Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\sin^2 60^0 + \tan 45^0 - \text{cosec}^2 45^0$$ is

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

B
$$-\dfrac{1}{4}$$

C
$$\dfrac{1}{4}$$

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

Solution

$$\sin^2 60^0 + \tan 45^0 - \text{cosec}^2 45^0$$
$$=\left(\dfrac{\sqrt 3}{2}\right)^2+1-(\sqrt 2)^2=\dfrac34-1=\dfrac{-1}{4}$$
Question 2
1 / -0

The value of $$ 4(\sin^4 30^0 + \cos^4 60^0)-3 (\cos^2 45^0 \sin^2 90^0)$$ is

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$-2$$

Solution

$$4(\sin^430^0+\cos^460^0)-3\sin^290^0\cos^2(45^0)$$

$$=4(\dfrac{1}{16}+\dfrac{1}{16})-3(1)(\dfrac{1}{2})$$

$$=\dfrac{8}{16}-\dfrac{3}{2}$$

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

$$=-1$$
Hence answer is $$-1$$
Question 3
1 / -0

If $$a \cos\theta-b \sin\theta=c$$, then $$a \sin\theta+b\cos\theta$$ equals-

A
$$a^2+b^2+c^2$$

B
$$a^2-b^2-c^2$$

C
$$\sqrt {a^2+b^2+c^2}$$

D
$$\pm \sqrt {a^2+b^2-c^2}$$

Solution

$$ a\cos\theta - b\sin\theta = c$$

Let,$$a\sin\theta + b\cos\theta = x$$

Squaring and adding we get
$$a^2\cos^2 \theta+b^2 \sin^2 \theta-2ab\sin \theta \cos \theta+a^2 \sin^2 \theta+b^2 \cos^2 \theta+2ab\sin \theta \cos \theta=c^2+x^2$$

$$\Rightarrow a^2 + b^2 -c^2 = x^2 $$

$$x=\pm \sqrt{a^2+b^2-c^2}$$

Option D is the correct option
Question 4
1 / -0

The value of $$\sec$$ $$(90^0 - \theta) \sin \theta$$ is

A
$$\sec \theta$$

B
$$\csc \theta$$

C
$$ \tan \theta$$

D
$$1$$

Solution

$$\sec (90^0-\theta)\sin\theta$$
$$=\csc\theta\:\sin\theta$$
$$=\dfrac{1}{\sin\theta}\sin\theta$$
$$=1$$
Hence answer is D
Question 5
1 / -0

The value of $$\sin^2 60^0.\cos^2 30^0 + \tan 45^0. \cos 60^0 \sin 30^0$$ is

A
$$\dfrac{13}{16}$$

B
$$5$$

C
$$1$$

D
$$-\dfrac{1}{2}$$

Solution

$$\sin ^260^0\cos ^230^0+\tan45^0\cos 60^0\sin 30^0$$

$$=\sin ^260^0\sin ^260^0+(1)\cos ^260^0$$

$$=\sin ^460^0+\cos ^260^0$$

Substituting the values of $$\sin 60^0$$ and $$\cos 60^0$$ we get

$$(\dfrac{\sqrt{3}}{2})\:^4+(\dfrac{1}{2})\:^2$$

$$=\dfrac{9}{16}+\dfrac{1}{4}$$

$$=\dfrac{13}{16}$$

Hence answer is A
Question 6
1 / -0

The value of $$\sin^2q. \cos^2q (\sec^2q+\text cosec^2q)$$ is

A
$$2$$

B
$$4$$

C
$$1$$

D
$$0$$

Solution

Given: $$\sin^2q \cos^2q (\sec^2q+\text{cosec}^2q)$$
$$=\sin^2q \cos^2q (\dfrac {1}{\cos^2q}+\cfrac {1}{\sin^2q})$$

$$=\sin^2q \cos^2q(\dfrac{\sin^2q+\cos^2q}{\sin^2q\cos^2q})$$

$$=\sin^2q+\cos^2q=1$$
Question 7
1 / -0

The value of $$4 \cos^2 60^0 + 4 \tan^2 45^0 - \csc^2 30^0$$ is

A
$$0$$

B
$$2$$

C
$$1$$

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

Solution

$$4\cos^260^0+4\tan^245^0-\csc^230^0$$
$$=4(\dfrac{1}{2})\:^2+4(1)-2^2$$
$$=1+4-4$$
$$=1$$
Hence answer is C
Question 8
1 / -0

In the given figure, the side $$PQ$$ (in cm) is

A
$$25$$

B
$$50$$

C
$$75$$

D
$$80$$

Solution

In the given triangle
$$AB=BP$$
Hence, $$\triangle ABP$$ is an isosceles triangle
Therefore, $$\angle PAB=\angle APB$$
Hence, $$\angle BPQ=90^0-(\angle PAB+\angle APB)$$
$$=(90^0-15^0-15^0)=60^0$$
Consider $$\triangle BPQ$$
$$\cos60^0=\dfrac{1}{2}=\dfrac{PQ}{BP}$$
$$=\dfrac{PQ}{100}$$
Hence, $$PQ=50$$ cm
The answer is option B.
Question 9
1 / -0

The value of $$\sin^6q+\cos^6q+3 \sin^2q \cos^2q$$ is-

A
$$0$$

B
$$1$$

C
$$2\sin q \cos q$$

D
$$\sin^4q +\cos^4q$$

Solution

$$\sin^6q+\cos^6q+3 \sin^2q \cos^2q$$
$$\Rightarrow (\sin^2q+\cos^2q)(\sin^4q+\cos^4q-\sin^2q \cos^2q)+3 \sin^2q \cos^2q$$
$$\Rightarrow \sin^4q+\cos^4q+2\sin^2q \cos^2q$$
$$\Rightarrow (\sin^2q+\cos^2q)^2=1$$.
Question 10
1 / -0

The value of $$\sin$$ $$45^0 \cos 45^0 (\tan 45^0 + \cot 45^0)$$ is

A
$$2$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

$$\sin 45^0 \cos 45^0(\tan 45^0+\ cot 45^0)$$
$$=(\dfrac{1}{\sqrt{2}})(\dfrac{1}{\sqrt{2}})(1+1)$$
$$=\dfrac{1}{2}(2)$$
$$=1$$
Hence answer is B

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Re-Attempt Weekly Quiz Competition

Introduction to Trigonometry Test - 31

Result

Introduction to Trigonometry Test - 31

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SHARING IS CARING

Question 1

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

$$-\dfrac{1}{4}$$

$$\dfrac{1}{4}$$

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

Solution

Question 2

$$0$$

$$1$$

$$-1$$

$$-2$$

Solution

Question 3

$$a^2+b^2+c^2$$

$$a^2-b^2-c^2$$

$$\sqrt {a^2+b^2+c^2}$$

$$\pm \sqrt {a^2+b^2-c^2}$$

Solution

Question 4

$$\sec \theta$$

$$\csc \theta$$

$$ \tan \theta$$

$$1$$

Solution

Question 5

$$\dfrac{13}{16}$$

$$5$$

$$1$$

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

Solution

Question 6

$$2$$

$$4$$

$$1$$

$$0$$

Solution

Question 7

$$0$$

$$2$$

$$1$$

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

Solution

Question 8

$$25$$

$$50$$

$$75$$

$$80$$

Solution

Question 9

$$0$$

$$1$$

$$2\sin q \cos q$$

$$\sin^4q +\cos^4q$$

Solution

Question 10

$$2$$

$$1$$

$$3$$

$$4$$

Solution

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