Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 63

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Weekly Quiz Competition

Question 1

1 / -0

Which of the following pair is a correct trignometric inter-relationship?

(1) $$\cos { \theta } $$	(a) $$\cfrac { 1 }{ \tan { \theta } } $$
(2) $$\tan { \theta } $$	(b) $$\cfrac { 1 }{ co\sec { \theta } } $$
(3) $$\cot { \theta } $$	(c) $$\cfrac { 1 }{ \sec { \theta } } $$
(4) $$\sin { \theta } $$	(d) $$\cfrac { 1 }{ \cot { \theta } } $$

$$1-d,2-e,3-b,4-a$$

$$1-b,2-a,3-e,4-d$$

$$1-c,2-d,3-a,4-b$$

$$1-e,2-b,3-c,4-d$$

Solution

We can write $$\cos \theta$$ as $$\dfrac{1}{\dfrac{1}{\cos \theta}}=\dfrac{1}{\sec \theta}$$

$$\therefore \cos \theta=\dfrac{1}{\sec \theta}$$ .......... $$(1)$$

Now, $$\tan \theta=\dfrac{\sin \theta}{\cos \theta}=\dfrac{1}{\dfrac{\cos \theta}{\sin \theta}}=\dfrac{1}{\cot \theta}$$

$$\therefore \tan \theta=\dfrac{1}{\cot \theta}$$ ........ $$(2)$$

$$\cot{\theta} =\cfrac{1}{\tan{\theta}}$$ ............ $$(3)$$

$$\sin { \theta } =\dfrac{1}{\dfrac{1}{\sin \theta}}=\dfrac{1}{\text{cosec}{\theta}}$$ ...... $$(4)$$

Hence, the correct answer is $$1-c,2-d,3-a,4-b$$

Question 2
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If $$\sin { 7\theta } =\cos { 2\theta } $$ for acute angles $$7\theta$$ and $$2\theta$$, then $$\theta =$$

A
$$10$$

B
$$90$$

C
$$20$$

D
$$30$$

Solution

Given : $$\sin { 7\theta } =\cos { 2\theta } $$
$$\implies \sin { 7\theta } =\sin { (90^o-2\theta ) } $$
$$\implies 7\theta =90^o-2\theta $$
$$\implies 9\theta =90^o$$
$$\implies \theta =10^o$$
Hence, the answer is $$10^o$$.
Question 3
1 / -0

The value of $$\tan 26^{\circ}\tan 64^{\circ}$$ is

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

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

C
$$0$$

D
$$1$$

Solution

$$\tan { 26° } \tan { 64° } =\tan { \left( 90°-64° \right) } \tan { 64° } $$
                             $$=\cot { 64° } \tan { 64° } $$$$(\because \tan (90-x) =\cot x)$$
                             $$=1.$$$$(\because\tan x\cdot \cot x=1 )$$
Hence,  $$\tan { 26° } \tan { 64° } =1.$$
Question 4
1 / -0

The value of $$\sec 29^{\circ} - \text{cosec } 61^{\circ}$$

A
$$1$$

B
$$0$$

C
$$\sec 60^{\circ}$$

D
$$\text{cosec }29^{\circ}$$

Solution

$$\sec { 29° } -\text{cosec}\;61°$$
$$=\sec { \left( 90°-61° \right) } -\text{cosec}\;61°$$
$$=\text{cosec}\;61°-\text{cosec}\;61°$$
$$=0$$
Hence, the answer is $$0.$$
Question 5
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If $$\tan \theta + \cot \theta = 2$$, then the value of $$\tan^{2}\theta + \cot^{2}\theta$$ is ____________.

A
$$4$$

B
$$2$$

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

D
$$5$$

Solution

We know $$tan\theta=\dfrac{1}{cot\theta}$$,

So, $$(tan\theta+cot\theta)^{2}=tan^{2}\theta+cot^{2}\theta+2$$

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

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

$$\Rightarrow tan^{2}\theta+cot^{2}\theta=2$$

So, Option (B)
Question 6
1 / -0

If $$3x \csc 36^{\circ} = \sec 54^{\circ}$$, then the value of $$x$$ is

A
$$0$$

B
$$1$$

C
$$\dfrac {1}{3}$$

D
$$\dfrac {3}{4}$$

Solution

Given,
$$3x\csc\;36°=\sec { 54° } $$
$$\Rightarrow 3x\csc\left( 90°-54° \right) =\sec { 54° } $$
$$\Rightarrow 3x\sec { 54° } =\sec { 54° } $$
$$\Rightarrow 3x=1$$
$$\Rightarrow x=\dfrac { 1 }{ 3 } $$
Hence, the answer is $$\dfrac { 1 }{ 3 }$$
Question 7
1 / -0

If $$4\tan \theta = 3$$, then $$\dfrac {5\sin \theta + 3\cos \theta}{5\sin \theta - 3\cos \theta} = $$

A
$$9$$

B
$$5$$

C
$$10$$

D
$$4$$

Solution

$$4\tan \theta = 3$$
$$\tan \theta = \dfrac {3}{4}$$. So, $$\sin \theta/cos \theta = \dfrac {3}{4}$$

