Trigonometric Functions Test 14

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Trigonometric Functions Test 14

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Question 1
1 / -0

If $$2\sin^2\theta -5\sin \theta +2 > 0, \theta \in (0, 2\pi)$$, then $$\theta \in$$

A
$$\left(\dfrac{5\pi}{6}, 2\pi\right)$$

B
$$\left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$$

C
$$\left(0, \dfrac{\pi}{6}\right)$$

D
$$\left(\dfrac{\pi}{80}, \dfrac{\pi}{6}\right)$$

Solution

$$(2\sin \theta -1)(\sin \theta -2) > 0$$
$$\therefore \sin \theta < \dfrac{1}{2}=\sin \dfrac{\pi}{6}=\sin \left(\pi -\dfrac{\pi}{6}\right)=\sin \dfrac{5\pi}{6}$$
$$\therefore \theta \in \left(0, \dfrac{\pi}{6}\right)$$ or $$\theta \in \left(\dfrac{5\pi}{6}, 2\pi\right)$$
$$\therefore \theta \in \left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$$.
Question 2
1 / -0

If $$0\leq x\leq \pi$$ and $$81^{\sin^2x}+81^{\cos^2x}=30$$, then x is equal to.

A
$$\pi /6$$

B
$$\pi /3$$

C
$$5\pi /6$$

D
$$2\pi /3$$

E
All correct

Solution

Put $$\cos^2x=1-\sin^2x$$
If $$y=81^{\sin^2x}$$ then $$y+\dfrac{81}{y}=30$$
$$y^2-30y+81=0$$, $$y=27, 3$$
$$3^{4\sin^2x}=3^3, 3^1$$
$$\therefore \sin^2x=\dfrac{3}{4}, \dfrac{1}{4}$$
$$\therefore \sin x=\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}$$
as $$\sin x$$ is $$+$$ive in the given interval $$0 < x < \pi$$
$$\therefore x=\dfrac{\pi}{3}, \pi -\dfrac{\pi}{3}=\dfrac{2\pi}{3}, \dfrac{\pi}{6}, \pi -\dfrac{\pi}{6}=\dfrac{5\pi}{6}$$.
Question 3
1 / -0

Solve: $$2(\cos x+\cos 2x)+\sin 2x(1+2\cos x)=2\sin x, -\pi \leq x \leq \pi$$.

A
$$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$$.

B
$$-\pi, -\pi /2, \pi /3, \pi$$.

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

D
None of these

Solution

$$2(\cos x+2\cos^2 x-1)+2\sin x\cos x. (1+2\cos x)-2\sin x=0$$
or $$2(2\cos^2x+\cos x-1)+2\sin x(2\cos^2x+\cos x-1)=0$$
$$2(1+\sin x)(\cos x+1)(2\cos x-1)=0$$
We have to determine values of x s.t. $$-\pi \leq x\leq \pi$$
$$1+\sin x=0$$ $$\therefore \sin x=-1$$
$$\therefore x=2n\pi +\dfrac{3\pi}{2}$$ $$\therefore x=-\dfrac{\pi}{2}$$,
for $$n=-1$$ $$\because -\pi \leq x < \pi$$
$$\cos x=-1=\cos \pi$$ $$\therefore \cos x=1/2=\cos (\pi/3)$$
$$\therefore x=2n\pi \pm \pi/3$$
$$\therefore x=\pi /3$$, $$-\pi /3$$
Hence the values of x s.t. $$-\pi \leq x \leq \pi$$ are
$$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$$.
Question 4
1 / -0

Solve $$\tan \theta +\sec \theta =\sqrt{3}; 0\leq \theta \leq 2\pi$$.

A
$$\theta =2n\pi -\dfrac{\pi}{6}$$.

B
$$ \theta =n\pi +\dfrac{\pi}{6}$$.

C
$$ \theta =2n\pi +\dfrac{5\pi}{6}$$.

D
$$ \theta =2n\pi +\dfrac{\pi}{6}$$.

