Trigonometric Functions Test -5

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Trigonometric Functions Test -5

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Question 1
1 / -0

In a triangle ABC right angled at C, tan A and tan B satisfy the equation

A
abx²−(a²+b²)x − ab = 0

B
abx² − c²x + ab = 0

C
c²x² − abx + c² = 0

D
ax² − bx + a = 0

Solution

Using the given triangle we have tan A = a/b and tan B = b/a Also we have tan A . tan B = 1 and tan A + tan B = a/b + b /a = ( a² + b² )/ab = c² / ab . Hence x = tan A and x = tan B are two roots of the given quadratic equation.
Question 2
1 / -0

What is $$\tan ( 360^o - A)$$?

A
$$\tan A$$

B
$$-\tan A$$

C
$$\cot A$$

D
$$-\cot A$$

Solution

As $$360 ^{°} - A$$ lies in fourth quadrant and $$\tan $$ is negative in fourth quadrant

$$\Rightarrow \tan (360^{°} - A) =-\tan A$$
Question 3
1 / -0

Value of $$\theta (0 < \theta < 360^o )$$ which satisfy the equation $$ \text{cosec }\theta +2 =0$$ is:

A
$$210^o, 100^o$$

B
$$240^o, 300^o$$

C
$$210^o, 240^o$$

D
$$210^o, 330^o$$

Solution

Given: $$ \text{cosec }\theta +2 =0$$

$$\Rightarrow \sin \theta =-\dfrac {1}{2}$$

$$\Rightarrow \theta =\left (\pi +\dfrac {\pi} {6}\right), $$ $$\left(2\pi - \dfrac {\pi} {6}\right)$$

$$\Rightarrow \theta=\dfrac{7\pi}{6},\dfrac{11\pi}{6}$$

$$\Rightarrow \theta =210^{°},\,330^{°}$$
Question 4
1 / -0

Let [x] be the greatest integar function. Then the equation sinx = [1+ sinx] + [1 - cosx] has

A
one solution in $$\left[ {{{ - \pi } \over 2},{\pi \over 2}} \right]$$

B
one solution in $$\left[ {{\pi \over 2},\pi } \right]$$

C
one solution in R

D
no solution in R

Solution

Here,$$\sin { x } =[1+\sin { x } ]+[1-\cos { x } ]\\ \therefore \sin { x } =2+[\sin { x } ]+[-\cos { x } ]$$
Plotting graphs at certain intervals,it is concluded that no point exist which satisfies the equation.
$$\therefore$$ No solution for $$x\epsilon R$$
Question 5
1 / -0

If $${\tan ^2}\theta = 2\,{\tan ^2}\phi + 1$$, then the value of $$\cos \,2\theta + {\sin ^2}\phi \,is$$ is

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$0$$

Solution

$${\tan ^2}\theta = 2{\tan ^2}\phi + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( 1 \right)$$
$$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$$
$$\dfrac{{1 - \left( {2{{\tan }^2}\phi + 1} \right)}}{{1 + {{\tan }^2}\theta }}$$
$$\dfrac{{ - 2{{\tan }^2}\phi }}{{1 + \left( {2{{\tan }^2}\phi + 1} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,by\,\left( 1 \right)$$
$$\dfrac{{ - 2{{\tan }^2}\phi }}{{2{{\sec }^2}\phi }}$$
$$ - {\sin ^2}\phi $$
$$\cos 2\theta + {\sin ^2}\phi = 0$$
Question 6
1 / -0

