Trigonometric Functions Test 20

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Trigonometric Functions Test 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$cot\Theta =sin2\Theta (where\Theta \neq n\pi ,n\ is\ an\ integer)\Theta $$=

A
45$$^{0}$$ and 60$$^{0}$$

B
45 $$^{0}$$ and 90 $$^{0}$$

C
only 45$$^{0}$$

D
only 90$$^{0}$$

Solution

$$\cot\theta =\sin 2\theta$$
$$\Rightarrow \dfrac{\cos\theta}{\sin\theta}=2\sin\theta \cos\theta$$
$$\Rightarrow \cos\theta =2\sin^2\theta \cos\theta$$
$$\Rightarrow \cos\theta -2\sin^2\theta \cos\theta =0$$
$$\Rightarrow \cos\theta(1-2\sin^2\theta)=0$$
$$\Rightarrow \cos\theta(\cos 2\theta)=0$$
$$\Rightarrow \cos\theta =0$$ or $$\cos 2\theta =0$$
$$\Rightarrow \theta =\dfrac{\pi}{2}$$ or $$2\theta =\dfrac{\pi}{2}$$
$$\Rightarrow \theta =\dfrac{\pi}{4}$$
$$\therefore \theta$$ can be $$45^o$$ or $$90^o$$.
Question 2
1 / -0

In a triangle ABC,
$$b\,cos\,(C+\theta)+c\,cos(B-\theta)=$$

A
$$a\,cos\,\theta$$

B
$$a\,sin\,\theta$$

C
$$a\,tan\,\theta$$

D
$$a\,cot\,\theta$$

Solution

In $$\triangle ABC,A+B+C=\pi$$ and $$a=2 R\sin A,b=2 R\sin B,c=2 R \sin C$$
$$b\cos (C+\theta)+c\cos (B-\theta)=R(2\sin B\cos (C+\theta)+2\sin C\cos (B-\theta))$$
$$=R(\sin (B+C+\theta)+\sin (B-C-\theta)+\sin (B+C-\theta)+\sin (C-B+\theta))$$
$$=R(\sin (\pi-(A-\theta))+\sin (\pi-(A+\theta))+\sin (B-C-\theta)-\sin (B-C-\theta))$$
$$=R(\sin (A-\theta)+\sin (A+\theta))$$
$$=R(2\sin A\cos \theta)$$
$$=(2 R\sin A)\cos \theta$$
$$=a\cos \theta$$
Question 3
1 / -0

If $$\cos\ \theta+\sin\ \theta=a$$, then $$\sin\ 2\theta=$$

A
$$a^{2}+1$$

B
$$a^{2}-1$$

C
$$a^{2}$$

D
$$1$$

Solution

$$\cos{\theta}+\sin{\theta}=a$$
Squaring both sides, we get
$${\cos}^{2}{\theta}+{\sin}^{2}{\theta}+2\cos{\theta}\sin{\theta}={a}^{2}$$
$$\Rightarrow\,1+2\cos{\theta}\sin{\theta}={a}^{2}$$ since $${\cos}^{2}{\theta}+{\sin}^{2}{\theta}=1$$
$$\Rightarrow\,\sin{2\theta}={a}^{2}-1$$
Question 4
1 / -0

If $$\sin^{2} x- \cos x=1/4$$, then the value of $$x$$ between $$0$$ and $$2\pi$$ are :

