Trigonometric Functions Test 19

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Trigonometric Functions Test 19

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Question 1
1 / -0

If $$(1+\tan{A}).(1+\tan{B})=2$$, then $$A+B$$ is

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{\pi}{3}$$

C
$$\dfrac{\pi}{4}$$

D
$$\dfrac{\pi}{6}$$

Solution

$$(1+\tan A)(1+\tan B)=2$$
$$\therefore \tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$$ ........... $$(1)$$
Now $$1+\tan A+\tan B+\tan A\tan B=2$$
$$\therefore \tan A+\tan B=1-\tan A\tan B$$
$$\therefore\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1$$ ........ $$(2)$$
$$\therefore$$ From $$(1)$$ & $$(2)$$
$$\tan (A+B)=1$$
$$\therefore A+B=\dfrac{\pi}{4}$$
Question 2
1 / -0

If $$3\sin { 2\theta } =2\sin { 3\theta }$$ and $$0<\theta <\pi$$, then value of $$\sin { \theta }$$ is

A
$$\dfrac{\sqrt{2}}{3}$$

B
`$$\dfrac{\sqrt{3}}{\sqrt{5}}$$

C
$$\dfrac{\sqrt{15}}{4}$$

D
$$\dfrac{\sqrt{2}}{\sqrt{5}}$$

Solution

$$3 \sin 2 \theta = 2 \sin 3 \theta$$
$$\rightarrow 3 \times 2 \sin \cos \theta = 2(3 \sin \theta - 4 \sin{3} \theta )$$
$$\rightarrow 6 \sin \theta \cos \theta = 6 \sin \theta - 8 \sin{3} \theta$$
$$\rightarrow \sin \theta (6 \cos \theta - 6+8 \sin^{2} \theta )=0$$
$$\therefore \sin \theta =0$$ or $$3 \cos \theta+ 4 \sin{2} \theta =3$$
$$\therefore \sin \theta =0$$ or $$ 3 \cos \theta + 4 (1- \cos{2} \theta)=3$$
$$\therefore \sin \theta =0$$ or $$3 \cos \theta -4 \cos{2} \theta +1 =0$$
$$\therefore 4 \cos^{2} \theta- 3 \cos \theta -1 =0$$
$$\therefore 4 \cos^{2} \theta - 4 \cos \theta + \cos \theta -1 =0$$
$$\therefore 4 \cos \theta (\cos \theta -1) +1 (\cos \theta -1)=0$$
$$\therefore (4 \cos \theta +1) (\cos \theta -1)=0$$
$$\therefore \cos \theta = \dfrac{-1}{4}$$ or $$ \cos \theta =1$$
$$\therefore \sin \theta = \sqrt{1-\left( \dfrac{-1}{4} \right)^{2}}$$
$$\therefore \sin \theta =\sqrt{1- \dfrac{1}{16}}$$
$$\therefore \sin \theta = \dfrac{\sqrt{15}}{4}$$
Question 3
1 / -0

State whether the following statement is true or false:
$$cosec{20^ \circ }\sec {20^ \circ } = $$

A
$$2$$

B
$$2 cosec {40^ \circ}$$

C
$$4$$

D
$$2\sin{45^ \circ} . cosec {40^ \circ}$$

Solution

$$csc 20 \sec 20$$

$$\implies \dfrac 1{\sin 20\cos 20}$$

$$\implies \dfrac 2{2\sin 20\cos 20}$$

$$\implies \dfrac 2{\sin 2(20)}$$

$$\implies \dfrac 2{\sin 40}$$

$$\implies 2{\csc 40}$$
Question 4
1 / -0

$${ cos }^{ 2 }2x+2{ cos }^{ 2 }x=1,x\epsilon \left( -\pi ,\pi \right)$$, then $$x$$ can take the values

A
$$\pm \dfrac { \pi }{ 2 }$$

B
$$\pm \dfrac { \pi }{ 4 }$$

C
$$\pm \dfrac { 3\pi }{ 8 }$$

D
Non of these

Solution

Given $$x\in (-\pi,\pi)\implies 2 x\in (-2\pi,2\pi)$$
$$\cos^2 2 x+2\cos^2 x=1$$
$$\implies \cos^2 2 x+2\cos^2 x-1=0$$
$$\implies \cos^2 2 x+(2\cos^2 x-1)=0$$
$$\implies \cos^2 2 x+\cos 2 x=0$$
$$\implies \cos 2 x(\cos 2 x+1)=0$$
So $$\cos 2 x=0$$ or $$-1$$
$$\implies 2 x=\pm \dfrac{\pi}{2},\pm \dfrac{3\pi}{2}$$ or $$\pm \pi$$
$$\implies x=\pm \dfrac{\pi}{4},\pm \dfrac{3\pi}{4}$$ or $$\pm\dfrac{\pi}{2}$$
Question 5
1 / -0