If $$(1 - \cos A)/2 = x$$, then the value of $$x$$ is

A
$$co{s^2}(A/2)$$

B
$$\sqrt {\sin (A/2)} $$

C
$$\sqrt {\cos (A/2)} $$

D
$${\sin ^2}(A/2)$$

Solution

$${\textbf{Step 1: Solve the given problem using the rules}}{\textbf{.}}$$
$$x = \dfrac{{1 - \cos A}}{2}$$
$$ \Rightarrow 1 - \cos A = 2{\sin ^2}\dfrac{A}{2}$$ $$\boldsymbol{\mathbf{\left[ {\because \cos A = 1 - 2{{\sin }^2}\dfrac{A}{2}} \right]}}$$
$$ \Rightarrow \dfrac{{1 - \cos A}}{2} = {\sin ^2}\dfrac{A}{2}$$
$$ \Rightarrow x = {\sin ^2}\dfrac{A}{2}$$
$${\textbf{Hence, the value of }}\mathbf{x}{\textbf{ is si}}{{\textbf{n}}^2}\mathbf{\dfrac{A}{2}. }$$
Question 7
1 / -0

The solution set of the equation $$4\sin \theta \cos \theta - 2\cos \theta - 2\sqrt 3 \sin \theta + \sqrt 3 = 0$$ in the interval $$(0,2\pi )$$ is-

A
$$\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}$$

B
$$\left\{ {\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}$$

C
$$\left\{ {\frac{\pi }{3},\frac{{5\pi }}{3},\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}$$

D
$$\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{11\pi }}{6}} \right\}$$

Solution

$$\begin{array}{l}4\sin \theta cos\theta - 2cos\theta - 2\sqrt {3\sin \theta } + \sqrt 3 = 0\\ = 2\cos \theta (2\sin \theta - 1) - \sqrt 3 \left( {2\sin \theta - 1} \right) = 0\\ = (2\cos \theta - \sqrt 3 )(2\sin \theta - 1) = 0\\ = 2\cos \theta - \sqrt 3 = 0\\ = \cos \theta = \frac{{\sqrt 3 }}{2}\end{array}$$ $$ = \theta = \frac{\pi }{6},\frac{{11\pi }}{6}$$
Same as,
$$\begin{array}{l}2\sin \theta = 1\\ = \sin \theta = \frac{1}{2}\\ = \theta = \frac{\pi }{6},\frac{{5\pi }}{6}\end{array}$$
Therefore:-
$$\theta = \frac{\pi }{6},\frac{{5\pi }}{6},\frac{{11\pi }}{6}$$
Question 8
1 / -0

A minimum value of $$\sin{x}\cos{2x}$$ is-

A
$$1$$

B
$$-1$$

C
$$-2/3\sqrt{6}$$

D
None of these

Solution

$$\cos x\cos 2x=2\sin x\sin 2x\implies \cos x=3\cos 3x\implies \sin x=\pm\displaystyle \frac{1}{\sqrt{6}}\implies$$ the minimum value of $$ \sin x\cos 2x$$ is $$\displaystyle \frac{-2}{3\sqrt{6}}$$
Question 9
1 / -0

If $$\cot\left( {a + \beta } \right) = 0,$$ then $$\sin\left( {a + 2\beta } \right)$$ can be

A
$$-\sin a$$

B
$$\sin \beta $$

C
$$\cos a$$

D
$$\cos \beta $$

Solution

Since $$\cot(α+β) = 0 ⇒ \cot(α+β) = \cot\ 90^\circ$$
Therefore, $$(α + β) = 90^\circ$$ ......(1)

Since,
$$\sin(α+2β)$$
$$= \sin[(α + β) + β ]$$
$$= \sin[90°+ β ]$$
$$= \cos β$$
Question 10
1 / -0

Select write the most appropriate answer from the given alternative in each following questions:-
The principal solution of tan $$x=-\sqrt{3}$$ are:

A
$$\frac{-\pi }{3}and \frac{2\pi }{3}$$

B
$$\frac{\pi }{3}and \frac{5\pi }{3}$$

C
$$\frac{2\pi }{3}and \frac{5\pi }{3}$$

D
$$\frac{2\pi }{3}and \frac{4\pi }{3}$$

Solution

$$ \tan x=-\sqrt{3}=\tan \left( -\frac{\pi }{3} \right) $$
$$ general\,solution $$
$$ x=n\pi -\frac{\pi }{3} $$
$$ principal\,solution $$
$$ for\,n=0 $$
$$ x=-\frac{\pi }{3} $$
$$ for\,n=1 $$
$$ x=\frac{2\pi }{3} $$
$$ Hence\,principal\,solution\,is:-\frac{\pi }{3}and\frac{2\pi }{3} $$
Question 11
1 / -0