A
$$\pi /3, 5\pi /3$$

B
$$\pi /3, -\pi/3$$

C
$$2 \pi/3, \pi/3$$

D
$$None\ of\ these$$

Solution

Given

$$\sin ^2x-\cos x=\dfrac 14$$

$$\implies 1-\cos ^2x-\cos x=\dfrac 14$$

$$\implies 4-4\cos ^2x-4\cos x=1$$

$$\implies 4\cos ^2x+4\cos x-3=0$$

$$\implies 4\cos ^2x+6\cos x-2\cos x-3=0$$

$$\implies 2\cos x(2\cos x+3)-1(2\cos x+3)=0$$

$$\implies (2\cos x-1)(2\cos x+3)=0$$

$$\implies \cos x=\dfrac 12$$ and $$ \cos x\neq -\dfrac 32$$

$$\implies x=\dfrac \pi 3,\dfrac {5\pi}3$$
Question 5
1 / -0

$$\dfrac{1+cot\alpha+cosec\alpha}{1-cot\alpha+cosec \alpha}=$$

A
$$\dfrac{sin\alpha}{1+cos\alpha}$$

B
$$\dfrac{sin\alpha}{1-cos\alpha}$$

C
$$\dfrac{1+cos\alpha}{sin\alpha}$$

D
$$\dfrac{1-sin\alpha}{cos\alpha}$$

Solution
Question 6
1 / -0

If $$\sin ^ { - 1 } x = y ,$$ then

A
$$0 \leq y \leq \pi$$

B
$$- \frac { \pi } { 2 } \leq y \leq \frac { \pi } { 2 }$$

C
$$0 < y < \pi$$

D
$$- \frac { \pi } { 2 } < y < \frac { \pi } { 2 }$$

Solution

$$\begin{array}{l} { \sin ^{ -1 } }x=y \\ or\, \, y={ \sin ^{ -1 } }x \end{array}$$
we know that
Range of principle Value of $$\sin $$ is
$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$
So, $$ - \frac{\pi }{2} \le y \le \frac{\pi }{2}$$
Question 7
1 / -0

The equation $$\sin x + \sin y + \sin z = -3$$ for $$0 \le x \le 2\pi, 0 \le y \le 2\pi, 0 \le z \le 2\pi$$, has

A
One solution

B
Two sets of solutions

C
Four sets of solutions

D
No solution

Solution

$$sinx+siny+sinz=-3$$
only possible then $$x=y=z={ sin }^{ -1 }\left( -1 \right) $$
So, $$x=y=z=3\Pi /2$$
for $$0\le x,y,x\le 2\pi $$
So, Only one solution is possible.
Question 8
1 / -0

Number of solutions of the equation $$4\sin^{2}x+\tan^{2}x+\cot^{2}x+\csc^{2}=6$$ in $$[0, \pi]$$ is

A
$$2$$

B
$$8$$

C
$$0$$

D
$$4$$

Solution
Question 9
1 / -0

Number of solution of $$\tan ( 2 x ) = \tan ( 6 x ) \operatorname { in } ( 0,3 \pi )$$ is

A
$$4$$

B
$$5$$

C
$$3$$

D
none of these

Solution
Question 10
1 / -0

Solution of $$7 \sin ^ { 2 } x + 3 \cos ^ { 2 } x = 4$$ is

A
$$n \pi \pm \frac { \pi } { 2 }$$

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

C
$$n \pi \pm \frac { \pi } {3 }$$

D
$$n \pi \pm \frac { \pi } { 6 }$$

Solution

$$\\7sin^2x+3cos^2x=4\\4sin^2x+3sin^2x+3cos^2x=4\\4sin^2x+3=4\\4sin^2x=1\\sin^2x=(\frac{1}{4})\\sinx=\pm (\frac{1}{2})\\\therefore\>x=n\pi\pm (\frac{\pi}{6})$$

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Trigonometric Functions Test 20

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Trigonometric Functions Test 20

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SHARING IS CARING

Question 1

45$$^{0}$$ and 60$$^{0}$$

45 $$^{0}$$ and 90 $$^{0}$$

only 45$$^{0}$$

only 90$$^{0}$$

Solution

Question 2

$$a\,cos\,\theta$$

$$a\,sin\,\theta$$

$$a\,tan\,\theta$$

$$a\,cot\,\theta$$

Solution

Question 3

$$a^{2}+1$$

$$a^{2}-1$$

$$a^{2}$$

$$1$$

Solution

Question 4

$$\pi /3, 5\pi /3$$

$$\pi /3, -\pi/3$$

$$2 \pi/3, \pi/3$$

$$None\ of\ these$$

Solution

Question 5

$$\dfrac{sin\alpha}{1+cos\alpha}$$

$$\dfrac{sin\alpha}{1-cos\alpha}$$

$$\dfrac{1+cos\alpha}{sin\alpha}$$

$$\dfrac{1-sin\alpha}{cos\alpha}$$

Solution

Question 6

$$0 \leq y \leq \pi$$

$$- \frac { \pi } { 2 } \leq y \leq \frac { \pi } { 2 }$$

$$0 < y < \pi$$

$$- \frac { \pi } { 2 } < y < \frac { \pi } { 2 }$$

Solution

Question 7

One solution

Two sets of solutions

Four sets of solutions

No solution

Solution

Question 8

$$2$$

$$8$$

$$0$$

$$4$$

Solution

Question 9

$$4$$

$$5$$

$$3$$

none of these

Solution

Question 10

$$n \pi \pm \frac { \pi } { 2 }$$

$$n \pi \pm \frac { \pi } { 4 }$$

$$n \pi \pm \frac { \pi } {3 }$$

$$n \pi \pm \frac { \pi } { 6 }$$

Solution

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