{ \cos }^{ 2 }\theta $$

$$ ={ r }^{ 2 }{ \sin }^{ 2 }\theta \left( { \cos }^{ 2 }\phi +{ \sin }^{ 2 }\phi \right) +{ r }^{ 2 }{ \cos }^{ 2 }\theta $$
$$ ={ r }^{ 2 }{ \sin }^{ 2 }\theta +{ r }^{ 2 }{ \cos }^{ 2 }\theta $$

$$ ={ r }^{ 2 }\left( { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta \right) $$

$$={ r }^{ 2 }$$
Question 3
1 / -0

If $$\sin { A } =a\cos { B } $$ and $$\cos { A } =b\sin { B } $$, then $$\left( { a }^{ 2 }-1 \right) \tan ^{ 2 }{ A } +\left( 1-{ b }^{ 2 } \right) \tan ^{ 2 }{ B } $$ is equal to

A
$$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 } }{ { b }^{ 2 } } $$

B
$$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } } $$

C
$$\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } } $$

D
$$\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 } } $$

Solution

Given, $$\sin { A } =a\cos { B } $$ and $$\cos { A } =b\sin { B } ,$$
On squaring and adding, we get
$${ a }^{ 2 }\cos ^{ 2 }{ B } +{ b }^{ 2 }\sin ^{ 2 }{ B } =1$$
$$\Rightarrow \displaystyle \cos ^{ 2 }{ B } =\frac { 1-{ b }^{ 2 } }{ { a }^{ 2 }-{ b }^{ 2 } } $$ and $$\displaystyle \sin ^{ 2 }{ B } =\frac { { a }^{ 2 }-1 }{ { a }^{ 2 }-{ b }^{ 2 } } ,$$
$$\Rightarrow \displaystyle \tan ^{ 2 }{ B } =\frac { { a }^{ 2 }-1 }{ 1-{ b }^{ 2 } } $$
Also, $$\displaystyle\tan ^{ 2 }{ A } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \cot ^{ 2 }{ B } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } .\frac { 1-{ b }^{ 2 } }{ { a }^{ 2 }-1 } $$
Therefore, $$\displaystyle\left( { a }^{ 2 }-1 \right) \tan ^{ 2 }{ A } +\left( 1-{ b }^{ 2 } \right) \tan ^{ 2 }{ B } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( 1-{ b }^{ 2 } \right) +{ a }^{ 2 }-1=\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { b }^{ 2 } } $$
Question 4
1 / -0

If $$\tan A+\cot A=4$$, then $$\tan^2A+\cot^2A$$ is equal to

A
$$10$$

B
$$11$$

C
$$12$$

D
$$14$$

Solution

Given, $$\tan A+\cot A=4$$
On squaring, we get
$$(\tan A+\cot A)^2=4^2 $$
$$\Rightarrow \tan^2A+\cot^2A+2 \tan A \cot A=16$$
$$\Rightarrow \tan^2A+\cot^2A+2 (1) = 16$$
$$\Rightarrow \tan^2A+\cot^2A+2 =16$$
$$\Rightarrow \tan^2A+\cot^2A =14$$
Question 5
1 / -0

If $$\displaystyle a\sec \alpha - c\tan \alpha =d$$ and $$\displaystyle b\sec \alpha -d\tan \alpha =c$$ $$\ \ \ then$$

A
$$a^{2}+b^{2}=b^{2}+d^{2}$$

B
$$a^{2}+d^{2}=b^{2}+c^{2}$$

C
$$a^{2}+b^{2}=c^{2}+d^{2}$$

D
none of these
Question 6
1 / -0

If $$\displaystyle \frac{2\sin \alpha }{1+\sin \alpha +\cos \alpha }=\lambda$$ then $$\displaystyle \frac{1+\sin \alpha -\cos \alpha }{1+\sin \alpha }$$ is equal to