The number of solutions to the equation $$sin\ 5x+\ sin\ 3x\ +\ sin\ x=0$$ for $$\ 0\leq x\leq \pi$$ is

A
1

B
2

C
3

D
None of these

Solution

$$ sin x+sin3x+sin5x = 0 x\in [0,\pi ]$$

$$ (sinx+sin5x)+sin3x = 0$$

$$ 2sin 3x cos2x + sin3x = 0$$ $$\displaystyle [\because sinx+siny = 2\left | sin \dfrac{x+y}{2} \right | \left | cos\dfrac{x-y}{2} \right |]$$

$$ sin 3x (2cos2x+1) = 0$$

$$ \Rightarrow sin 3x = 0 , 2cos2x +1 = 0$$

$$ 3x = n\pi$$ $$\quad|$$ , $$cos 2x = \frac{-1}{2}$$

$$ x = \dfrac{n\pi }{3}$$$$\quad|$$ $$ 2x =2n\pi \pm \dfrac{2\pi }{3} [\because cos2\dfrac{2\pi }{3} = \dfrac{-1}{2}]$$

$$ \quad \quad \quad \quad \, | x = n\pi \pm \dfrac{\pi }{3}$$

For $$ x\in [ 0,\pi ], x = 0, \dfrac{\pi }{3}, \dfrac{2\pi }{3}, \pi $$ [ 4 solutions]

$$ \therefore $$ None of these is correct option
Question 6
1 / -0

If $$\cos^{2}x=\sin x$$, then the value of $$\cos^{2}x(1+\cos^{2}x)$$ is equal to

A
$$-1$$

B
$$1$$

C
$$0$$

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

Solution

Given $$\cos^2 x=\sin x$$
Squaring on both sides
$$\implies (\cos^2 x)^2=\sin^2 x$$
$$\implies \cos^4 x=1-\cos^2 x$$
$$\implies \cos^4 x+\cos^2 x=1$$
$$\implies \cos^2 x(1+\cos^2 x)=1$$
Question 7
1 / -0

If $$\cos2x\ +\ 2\cos\ x\ =\ 1$$, then $$\sin^{2}x(2-\cos^{2}x)$$ is

A
$$1$$

B
$$2$$

C
$$4$$

D
none of these

Solution

Solution:- (A) $$1$$
$$\cos{2x} + 2 \cos{x} = 1$$
$$2 \cos^{2}{x} - 1 + 2 \cos{x} = 1 \; \left( \because \cos{2x} = 2 \cos^{2}{x} - 1 \right)$$
$$\Rightarrow 2 \cos^{2}{x} + 2 \cos{x} - 2 = 0$$
$$\Rightarrow \cos^{2}{x} + \cos{x} = 1 ..... \left( 1 \right)$$
$$\Rightarrow 1 - \cos^{2}{x} = \cos{x}$$
$$\Rightarrow \sin^{2}{x} = \cos{x} ..... \left( 2 \right)$$
Now,
$$\sin^{2} \left( 2 - \cos^{2}{x} \right)$$
$$= \sin^{2}{x} \left( 1 + \sin^{2}{x} \right)$$
$$= \cos{x} \left( 1 + \cos{x} \right) \; \left[ \text{From} \left( 2 \right) \right]$$
$$= \cos{x} + \cos^{2}{x}$$
$$= 1 \; \left[ \text{From} \left( 1 \right) \right]$$
Hence $$\sin^{2} \left( 2 - \cos^{2}{x} \right) = 1$$.
Question 8
1 / -0