Solution

$$\tan\theta +sec\theta =\sqrt{3}$$
$$\dfrac{1+\sin\theta}{\cos\theta}=\sqrt{3}$$
$$\dfrac{\left(\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}\right)^2}{\cos^2\dfrac{\theta}{2}-\sin^2\dfrac{\theta}{2}}=\sqrt{3}$$
or $$\dfrac{\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}}{\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}}=\sqrt{3}$$ or $$\dfrac{1+t}{1-t}=\sqrt{3}$$
or $$\tan\left(\dfrac{\pi}{4} +\dfrac{\theta}{2}\right)=\tan\dfrac{\pi}{3}$$
$$\therefore \dfrac{\theta}{2}+\dfrac{\pi}{4}=n\pi +\dfrac{\pi}{3}$$
$$\therefore \theta =2n\pi +\dfrac{\pi}{6}$$.
Question 5
1 / -0

A balloon is observed simultaneously from three points A B and C, on a straight road directly under it. The angular elevation at B is twice of what it is at A and the angular elevation at C is thrice of what it is at A. If the distance between A and B is 200 meters and the distance between B and C is 100 meters, then find the height of the balloon.

A
$$50 \sqrt{3}$$m

B
$$50 $$m

C
$$150 \sqrt{3}$$m

D
$$100 \sqrt{3}$$m

Solution

$$x = h \cot 3 \alpha $$ ...............(i)
$$(x + 100) = h \cot 2 \alpha $$ ............ (ii)
$$(x + 300) = h \cot \alpha $$ ............(iii)

From (i) and (ii)
$$-100 = h (\cot 3 \alpha - \cot 2 \alpha) = h\dfrac{(sin2\alpha \ cos3\alpha - cos2\alpha \ sin3\alpha)}{sin3\alpha \ sin2\alpha} = \dfrac{sin(3\alpha - 2\alpha)}{sin3\alpha \ sin2\alpha}$$
Similarly,
From (ii) and (iii)
$$-200 = h (\cot 2 \alpha - \cot \alpha) = \dfrac{sin(2\alpha - \alpha)}{sin2\alpha \ sin\alpha} $$

$$100 = h \left (\dfrac{\sin \alpha}{\sin 3 \alpha \sin 2 \alpha} \right ) $$ on solving

$$200 = h \left (\dfrac{\sin \alpha}{\sin 2 \alpha \sin \alpha} \right ) $$ on solving

On dividing the above equations we get,

$$\dfrac{\sin 3 \alpha}{\sin \alpha} = \dfrac{200}{100} \Rightarrow \dfrac{\sin 3 \alpha}{\sin \alpha} = 2 $$ ......... (3)

We know that: $$sin3\alpha = 3sin\alpha - 4sin^3\alpha$$

From eq (3) we get $$\Rightarrow 3 \sin \alpha - 4 \sin^{3} \alpha - 2 \sin \alpha = 0$$

$$\Rightarrow 4 \sin^{3} \alpha - \sin \alpha = 0 \Rightarrow \sin \alpha = 0 $$ or $$ \sin^{2} \alpha = \dfrac{1}{4} $$

$$\sin^{2} \alpha = \dfrac{1}{4} = \sin^{2} \left (\dfrac{\pi}{6} \right ) \Rightarrow \alpha = \dfrac{\pi}{6} $$

Hence
$$h = 200 \ sin2\alpha= 200 \sin \dfrac{\pi}{3} = 200 \dfrac{\sqrt{3}}{2} = 100 \sqrt{3} $$

So the height of the balloon $$100 \sqrt{3} $$
Question 6
1 / -0

If $$P\left( 4 \right) = 3$$ and $$\displaystyle {\sin ^6}x + {\cos ^6}x = {a \over b}\left( {a,b \in N} \right)$$ and $$a,b$$ are relatively prime, then $$a + b$$ is equal to

