Trigonometric Functions Test -7

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Trigonometric Functions Test -7

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statements is incorrect ?

A
sec θ = 1/2

B
tan θ = 1

C
cos θ = 1

D
sinθ = − 1/5

Solution

secθ = 1/2 ⇒ cosθ = 2

which is not possible since cosθcosθ is a periodic function whose value oscillates between -1 and 1
Question 2
1 / -0

For $$x \in (0, \pi)$$, the equation $$\sin x + 2 \sin 2 x - \sin 3x =3$$, has

A
infinitely many solutions

B
three solutions

C
one solution

D
no solution

Solution

$$\sin x + 2 \sin2x - \sin 3 x = 3$$
$$\sin x+4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 3$$
$$\sin x [-2 + 4 \cos x + 4 (1 - \cos^2 x)] =3$$
$$\sin x [2 - (4 \cos^2 x - 4 \cos x +1)+1] =3$$
$$\sin x [3 - (2 \cos x - 1)^2] =3$$
$$\Rightarrow \sin x = 1$$ and $$2 \cos x - 1 =0$$
$$\Rightarrow x = \dfrac{\pi}{2}$$ and $$x = \dfrac{\pi}{3}$$
Which is not possible at same time.

Hence, no solution.
Question 3
1 / -0

If $$0^{\circ}\leq \theta\leq 90^{\circ}$$ and $$\sqrt{3} tan\theta - sec\theta=1$$, then $$\theta$$ has the value

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

$$\sqrt{3} tan \theta -1 =sec \theta$$
Squaring and substituting $$sec^2 \theta =1 + tan^2 \theta$$
$$2 tan^2 \theta -2 \sqrt{3} tan \theta =0$$ $$\therefore 2 tan \theta (tan \theta - \sqrt{3}) =0$$
$$\theta = 60^{\circ}$$ is one of the values.
Question 4
1 / -0

The of value of $$\quad \sin60^{\small\circ}\cos30^{\small\circ} + \cos60^{\small\circ}\sin30^{\small\circ}$$ is equal to

A
$$\sin 90^0$$

B
$$\sin 45^0$$

C
$$\sin 180^0$$

D
$$\sin 30^0$$

Solution

$$sin60^{0}cos30^{0}+cos60^{0}sin30^{0}$$
$$=\frac{3}{4}+\frac{1}{4}$$
$$=1$$
$$=sin90^{0}$$

$$4cos^{3}30^{0}-3cos30^{0}$$
$$=\dfrac{3\sqrt{3}}{2}-\dfrac{3\sqrt{3}}{2}$$
$$=0$$
$$=cos90^{0}$$
Question 5
1 / -0

Which one of the following relations is true ?

A
$$\cos^2 \theta - \sin^2 \theta=1$$

B
$$\text{cosec}^2 \theta - \sec^2 \theta=1$$

C
$$\cot^2 \theta - \tan^2 \theta=1$$

D
$$\sec^2 \theta - \tan^2 \theta=1$$

Solution

(A) L.H.S $$={ \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta$$
$$=1-2 { \sin }^{ 2 }\theta$$ is never $$=1$$.
So this statement is not correct.

(B) L.H.S $$={ \text{cosec} }^{ 2 }\theta -{ \sec }^{ 2 }\theta$$
$$=\dfrac { 1 }{ { \cos }^{ 2 }\theta } -\dfrac { 1 }{ { \sin }^{ 2 }\theta } =\dfrac { { \sin }^{ 2 }\theta -{ \cos }^{ 2 }\theta }{ { \sin }^{ 2 }\theta } \neq 1\neq R.H.S$$
$$\therefore$$ This statement is not correct.

(C) L.H.S $$={ \cot }^{ 2 }\theta -{ \tan }^{ 2 }\theta$$
$$ =\dfrac { { \cos }^{ 2 }\theta }{ { \sin }^{ 2 }\theta } -\dfrac { { \sin }^{ 2 }\theta }{ { \cot }^{ 2 }\theta } =\dfrac { { \cos }^{ 4 }\theta -{ { \sin }^{ 4 } }\theta }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta }$$

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

$$ =\dfrac { { \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } \neq 1\neq R.H.S$$
So this statement is not correct.

