Trigonometric Functions Test -3

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Trigonometric Functions Test -3

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Question 1
1 / -0

$$\displaystyle \frac{\sin(\theta+A)}{\sin(\theta+B)}=\sqrt{\frac{\sin 2A}{\sin 2B}}$$ then $$\tan^{2}\theta=$$

A
tanA

B
tanB

C
tanA+tanB

D
tanAtanB

Solution

$$\cfrac { sin\left( \theta +A \right) }{ sin\left( \theta +B \right) } =\sqrt { \cfrac { sin2A }{ sin2B } } $$
Squaring both sides, we get
$$\cfrac { { sin }^{ 2 }\left( \theta +A \right) }{ { sin }^{ 2 }\left( \theta +B \right) } =\cfrac { sin2A }{ sin2B } \\ \Rightarrow \cfrac { { sin }^{ 2 }\theta { cos }^{ 2 }A+{ cos }^{ 2 }\theta { sin }^{ 2 }A }{ { sin }^{ 2 }\theta { cos }^{ 2 }B+{ cos }^{ 2 }\theta { sin }^{ 2 }B } =\cfrac { 2sin^2Acos^2A }{ 2sin^2Bcos^2B } \\ \Rightarrow { sin }^{ 2 }\theta { cos }^{ 2 }AsinBcosB+{ cos }^{ 2 }\theta { sin }^{ 2 }AsinBcosB={ sin }^{ 2 }\theta { cos }^{ 2 }BsinAcosA+{ cos }^{ 2 }\theta { sin }^{ 2 }BsinAcosA\\ \Rightarrow { sin }^{ 2 }\theta { cos }^{ 2 }AsinBcosB-{ sin }^{ 2 }\theta { cos }^{ 2 }BsinAcosA={ cos }^{ 2 }\theta { sin }^{ 2 }BsinAcosA-{ cos }^{ 2 }\theta { sin }^{ 2 }AsinBcosB\\ \Rightarrow { sin }^{ 2 }\theta cosAcosB\left( cosAsinB-sinAcosB \right) ={ cos }^{ 2 }\theta sinAsinB\left( cosAsinB-sinAcosB \right) \\ \Rightarrow \cfrac { { sin }^{ 2 }\theta }{ { cos }^{ 2 }\theta } =\cfrac { sinAsinB }{ cosAcosB } \quad \Rightarrow { tan }^{ 2 }\theta =tanAtanB$$
Question 2
1 / -0

lf $$\alpha$$ and $$\beta$$ are two different solutions lying between $$\displaystyle \frac{-\pi}{2}$$ and $$\displaystyle \frac{\pi}{2}$$ of the equation $$2t\mathrm{a}n\theta+\mathrm{S}\mathrm{e}\mathrm{c}\theta=2$$ then $$ t\mathrm{a}n\alpha+t\mathrm{a}n\beta$$

A
0

B
1

C
4/3

D
8/3

Solution

The given equation is $$2\tan{\theta}+\sec{\theta}=2$$
$$\Rightarrow \sec{\theta}=2-2\tan{\theta}$$
Squaring on both sides, we get
$${\sec}^{2}{\theta}=4{\left(1-\tan{\theta}\right)}^{2}$$
$${\sec}^{2}{\theta}=4\left(1+{\tan}^{2}{\theta}-2\tan{\theta}\right)$$
$$1+{\tan}^{2}{\theta}=4\left(1+{\tan}^{2}{\theta}\right)-8\tan{\theta}$$
$$3{\tan}^{2}{\theta}-8\tan{\theta}+3=0$$
Since $$\alpha$$ and $$\beta$$ are the roots of the equation,
$$\therefore \tan{\alpha}+\tan{\beta}=\dfrac{8}{3}$$
Question 3
1 / -0

If $$\displaystyle \frac{\cos(A+B)}{\cos(A-B)}+\frac{\cos(C-D)}{\cos(C+D)}=0$$, then $$\cot A\ cot B\cot C\, cot D$$ is equal to

