Trigonometric Functions Test 13

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Trigonometric Functions Test 13

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Weekly Quiz Competition

Question 1
1 / -0

Solve:
$$\dfrac{\cot \theta+\text{cosec }\theta-1}{\cot\theta-\text{cosec }\theta+1}$$

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

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

C
$$\dfrac{1-\cos\theta}{\sin\theta}$$

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

Solution

$$\dfrac{\cot \theta+ \text{cosec }\theta-1}{\cot \theta-\text{cosec }\theta+1}$$

$$=\dfrac{\cot \theta+\text{cosec }\theta-( \text{cosec}^2 \theta-\cot^2\, \theta)}{\cot \theta-\text{cosec }\theta+1}$$

$$=\dfrac{(\cot \theta+\text{cosec }\theta)[1+\cot \theta-\text{cosec }\theta]}{1+\cot \theta-\text{cosec }\theta}$$

$$=\cot \theta+\text{cosec }\theta$$

$$=\dfrac{\cos\,\theta}{\sin \theta}+\dfrac{1}{\sin \theta}$$

$$=\dfrac{1+\cos\,\theta}{\sin\,\theta}$$
Question 2
1 / -0

If $$\cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta } $$, then $$\cos { \theta } +\sin { \theta } $$ is

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

B
$$\sqrt { 2 } \sin { \theta } $$

C
$$0$$

D
$$1$$

Solution

$$\rightarrow { \left( \sin \theta +\cos \theta \right) }^{ 2 }+{ \left( \sin \theta -\cos \theta \right) }^{ 2 }=2$$
$$\rightarrow { \left( \sin \theta +\cos \theta \right) }^{ 2 }+2{ \sin }^{ 2 }\theta =2\Rightarrow \sin \theta +\cos \theta =\sqrt { 2-2{ \sin }^{ 2 }\theta } $$
$$\rightarrow \left( \sin \theta +\cos \theta \right) =\sqrt { 2 } \cos \theta $$
Question 3
1 / -0

$$\tan { \theta } \left( 1-\cot ^{ 2 }{ \theta } \right) $$ is equal to

A
$$\cot { \theta } \left( 1-\tan ^{ 2 }{ \theta } \right) $$

B
$$\cot { \theta } \left( \tan ^{ 2 }{ \theta } -1 \right) $$

C
$$\cot { \theta } \tan ^{ 2 }{ \theta } $$

D
$$\tan { \theta } co\sec ^{ 2 }{ \theta } $$

Solution

$$\rightarrow \tan\theta \left( 1-{ cot }^{ 2 }\theta \right) $$
$$\rightarrow \dfrac { 1 }{ \cot\theta } \left( 1-\dfrac { 1 }{ { \tan }^{ 2 }\theta } \right) $$
$$\rightarrow \dfrac { 1 }{ \cot\theta .{ \tan }^{ 2 }\theta } \left( { \tan }^{ 2 }\theta -1 \right) $$
$$\rightarrow \cot\theta \left( { \tan }^{ 2 }\theta -1 \right) $$
Question 4
1 / -0

The value of
$$\cos { \left( \pi /5 \right) } \cos { \left( 2\pi /5 \right) } \cos { \left( 4\pi /5 \right) } \cos { \left( 8\pi /5 \right) } $$ is

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

B
$$0$$

C
$$\dfrac{-1}{8}$$

D
$$\dfrac{-1}{16}$$

Solution

$$\cos { \left( \pi /5 \right) } \cos { \left( 2\pi /5 \right) } \cos { \left( 4\pi /5 \right) } \cos { \left( 8\pi /5 \right) } =\cfrac { 1 }{ { 2 }^{ 4 }\sin { \left( \pi /5 \right) } } \sin { \cfrac { { 2 }^{ 4 }\pi }{ 5 } } $$
$$=-\cfrac { 1 }{ 16 } \left[ \because\quad \sin { \cfrac { 16\pi }{ 5 } =\sin { \left( 3\pi +\cfrac { \pi }{ 5 } \right) } =-\sin { \cfrac { \pi }{ 5 } } } \right] $$
Question 5
1 / -0

If $$2\cos A+3\cos B+5\cos C$$$$=2\sin A+3\sin B+5\sin C=0$$ then
$$8\cos 3A+27\cos 3B+125\cos C$$$$=k\cos(A+B+C)$$ then $$k=$$