A
$$\displaystyle \frac{1}{\lambda }$$

B
$$\displaystyle \lambda$$

C
$$\displaystyle 1-\lambda$$

D
$$\displaystyle 1+\lambda$$

Solution

$$\lambda =\cfrac { 2sin\alpha }{ 1+sin\alpha +cos\alpha } \\ =\cfrac { 2sin\alpha }{ \left( 1+sin\alpha \right) +\left( cos\alpha \right) } \times \cfrac { \left( 1+sin\alpha \right) -\left( cos\alpha \right) }{ \left( 1+sin\alpha \right) -\left( cos\alpha \right) } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha \right) }{ { \left( 1+sin\alpha \right) }^{ 2 }-{ \left( cos\alpha \right) }^{ 2 } } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha \right) }{ 1+{ sin }^{ 2 }\alpha +2sin\alpha -{ cos }^{ 2 }\alpha } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha \right) }{ 1+{ sin }^{ 2 }\alpha +2sin\alpha -1+{ sin }^{ 2 }\alpha } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha \right) }{ 2{ sin }^{ 2 }\alpha +2sin\alpha } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha \right) }{ 2{ sin }\alpha \left( 1++sin\alpha \right) } \\ =\cfrac { \left( 1+sin\alpha -cos\alpha \right) }{ \left( 1+sin\alpha \right) } $$
Question 7
1 / -0

If $$\displaystyle \sin\theta + cosec \theta= 2,$$ then the value of $$\displaystyle \sin ^{8}\theta + cosec ^{8}\theta$$ is equal to

A
$$2$$

B
$$\displaystyle 2^{8}$$

C
$$\displaystyle 2^{4}$$

D
none of these

Solution

$$sin\theta +cosec\theta =2\\ \Rightarrow sin\theta +\cfrac { 1 }{ sin\theta } =2$$
Squaring both sides
$${ \left( sin\theta +\cfrac { 1 }{ sin\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { sin }^{ 2 }\theta +\cfrac { 1 }{ { sin }^{ 2 }\theta } +2sin\theta \cfrac { 1 }{ sin\theta } =4\\ \Rightarrow { sin }^{ 2 }\theta +\cfrac { 1 }{ { sin }^{ 2 }\theta } =2$$
Again squaring both sides
$${ \left( { sin }^{ 2 }\theta +\cfrac { 1 }{ { sin }^{ 2 }\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { sin }^{ 4 }\theta +\cfrac { 1 }{ { sin }^{ 4 }\theta } =2$$
Similarly,
$${ sin }^{ 8 }\theta +\cfrac { 1 }{ { sin }^{ 8 }\theta } =2$$
$$\Rightarrow { sin }^{ 8 }\theta +{ cosec }^{ 8 }\theta =2$$
Question 8
1 / -0

If $$\displaystyle \frac{\sin A}{\sin B}= \frac{\sqrt{3}}{2}$$and $$\displaystyle \frac{\cos A}{\cos B}= \frac{\sqrt{5}}{2},0< A, B< \dfrac{\pi}{2},$$ then find the value of $$\displaystyle \tan A+\tan B$$

A
$$\displaystyle \frac{\sqrt{5}-\sqrt{3}}{\sqrt{3}}$$

B
$$\displaystyle \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}}$$

C
$$\displaystyle \frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$$

D
$$\displaystyle \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}}$$

Solution

Given $$\displaystyle \frac{\sin A}{\sin B}= \frac{\sqrt{3}}{2}$$and $$\displaystyle \frac{\cos A}{\cos B}= \frac{\sqrt{5}}{2}$$
$$\Rightarrow \displaystyle \sin A= \frac{\sqrt{3}}{2}\sin B$$ and $$\displaystyle \cos A= \frac{\sqrt{5}}{2}\cos B$$
Squaring and adding both equations,
$$\displaystyle \sin^2A+\cos^2A=\frac{3}{4}\sin^2B+\frac{5}{4}\sin^2B$$
$$\Rightarrow 3\sin^2B+5\cos^2B=4\Rightarrow \cos^2B =\cfrac{1}{2}\Rightarrow \tan^2B =1\Rightarrow \tan B=1 $$
Now using first equation, $$\displaystyle \sin A=\frac{\sqrt{3}}{2\sqrt{2}}\Rightarrow \tan A=\frac{\sqrt{3}}{\sqrt{5}}$$
Hence $$\displaystyle \tan A+\tan B = \frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$$
Question 9
1 / -0