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

Given $$\displaystyle \dfrac{\cos (A+B)}{\cos (A-B)}+\dfrac{\cos (C-D)}{\cos (C+D)}=0 $$

$$\Rightarrow \dfrac{1-\tan A \tan B}{1+\tan A \tan B}+\dfrac{1+\tan C \tan D}{1-\tan C \tan D}=0$$
$$[\cos (A+B)=\cos A \cos B-\sin A \sin B , \cos (A-B)=\cos A \cos B+\sin A \sin B]$$

$$\displaystyle =\dfrac{1-\tan A \tan B-\tan C \tan D+\tan A \tan B \tan C \tan D +1+\tan A \tan B+\tan C \tan D +\tan A \tan B \tan C \tan D}{(1+\tan A \tan B)(1-\tan C \tan D)}$$

$$\displaystyle \Rightarrow \dfrac{2(1+\tan A \tan B \tan C \tan D)}{(1+\tan A \tan B)(1-\tan C \tan D)}=0$$

$$ \Rightarrow\tan A \tan B \tan C \tan D=-1$$

$$\Rightarrow \cot A \cot B \cot C \cot D =-1$$
Question 4
1 / -0

If $$y\displaystyle \tan\left (\theta-\dfrac {\pi}{6}\right ) = x\tan \left (\theta + \dfrac {2\pi}{3}\right ) $$ , then $$\dfrac {y+x}{y-x}=$$

A
$$\cos {2\theta}$$

B
$$2\cos {2\theta}$$

C
$$\sin {2\theta}$$

D
$$2\sin {2\theta}$$

Solution

$$\dfrac{y}{x}=\dfrac{tan(\theta+\dfrac{2\pi}{3})}{tan(\theta-\dfrac{\pi}{6})}$$

$$\dfrac{y+x}{y-x}=\dfrac{tan(\theta+\dfrac{2\pi}{3})+tan(\theta-\dfrac{\pi}{6})}{tan(\theta+\dfrac{2\pi}{3}) -tan(\theta-\dfrac{\pi}{6})}$$

$$=\dfrac{sin(\theta+\dfrac{2\pi}{3}+\theta-\dfrac{\pi}{6})}{sin(\theta+\dfrac{2\pi}{3}-\theta+\dfrac{\pi}{6})}$$

$$=\dfrac{sin(2\theta+\dfrac{\pi}{2})}{sin(\dfrac{5\pi}{6})}$$

$$=\dfrac{cos2\theta}{sin(\pi-\dfrac{\pi}{6})}$$

$$=cosec(\dfrac{\pi}{6}).cos2\theta$$

$$=2cos2\theta$$.
Question 5
1 / -0

$$\cos\alpha\sin(\beta-\gamma)+\cos\beta\sin(\gamma-\alpha)+\cos\gamma\sin(\alpha-\beta)= $$

A
$$0$$

B
1/2

C
1

D
$$ 4\cos\alpha\cos\beta\cos\gamma$$

Solution

$$cos \alpha sin (\beta -\gamma )+cos \beta (sin (\gamma -\alpha ))+cos \gamma sin (\alpha -\beta )$$

$$=cos \alpha (sin \beta cos \gamma -sin \gamma cos \beta ) +cos \beta (sin \gamma cos \alpha -sin \alpha cos \gamma )+cos \gamma (sin \alpha cos \beta -sin \beta cos \alpha )$$

$$=0 [\because sin (a-b)=sin a cos b-sin b cos a]$$
Question 6
1 / -0

$$\displaystyle \sum \cos \alpha \sin (\beta - \gamma) =$$

A
$$1$$

B
$$4 \cos \alpha \cos \beta \cos \gamma$$

C
$$0$$

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

Solution

$$\displaystyle \sum \cos \alpha \sin (\beta -\gamma) =\cos \alpha \sin (\beta -\gamma)+\cos \beta \sin (\gamma -\alpha)+\cos \gamma \sin (\alpha -\beta)$$
$$=\cos \alpha \sin \beta \cos \gamma -\cos \alpha \cos \beta \sin \gamma +\cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha +\cos \gamma \sin \alpha \cos \beta -\cos \gamma \cos \beta \sin \alpha$$
$$=0$$
Question 7
1 / -0

$$\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$$. Find the value of $$\sin 3A+\sin 3B$$