A
$$70$$

B
$$80$$

C
$$90$$

D
$$60$$

Solution

$$\Rightarrow$$$$2\cos { A } +3\cos { B } +5\cos { C } =2\sin { A } +3\sin { B } +5\sin { C } =0$$

$$\Rightarrow$$$$ \cos { \theta } =\cfrac { 1 }{ 2 } ({ e }^{ j\theta }+{ e }^{ -j\theta })$$

$$\Rightarrow$$$$ \sin { \theta } =\cfrac { 1 }{ 2j } ({ e }^{ j\theta }-{ e }^{ -j\theta })$$

$$ \therefore 2\cfrac { 1 }{ 2 } ({ e }^{ jA }+{ e }^{ -jA })+3\cfrac { 1 }{ 2 } ({ e }^{ jB }+{ e }^{ -jB })+5\cfrac { 1 }{ 2 } ({ e }^{ jC }+{ e }^{ -jC })=0$$

$$\Rightarrow$$$$ 2({ e }^{ jA }+{ e }^{ -jA })+3({ e }^{ jB }+{ e }^{ -jB })+5({ e }^{ jC }+{ e }^{ -jC })=0\quad -(i)$$

Also,$$\\ 2({ e }^{ jA }-{ e }^{ -jA })+3({ e }^{ jB }-{ e }^{ -jB })+5({ e }^{ jC }-{ e }^{ -jC })=0\quad -(ii)$$

$$\Rightarrow$$As$$(i)-(ii)$$

$$\Rightarrow$$$$ 4{ e }^{ -jA }+6{ e }^{ -jB }+10{ e }^{ -jC }=0$$

$$\Rightarrow$$$$ 2{ e }^{ -jA }+3{ e }^{ -jB }+5{ e }^{ -jC }=0$$

Cubing on both sides.
$$\Rightarrow$$$$8{ e }^{ -j3A }+27{ e }^{ -j3B }+125{ e }^{ -j3C }+3({ e }^{ -jA }\times 2\times 3{ e }^{ -jB }.5{ e }^{ -jC })=0$$

$$ \therefore 8{ e }^{ -j3A }+27{ e }^{ -j3B }+125{ e }^{ -j3C }=90{ e }^{ -jA }{ e }^{ -jB }{ e }^{ -jC }$$

$$ \therefore k=90$$
Question 6
1 / -0

If $$\sin { \alpha } =12/13\left( 0<\alpha <\pi /2 \right)$$ and
$$\cos { \beta } =-\dfrac { 3 }{ 5 } \left( \pi <\beta <\dfrac { 3 }{ 2 } \pi \right) $$, the value of $$\sin { \left( \alpha +\beta \right) }$$ is

A
$$\dfrac{-56}{65}$$

B
$$\dfrac{16}{65}$$

C
$$\dfrac{56}{65}$$

D
$$\dfrac{-16}{65}$$

Solution

Ans. $$(a)$$
Since $$0<\alpha <\pi$$ we have $$\sin { \alpha }$$ =$$\dfrac{12}{13}$$
$$\Rightarrow \cos { \alpha } =\surd \left[ 1-{ \left( 12/13 \right) }^{ 2 } \right] $$=$$\dfrac{5}{13}$$
And since $$\pi <\beta <3\pi /2$$, we have $$\cos { \beta } =-3/5\Rightarrow \sin { \beta } =-4/5$$
Hence $$\sin { \left( \alpha +\beta \right) } =\sin { \alpha } \cos { \beta } +\cos { \alpha } \sin { \beta }$$
$$=\dfrac { 12 }{ 13 } \left( -\dfrac { 3 }{ 5 } \right) +\left( \dfrac { 5 }{ 13 } \right) \left( -\dfrac { 4 }{ 5 } \right) =-\dfrac { 56 }{ 65 } $$
Question 7
1 / -0

If $$\sin \left(\sin^{-1}\dfrac{1}{5}+\cos ^{-1}x\right)=1$$, then find the value of $$x$$.

A
$$-\dfrac{1}{5}$$

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

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

D
$$\dfrac{1}{5}$$

Solution

We have,

$$ \sin \left( {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x \right)=1 $$

$$ {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x={{\sin }^{-1}}1 $$

$$ {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2} $$

$$ {{\sin }^{-1}}\dfrac{1}{5}=\dfrac{\pi }{2}-{{\cos }^{-1}}x $$

$$ {{\sin }^{-1}}\dfrac{1}{5}={{\sin }^{-1}}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \left( {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \right) $$

$$ x=\dfrac{1}{5} $$

Hence, this is the answer.
Question 8
1 / -0

The maximum value of the expression $$\dfrac { 1 }{ \sin ^{ 2 }{ \theta } +3\sin { \theta } \cos { \theta } +5\cos ^{ 2 }{ \theta } } $$ is