The solutions of the equation $$\sin x+3\sin 2x+\sin 3x=\cos x+3\cos 2x+\cos 3x$$ in the interval $$0\le x\le 2\pi$$, are

A
$$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{2\pi}{3}$$

B
$$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}$$

C
$$\dfrac{4\pi}{3}, \dfrac{9\pi}{3}, \dfrac{2\pi}{3}, \dfrac{13\pi}{8}$$

D
$$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{3}, \dfrac{4\pi}{3}$$

Solution

$$ \sin x+3\sin 2x+\sin 3x=\cos x+3\cos 2x+\cos 3x $$
$$ formula\,used $$
$$ \sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2} $$
$$ \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2} $$
$$ \cos \left( -\theta \right)=\cos \theta $$
$$ \Rightarrow 2\sin 2x\cos x+3\sin 2x=2\cos 2x\cos x+3\cos 2x $$
$$ \Rightarrow \sin 2x\left[ 2\cos x+3 \right]=\cos 2x\left[ 2\cos x+3 \right] $$
$$ \Rightarrow \sin 2x=\cos 2x $$
$$ \Rightarrow \tan 2x=1=\tan \frac{\pi }{4} $$
$$ \therefore 2x=n\pi +\frac{\pi }{4} $$
$$ x=\frac{n\pi }{2}+\frac{\pi }{8} $$
$$ 0\le x\le 2\pi $$
$$ \therefore \,x=\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi}8 $$

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Trigonometric Functions Test -5

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Question 1

abx2−(a2+b2)x − ab = 0

abx2 − c2x + ab = 0

c2x2 − abx + c2 = 0

ax2 − bx + a = 0

Solution

Question 2

$$\tan A$$

$$-\tan A$$

$$\cot A$$

$$-\cot A$$

Solution

Question 3

$$210^o, 100^o$$

$$240^o, 300^o$$

$$210^o, 240^o$$

$$210^o, 330^o$$

Solution

Question 4

one solution in $$\left[ {{{ - \pi } \over 2},{\pi \over 2}} \right]$$

one solution in $$\left[ {{\pi \over 2},\pi } \right]$$

one solution in R

no solution in R

Solution

Question 5

$$1$$

$$2$$

$$-1$$

$$0$$

Solution

Question 6

$$co{s^2}(A/2)$$

$$\sqrt {\sin (A/2)} $$

$$\sqrt {\cos (A/2)} $$

$${\sin ^2}(A/2)$$

Solution

Question 7

$$\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}$$

$$\left\{ {\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}$$

$$\left\{ {\frac{\pi }{3},\frac{{5\pi }}{3},\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}$$

$$\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{11\pi }}{6}} \right\}$$

Solution

Question 8

$$1$$

$$-1$$

$$-2/3\sqrt{6}$$

None of these

Solution

Question 9

$$-\sin a$$

$$\sin \beta $$

$$\cos a$$

$$\cos \beta $$

Solution

Question 10

$$\frac{-\pi }{3}and \frac{2\pi }{3}$$

$$\frac{\pi }{3}and \frac{5\pi }{3}$$

$$\frac{2\pi }{3}and \frac{5\pi }{3}$$

$$\frac{2\pi }{3}and \frac{4\pi }{3}$$

Solution

Question 11

$$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{2\pi}{3}$$

$$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}$$

$$\dfrac{4\pi}{3}, \dfrac{9\pi}{3}, \dfrac{2\pi}{3}, \dfrac{13\pi}{8}$$

$$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{3}, \dfrac{4\pi}{3}$$

Solution

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abx²−(a²+b²)x − ab = 0

abx² − c²x + ab = 0

c²x² − abx + c² = 0

ax² − bx + a = 0