(D) R.H.S $$={ \sec }^{ 2 }\theta -{ \tan }^{ 2 }\theta$$
$$=\dfrac { 1 }{ { \cos }^{ 2 }\theta } -\dfrac { { \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =\dfrac { 1-{ \sin }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =\dfrac { { \cos }^{ 2 }\theta }{ { \cos }^{ 2 }\theta } =1=L.H.S$$
This statement is correct.
Question 6
1 / -0

The value of $$5\tan ^{ 2 }{ \theta } -5\sec ^{ 2 }{ \theta } $$ is :

A
$$1$$

B
$$-5$$

C
$$0$$

D
$$5$$

Solution

$$5{ \tan }^{ 2 }\theta -5{ \sec }^{ 2 }\theta \\ =5({ \tan }^{ 2 }\theta- { \sec }^{ 2 }\theta )$$

We know that
$$ { \tan }^{ 2 }\theta +1={ \sec }^{ 2 }\theta \\ \Rightarrow { \tan }^{ 2 }\theta -{ \sec }^{ 2 }\theta =-1\\ \therefore 5(-1)=-5$$
Question 7
1 / -0

$$\displaystyle \frac { \sin { A } }{ 1+\cos { A } } +\frac { \sin { A } }{ 1-\cos { A } } $$ is equal to :

A
$$\displaystyle \sin { A } $$

B
$$\displaystyle 2cosecA$$

C
$$\displaystyle \cos { A } $$

D
None of these

Solution

$$\displaystyle \frac { \sin { A } }{ 1+\cos { A } } +\frac { \sin { A } }{ 1-\cos { A } } $$
$$\displaystyle =\sin { A\left[ \frac { 1-\cos { A } +1+\cos { A } }{ \left( 1+\cos { A } \right) \left( 1-\cos { A } \right) } \right] } $$
$$\displaystyle =\sin { A } \left( \frac { 2 }{ { sin }^{ 2 }A } \right) =\frac { 2 }{ \sin { A } } =2cosecA$$
Question 8
1 / -0

If $$0^{\circ} \leq \theta \leq 90^{\circ}$$ and $$\sqrt{2} tan \theta - sec \theta =0$$, then the value of $$(\sqrt{2} sin \theta + 2 tan \theta)$$ is -

A
$$\sqrt{2}+2$$

B
$$\frac{\sqrt{2}}{2} + \frac{2}{\sqrt{3}}$$

C
3

D
2

Solution

$$\sqrt{2} \tan \theta = \sec \theta\ or\ 2 \tan^2 \theta = \tan^2 \theta +1$$
$$\therefore \tan^2 \theta =1 \therefore \theta = 45^{\circ}$$
$$\sqrt{2} \sin \theta + 2 \tan \theta = \sqrt{2} \times \frac{1}{\sqrt{2}} + 2 \times 1 =3$$
Question 9
1 / -0

Which is true for all values of $$\theta$$?

A
$$\sin^2 \theta -\cos^2\theta=1$$

B
$$\text{cosec}^2\theta- \cot^2 \theta=1$$

C
$$\text{cosec}^2 \theta - \cos^2 \theta=1$$

D
$$\sec^2 \theta - \sin^2 \theta=1$$

Solution

We need to verify every option
$$(A)$$
Consider, $$\sin^2 \theta - \cos^2 \theta$$
$$=1-\cos^2 \theta-\cos^2 \theta$$
$$=1-2\cos^2 \theta$$
$$=-(2\cos^2 \theta-1)=-\cos 2\theta$$
Now, $$-\cos 2\theta=1$$ if $$\cos 2\theta=-1\implies 2\theta=\cos^{-1}(-1)=(2n+1)\pi$$
$$\implies \theta=\dfrac{(2n+1)\pi}{2}, n\in \mathbb{Z}$$
Thus, it is not true for every $$\theta$$