Hence, $$\dfrac {5\left (\dfrac {sin \theta}{cos \theta}\right ) + 3}{5\left (\dfrac {sin \theta}{cos \theta}\right ) - 3} $$.
$$\dfrac {5\left (\dfrac {3}{4}\right ) + 3}{5\left (\dfrac {3}{4}\right ) - 3} $$.
$$=9$$
Question 8
1 / -0

If $$ cos\, 25^0 +sin \, 25^0=k$$, then $$cos\, 20^0$$ is equal to

A
$$\dfrac{k}{\sqrt{2}}$$

B
$$-\dfrac{k}{\sqrt{2}}$$

C
$$\pm \dfrac{k}{\sqrt{2}}$$

D
none of these

Solution

$$\cos 25^{\circ}+\sin 25^{\circ}=k$$
$$\dfrac{1}{\sqrt{2}}\cos 25^{\circ}+\dfrac{1}{\sqrt{2}}\sin 25^{\circ}=\dfrac{k}{\sqrt{2}}$$
$$\sin (45^{\circ}+25^{\circ})=\dfrac{k}{\sqrt{2}}$$
$$\sin 70^{\circ}=\dfrac{k}{\sqrt{2}}$$
$$\sin (90^{\circ}-20^{\circ})=\dfrac{k}{\sqrt{2}}$$
$$\cos 20^{\circ}=\dfrac{k}{\sqrt{2}}$$
Question 9
1 / -0

Choose the correct answers from the alternatives given.
If $$0^{\circ} \leq A \leq 90^{\circ}$$. the simplified form of the given expression $$\sin A \cos A (\tan A- \cot A)$$ is

A
$$1 -$$ $$\cos ^2A$$

B
$$1 - 2$$$$\sin^2A$$

C
$$2\sin^2A -1$$

D
$$1$$

Solution

$$\sin A\cos A(\tan A-\cot A)=\sin A\cos A\bigg(\dfrac{\sin A}{\cos A}-\dfrac{\cos A}{\sin A}\bigg)$$
$$=\sin A\cos A\times \dfrac{\sin A}{\cos A}-\sin A\cos A\times \dfrac{\cos A}{\sin A}$$
$$=\sin^2 A-\cos ^2 A$$
$$=\sin^2 A-(1-\sin^2 A)$$
$$=2\sin^2 A-1$$
Question 10
1 / -0

Evaluate: $$\displaystyle\frac{-\tan\theta \cot(90^o-\theta)+\sec\theta cosec(90^o-\theta)+\sin^235^o+\sin^255^o}{\tan 10^o\tan 20^o\tan 30^o \tan 70^o \tan 80^o}$$.

A
$$\sqrt{3}$$

B
$$2\sqrt{3}$$

C
$$\displaystyle\frac{\sqrt{3}}{2}$$

D
$$1$$

Solution

as we know $$ cot(90^0 $$ $$-$$ $$ \theta $$ $$ ) $$ $$ =$$ $$ tan $$ $$ \theta $$ and $$ cosec(90^0 $$ $$-$$ $$ \theta $$ $$ ) $$ $$ =$$ $$ sec $$ $$ \theta $$
Numerator $$ sec ^2 $$ $$ \theta$$ $$ -$$ $$ tan ^ 2 $$ $$ \theta $$ $$ =1$$ and $$ sin 55^0 =$$ $$ cos 35^0$$
So using the above mentioned identity the numerator outcome is $$ 2 $$ and the denominator $$ tan 80^0 = $$ $$ cot 10^0 $$ and $$ tan 10^0 * cot 10^0 = 1$$ similarly for $$ tan 20^0 cot 20^0 = 1$$
so denominator is $$ tan30^0 = 1 / $$ $$ \sqrt{3} $$
Hence answer is $$ 2* $$ $$ \sqrt{3} $$
Hence option B is correct.

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 63

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Introduction to Trigonometry Test - 63

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SHARING IS CARING

Question 1

$$1-d,2-e,3-b,4-a$$

$$1-b,2-a,3-e,4-d$$

$$1-c,2-d,3-a,4-b$$

$$1-e,2-b,3-c,4-d$$

Solution

Question 2

$$10$$

$$90$$

$$20$$

$$30$$

Solution

Question 3

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

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

$$0$$

$$1$$

Solution

Question 4

$$1$$

$$0$$

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

$$\text{cosec }29^{\circ}$$

Solution

Question 5

$$4$$

$$2$$

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

$$5$$

Solution

Question 6

$$0$$

$$1$$

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

$$\dfrac {3}{4}$$

Solution

Question 7

$$9$$

$$5$$

$$10$$

$$4$$

Solution

Question 8

$$\dfrac{k}{\sqrt{2}}$$

$$-\dfrac{k}{\sqrt{2}}$$

$$\pm \dfrac{k}{\sqrt{2}}$$

none of these

Solution

Question 9

$$1 -$$ $$\cos ^2A$$

$$1 - 2$$$$\sin^2A$$

$$2\sin^2A -1$$

$$1$$

Solution

Question 10

$$\sqrt{3}$$

$$2\sqrt{3}$$

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

$$1$$

Solution

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