If $$\displaystyle 5\tan\theta = 4 $$ then $$\displaystyle \frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }$$ is equal to

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{6}$$

D
$$6$$

Solution

$$\cfrac { 5sin\theta -3cos\theta }{ 5sin\theta +2cos\theta } =\cfrac { 5tan\theta -3 }{ 5tan\theta +2 } $$ [dividing num and denominator by $$\cos\theta$$]
Substituting $$tan\theta =\cfrac { 4 }{ 5 } $$, we get
$$\cfrac { 5\cfrac { 4 }{ 5 } -3 }{ 5\cfrac { 4 }{ 5 } +2 } =\cfrac { 1 }{ 6 } $$
Question 10
1 / -0

If $$\displaystyle \sin x=2\cos y,\sqrt 6 \sin y=\tan z$$ and $$\displaystyle 2\sin z=\sqrt 3 \cos x,u,v,w $$ denotes, respectively,$$\displaystyle \sin ^{2}x, \sin ^{2}y, \sin ^{2}z,$$ then the value of $$u+v+w$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
None of these

Solution

$$u+v+w=\sin ^{2}(x)+\sin ^{2}y+\sin ^{2}z$$

$$=\sin ^{2}(x)+\sin ^{2}y+\dfrac{3}{4}\cos ^{2}x$$

$$=[\sin ^{2}(x)+\dfrac{3}{4}\cos ^{2}x]+\sin ^{2}y$$

$$=[\dfrac{\sin ^{2}(x)}{4}+\dfrac{3\sin ^{2}(x)}{4}+\dfrac{3}{4}\cos ^{2}x]+\sin ^{2}(y)$$

$$=\dfrac{\sin ^{2}(x)}{4}+\dfrac{3}{4}+1-\cos ^{2}y$$

$$=\dfrac{\sin ^{2}(x)}{4}+\dfrac{3}{4}+1-\dfrac{\sin ^{2}(x)}{4}$$

$$=\dfrac{7}{4}$$

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 32

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Introduction to Trigonometry Test - 32

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SHARING IS CARING

Question 1

$$-1$$

$$0$$

$$1$$

$$\cfrac {1}{2}$$

Solution

Question 2

$$\displaystyle \theta ,\phi$$

$$\displaystyle r,\theta$$

$$\displaystyle r,\phi$$

$$r$$

Solution

Question 3

$$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 } }{ { b }^{ 2 } } $$

$$\displaystyle \frac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } } $$

$$\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } } $$

$$\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 } } $$

Solution

Question 4

$$10$$

$$11$$

$$12$$

$$14$$

Solution

Question 5

$$a^{2}+b^{2}=b^{2}+d^{2}$$

$$a^{2}+d^{2}=b^{2}+c^{2}$$

$$a^{2}+b^{2}=c^{2}+d^{2}$$

none of these

Question 6

$$\displaystyle \frac{1}{\lambda }$$

$$\displaystyle \lambda$$

$$\displaystyle 1-\lambda$$

$$\displaystyle 1+\lambda$$

Solution

Question 7

$$2$$

$$\displaystyle 2^{8}$$

$$\displaystyle 2^{4}$$

none of these

Solution

Question 8

$$\displaystyle \frac{\sqrt{5}-\sqrt{3}}{\sqrt{3}}$$

$$\displaystyle \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}}$$

$$\displaystyle \frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$$

$$\displaystyle \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}}$$

Solution

Question 9

$$0$$

$$1$$

$$\dfrac{1}{6}$$

$$6$$

Solution

Question 10

$$2$$

$$3$$

$$4$$

None of these

Solution

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