$$(B)$$
Consider, $$\text{cosec}^2\theta-\cot^2\theta$$
$$=\dfrac{1}{\sin^2 \theta}-\dfrac{\cos^2\theta}{\sin^2 \theta}$$
$$=\dfrac{1-\cos^2\theta}{\sin^2 \theta}=\dfrac{\sin^2 \theta}{\sin^2 \theta}=1$$
$$\therefore \text{cosec}^2\theta-\cot^2\theta=1,$$ For every value of $$\theta$$

$$(C)$$
$$\text{cosec}^2\theta-\cos^2 \theta=1\implies 1-\sin^2\theta \cos^2\theta=\sin^2\theta$$
$$\implies 1-\sin^2\theta=\sin^2\theta \cos^2\theta$$
$$\implies \sin^2\theta=1 \implies \theta=\dfrac{(2n+1)\pi}{2}$$
Thus, $$\text{cosec}^2\theta-\cos^2 \theta=1$$ is true for $$\theta=\dfrac{(2n+1)\pi}{2}$$ but not for all values of $$\theta$$.

$$(D)$$
$$\sec^2\theta-\sin^2\theta=1\implies \dfrac{1}{\cos^2\theta}-\sin^2\theta=1$$
$$\implies1-\sin^2\theta \cos^2\theta=\cos^2\theta$$
$$\implies \sin^2\theta=\sin^2\theta \cos^2\theta \implies \cos^2\theta=1$$
$$\implies \cos \theta=\pm 1 \implies \theta=n\pi, n\in \mathbb{Z}$$
$$\sec^2\theta-\sin^2\theta=1$$ for $$\theta\in \mathbb{Z}$$ but not for all values of $$\theta$$.
Hence, option B is correct.
Question 10
1 / -0

The value of $$ 1+ \cot^2 A$$ is

A
$$\cos^2 A$$

B
$$\sec^2 A$$

C
$$\tan^2 A$$

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

Solution

We know that,
$$\cot A= \dfrac{\text{base}}{\text{perpendicular}} = \dfrac bp$$

$$1+\cot^2A=1+\dfrac{b^2}{p^2}=\dfrac{b^2+p^2}{p^2}=\dfrac{h^2}{p^2}=\text{cosec}^2A$$
Question 11
1 / -0

$$sin^2x+cos^2x=$$

A
0

B
$$\sqrt{2}$$

C
1

D
-1

Solution

Identity used:
$$\sin^2\theta+\cos^2\theta=1$$
Option C is correct.

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Trigonometric Functions Test -7

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Question 1

sec θ = 1/2

tan θ = 1

cos θ = 1

sinθ = − 1/5

Solution

Question 2

infinitely many solutions

three solutions

one solution

no solution

Solution

Question 3

$$30^{\circ}$$

$$45^{\circ}$$

$$60^{\circ}$$

$$90^{\circ}$$

Solution

Question 4

$$\sin 90^0$$

$$\sin 45^0$$

$$\sin 180^0$$

$$\sin 30^0$$

Solution

Question 5

$$\cos^2 \theta - \sin^2 \theta=1$$

$$\text{cosec}^2 \theta - \sec^2 \theta=1$$

$$\cot^2 \theta - \tan^2 \theta=1$$

$$\sec^2 \theta - \tan^2 \theta=1$$

Solution

Question 6

$$1$$

$$-5$$

$$0$$

$$5$$

Solution

Question 7

$$\displaystyle \sin { A } $$

$$\displaystyle 2cosecA$$

$$\displaystyle \cos { A } $$

None of these

Solution

Question 8

$$\sqrt{2}+2$$

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

3

2

Solution

Question 9

$$\sin^2 \theta -\cos^2\theta=1$$

$$\text{cosec}^2\theta- \cot^2 \theta=1$$

$$\text{cosec}^2 \theta - \cos^2 \theta=1$$

$$\sec^2 \theta - \sin^2 \theta=1$$

Solution

Question 10

$$\cos^2 A$$

$$\sec^2 A$$

$$\tan^2 A$$

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

Solution

Question 11

0

$$\sqrt{2}$$

1

-1

Solution

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