$$\dfrac{{\sec 8{\text{A}} - 1}}{{\sec 4{\text{A}} - 1}} = $$

A
$$\dfrac{{\tan 2{\text{A}}}}{{\tan 8{\text{A}}}}$$

B
$$\dfrac{{\tan 8{\text{A}}}}{{\tan 2{\text{A}}}}$$

C
$$\dfrac{{{\text{cot8A}}}}{{\cot 2{\text{A}}}}$$

D
$$\dfrac{{{\text{tan6A}}}}{{\tan 2{\text{A}}}}$$

Solution

Solution :-
$$\displaystyle \frac{sec8A-1}{sec4A-1} = \frac{1/cos8A-1}{1/cos4A-1} $$
$$\displaystyle \Rightarrow \frac{\frac{1-cos8A}{cos8A}}{\frac{1-cos4A}{cos4A}} = \frac{cos4A(1-cos8A)}{cos8A(1-cos4A)} $$
$$\displaystyle \Rightarrow \frac{cos4A(1-(1-2sin^{2}4A))}{cos8A(1-(1-2sin^{2}2A))} = \frac{cos4A.sin^{2}2A}{cos8A.sin^{2}2A} $$
$$\displaystyle \Rightarrow \frac{(2cos4Asin4A)sin4A}{(2cos8A\,sin^{2}2A)} = \frac{sin8Asin4A}{2cos8A.sin^{2}2A} $$
$$\displaystyle \Rightarrow tan8A(\frac{sin4A}{2sin^{2}A}) = tan8A.(\frac{2sin2A.cos2A}{2sin^{2}2A}) $$
$$\displaystyle \Rightarrow \frac{tan8A}{tan2A} $$
Question 9
1 / -0

If $$\tan \theta + \, \cot \theta = 4$$, then the value of $$\tan^3 \theta + \cot^3 \theta$$ is

A
$$52$$

B
$$16$$

C
$$7 \dfrac{9}{8}$$

D
$$27 \dfrac{1}{27}$$

Solution

$$\tan\theta +\cot\theta =4$$
$$(\tan\theta +\cot\theta )^{3}=4^{3}$$
$$\tan^{3}\theta +\cot^{3}\theta +(\tan\theta +\cot\theta )3\tan\theta \cot\theta =64 $$
$$\tan^{3}\theta +\cot^{3}\theta +(4)3=64 $$
$$\tan^{3}\theta +\cot^{3}\theta =64-12 $$
$$\tan^{3}\theta +\cot^{3}\theta =52$$
Question 10
1 / -0

The positive integer value of $$n>3$$ satisfing the equation
$$\frac{1}{{\sin \left( {\frac{\pi }{n}} \right)}} = \frac{1}{{\sin \left( {\frac{{2\pi }}{n}} \right)}} + \frac{1}{{\sin \left( {\frac{{3\pi }}{n}} \right)}}$$
is

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

Solution

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Trigonometric Functions Test 19

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Trigonometric Functions Test 19

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Question 1

$$\dfrac{\pi}{2}$$

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{4}$$

$$\dfrac{\pi}{6}$$

Solution

Question 2

$$\dfrac{\sqrt{2}}{3}$$

`$$\dfrac{\sqrt{3}}{\sqrt{5}}$$

$$\dfrac{\sqrt{15}}{4}$$

$$\dfrac{\sqrt{2}}{\sqrt{5}}$$

Solution

Question 3

$$2$$

$$2 cosec {40^ \circ}$$

$$4$$

$$2\sin{45^ \circ} . cosec {40^ \circ}$$

Solution

Question 4

$$\pm \dfrac { \pi }{ 2 }$$

$$\pm \dfrac { \pi }{ 4 }$$

$$\pm \dfrac { 3\pi }{ 8 }$$

Non of these

Solution

Question 5

1

2

3

None of these

Solution

Question 6

$$-1$$

$$1$$

$$0$$

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

Solution

Question 7

$$1$$

$$2$$

$$4$$

none of these

Solution

Question 8

$$\dfrac{{\tan 2{\text{A}}}}{{\tan 8{\text{A}}}}$$

$$\dfrac{{\tan 8{\text{A}}}}{{\tan 2{\text{A}}}}$$

$$\dfrac{{{\text{cot8A}}}}{{\cot 2{\text{A}}}}$$

$$\dfrac{{{\text{tan6A}}}}{{\tan 2{\text{A}}}}$$

Solution

Question 9

$$52$$

$$16$$

$$7 \dfrac{9}{8}$$

$$27 \dfrac{1}{27}$$

Solution

Question 10

$$7$$

$$8$$

$$9$$

$$10$$

Solution

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