A
$$0$$

B
$$2$$

C
$$1$$

D
$$-1$$

Solution

$$\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$$

$$\dfrac { 1 }{ 2 } \sin A+\dfrac { 1 }{ 2 } \sin B=\dfrac { \sqrt { 3 } }{ 2 } \cos B-\dfrac { \sqrt { 3 } }{ 2 } \cos A$$

$$\dfrac { 1 }{ 2 } \sin A+\dfrac { \sqrt { 3 } }{ 2 } \cos A=\dfrac { \sqrt { 3 } }{ 2 } \cos B-\dfrac { 1 }{ 2 } \sin B$$

$$\sin (A+{ 60 }^{ 0 })=\cos (B+{ 30 }^{ 0 })$$

$$\sin (A+{ 60 }^{ 0 })=\sin ({ 60 }^{ 0 }-B)$$

$$\Rightarrow A+{ 60 }^{ 0 }={ 60 }^{ 0 }-B$$

So, $$\sin 3A+\sin 3B=sin(-3B)+\sin 3B=0$$
Question 8
1 / -0

If $$\displaystyle \tan A=\cfrac{x\sin B}{1-x\cos B}$$
and $$\displaystyle \tan B=\cfrac{y\sin A}{1-y\cos A}$$, then $$\displaystyle \cfrac{\sin A}{\sin B}=$$

A
$$\dfrac{x}{y}$$

B
$$\dfrac{y}{x}$$

C
$$x+y$$

D
$$x-y$$

Solution

$$\displaystyle \tan A =\cfrac{\sin A }{cos A}=\cfrac{x \sin B}{1-x cos B}$$
$$\sin A -x \sin A cos B=x \sin B cos A$$
$$\displaystyle x=\cfrac{\sin A }{\sin (A+B)}$$
Similarly , $$\displaystyle y=\cfrac{\sin B}{\sin (A+B)}$$
$$\displaystyle \therefore \cfrac{\sin A }{\sin B}=\cfrac{x}{y}$$
Question 9
1 / -0

If $$\displaystyle \sin x\cos y=\frac{1}{4},3\tan x=4\tan y$$, then $$ \sin(x-y)=$$

A
$$\dfrac{1}{16}$$

B
$$\dfrac{7}{16}$$

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

D
$$\dfrac{3}{16}$$

Solution

Given $$\sin x \cos y=\cfrac{1}{4}$$ .......1
$$3\tan x=4\tan y$$
$$3 \sin x \cos y=4 \sin y \cos x$$
from 1
$$3\times \cfrac{1}{4}=4 \sin y \cos x$$
$$\sin y \cos x=\cfrac{3}{16}$$ ............2
$$\therefore \sin (x-y)=\sin x \cos y-\sin y \cos x$$
$$\displaystyle =\cfrac{1}{4}-\cfrac{3}{16}=\cfrac{1}{16}$$
Question 10
1 / -0

For all values of $$\theta,\cos\theta.\text{cosec }\theta\sqrt{\sec^2\theta-1}$$ is

A
$$1$$

B
$$\cos^2\theta$$

C
$$\cot\theta$$

D
$$\tan\theta$$

Solution

$$ { \cos }\theta\text{cosec }\theta \sqrt { { \sec }^{ 2 }\theta -1 }$$
$$={ \cos }\theta .\dfrac { 1 }{ \sin\theta } .\sqrt { { \tan }^{ 2 }\theta }$$

$$=\cot\theta .\tan\theta \\ =1 $$.

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Re-Attempt Weekly Quiz Competition

Trigonometric Functions Test -3

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Trigonometric Functions Test -3

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SHARING IS CARING

Question 1

tanA

tanB

tanA+tanB

tanAtanB

Solution

Question 2

0

1

4/3

8/3

Solution

Question 3

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 4

$$\cos {2\theta}$$

$$2\cos {2\theta}$$

$$\sin {2\theta}$$

$$2\sin {2\theta}$$

Solution

Question 5

$$0$$

1/2

1

$$ 4\cos\alpha\cos\beta\cos\gamma$$

Solution

Question 6

$$1$$

$$4 \cos \alpha \cos \beta \cos \gamma$$

$$0$$

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

Solution

Question 7

$$0$$

$$2$$

$$1$$

$$-1$$

Solution

Question 8

$$\dfrac{x}{y}$$

$$\dfrac{y}{x}$$

$$x+y$$

$$x-y$$

Solution

Question 9

$$\dfrac{1}{16}$$

$$\dfrac{7}{16}$$

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

$$\dfrac{3}{16}$$

Solution

Question 10

$$1$$

$$\cos^2\theta$$

$$\cot\theta$$

$$\tan\theta$$

Solution

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