A
$$2$$

B
$$3$$

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

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

Solution

Ans. Ans. $$(a)$$.
$$f\left( \theta \right) =\dfrac { 1 }{ \sin ^{ 2 }{ \theta } +3\sin \theta \cos \theta +5\cos ^{ 2 }{ \theta } }$$
$$=\dfrac { 1 }{ \dfrac { 1-\cos { 2\theta } }{ 2 } +\dfrac { 3 }{ 2 } \sin { 2\theta } +\dfrac { 5\left( 1+\cos { 2\theta } \right) }{ 2 } } $$
$$=\dfrac{ 2 }{ 6+3\sin 2\theta+4 \cos 2\theta }$$
Hence, $$\left[ f\left( \theta \right) \right] _{ maximum }=\dfrac { 2 }{ 6-5 } =2$$
Question 9
1 / -0

The number of solutions of the pair of equations.
$$2\sin^2\theta -\cos 2\theta =0$$
$$2\cos^2\theta -3\sin \theta =0$$
in the interval $$[0, 2\pi]$$ is?

A
Zero

B
One

C
Two

D
Four

Solution

From $$1$$st, $$2\sin^2\theta -(1-2\sin^2\theta)=0$$
$$\therefore 4\sin^2\theta =1$$ or $$\sin\theta =\pm 1/2$$
$$\therefore \theta =\pi/6, \pi +\pi/6$$ in $$[0, 2\pi]$$
Also from $$2$$nd equation, we have
$$2(1-\sin^2\theta)-3\sin\theta =0$$
or $$2\sin^2\theta +3\sin\theta -2=0$$
or $$(\sin \theta +2)(2\sin \theta -1)=0$$
$$\therefore \sin\theta =1/2$$ as $$-2$$ is rejected
$$\therefore \sin\theta =1/2$$ which in already included in $$1$$st. Hence there are only two solutions $$\Rightarrow$$ (c).
Question 10
1 / -0

If $$\alpha, \beta, \gamma, \delta$$ are the smallest $$+$$ive angles in ascending order of magnitude which have their sines equal to a $$+$$ive quantity $$\lambda$$ then the value of $$4\sin \dfrac{\alpha}{2}+3\sin \dfrac{\beta}{2}+2\sin \dfrac{\gamma}{2}+\sin \dfrac{\delta}{2}=$$.

A
$$2\sqrt{1-\lambda}$$

B
$$2\sqrt{1+\lambda}$$

C
$$2\sqrt{\lambda}$$

D
$$2\sqrt{\lambda +2}$$

Solution

Ans. (b)
$$\sin\alpha =\lambda$$
$$\theta =n\pi +(-1)^n\alpha$$
For $$+$$ive values, $$n=0, 1, 2, 3$$
$$\therefore \alpha =\alpha, \beta =\pi -\alpha, \gamma =2\pi +\alpha , \delta =3\pi -\alpha$$
$$\therefore E=4\sin\dfrac{\alpha}{2}+3\sin\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)+2\sin\left(\pi +\dfrac{\alpha}{2}\right)+\sin\left(\dfrac{3\pi}{2}-\dfrac{\alpha}{2}\right)$$
$$=4\sin\dfrac{\alpha}{2}+3\cos\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}-\cos\dfrac{\alpha}{2}$$
$$=2\left(\cos\dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)=2\sqrt{\left(\cos \dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)^2}$$
$$=2\sqrt{1+\sin\alpha}=2\sqrt{1+\lambda}$$.

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Trigonometric Functions Test 13

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Question 1

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

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

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

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

Solution

Question 2

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

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

$$0$$

$$1$$

Solution

Question 3

$$\cot { \theta } \left( 1-\tan ^{ 2 }{ \theta } \right) $$

$$\cot { \theta } \left( \tan ^{ 2 }{ \theta } -1 \right) $$

$$\cot { \theta } \tan ^{ 2 }{ \theta } $$

$$\tan { \theta } co\sec ^{ 2 }{ \theta } $$

Solution

Question 4

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

$$0$$

$$\dfrac{-1}{8}$$

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

Solution

Question 5

$$70$$

$$80$$

$$90$$

$$60$$

Solution

Question 6

$$\dfrac{-56}{65}$$

$$\dfrac{16}{65}$$

$$\dfrac{56}{65}$$

$$\dfrac{-16}{65}$$

Solution

Question 7

$$-\dfrac{1}{5}$$

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

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

$$\dfrac{1}{5}$$

Solution

Question 8

$$2$$

$$3$$

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

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

Solution

Question 9

Zero

One

Two

Four

Solution

Question 10

$$2\sqrt{1-\lambda}$$

$$2\sqrt{1+\lambda}$$

$$2\sqrt{\lambda}$$

$$2\sqrt{\lambda +2}$$

Solution

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