A
$$7$$

B
$$23$$

C
$$16$$

D
$$9$$

Solution
Question 7
1 / -0

The equation $${\sin ^6}x + {\cos ^6}x = {a^2}$$ has real solution if

A
$$a \in \left( { - 1,1} \right)$$

B
$$a \in \left( { - 1, - {1 \over 2}} \right)$$

C
$$\eqalign{
& a \in \left( { - {1 \over 2},{1 \over 2}} \right) \cr
& a \in \left( {{1 \over 2},1} \right) \cr} $$

D
$$a \in \left( {{1 \over 2},1} \right)$$

Solution

$$a^{2}=\sin^{6}x+\cos^{6}x$$
$$\Rightarrow a^{2}=(\sin^{2}x+\cos^{2}x)(\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x)$$
$$\Rightarrow a^{2}=((\sin^{2}x+\cos^{2}x)^{2}-3\sin^{2}x \cos^{2}x)$$ as($$\sin^{2}x+\cos^{2}x=1)$$
$$\Rightarrow a^{2}=(1-3\sin^{2}x\cos^{2}x)$$
$$\Rightarrow a^{2}=(1-\cfrac{3}{4}\sin^{2}2x)$$
$$\sin^{2}2x \in [0,1]$$
$$\therefore 1-\cfrac{3}{4}\sin^{2}(2x) \in [\cfrac{1}{4},1]$$
$$\cfrac{1}{4}\le a^{2} \le 1$$
$$a \in [(-\infty,\cfrac{-1}{2}]\cup[\cfrac{1}{2}, \infty)]\cap[-1,1]$$
$$ \therefore a\in[-1,\cfrac{-1}{2}]\cup[\cfrac{1}{2},1]$$
Question 8
1 / -0

If $$\sec A + \tan A = m $$ and $$ \sec A - \tan A = n$$, find the value of $$\sqrt{mn}$$.

A
$$0$$

B
$$\pm 1$$

C
$$\pm 2$$

D
$$\pm 3$$

Solution

Use trigonometry formula:
$$sec^{2}\theta -tan^{2}\theta =1$$

$$secA+tanA=m$$ ......(1)
$$secA-tanA=n$$ .........(2)

multiply the equations (1) and (2),
$$\left ( secA+tanA \right )\left ( secA-tanA \right )=mn$$
$$sec^{2}A-tan^{2}A=mn$$
$$1=mn$$

Therefore, $$\sqrt{mn}=\pm 1$$
Question 9
1 / -0

The most general value of $$\theta $$ satisfying both the equations $$\sin \theta = \frac{1}{2},\tan \theta = \frac{1}{{\sqrt 3 }}\;is\;\left( {n \in I} \right)$$

A
$$2n\pi + \frac{\pi }{6}$$

B
$$\left( b \right)2n\pi - \frac{{7\pi }}{6}$$

C
$$\left( c \right)2n\pi + \frac{{5\pi }}{6}$$

D
None of these

Solution
Question 10
1 / -0

The expression $$\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$$ reduces to :

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

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

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

D
$$\dfrac {1+\cos A}{\cos A}$$

Solution

$$\sec^2\theta=1+\tan^2\theta$$
$$\Rightarrow \cfrac {\tan A+\sec A-1}{\tan A-\sec A+1}$$
$$\Rightarrow \cfrac {\tan A+\sec A-(\sec^2A-\tan^2A)}{\tan A-\sec A+1}$$
$$\Rightarrow \cfrac {(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$$
$$\Rightarrow \tan A+\sec A$$
$$\Rightarrow \cfrac {\sin A}{\cos A}+\cfrac {1}{\cos A}$$
$$\Rightarrow \cfrac {1+\sin A}{\cos A}$$

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Trigonometric Functions Test 14

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Question 1

$$\left(\dfrac{5\pi}{6}, 2\pi\right)$$

$$\left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$$

$$\left(0, \dfrac{\pi}{6}\right)$$

$$\left(\dfrac{\pi}{80}, \dfrac{\pi}{6}\right)$$

Solution

Question 2

$$\pi /6$$

$$\pi /3$$

$$5\pi /6$$

$$2\pi /3$$

All correct

Solution

Question 3

$$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$$.

$$-\pi, -\pi /2, \pi /3, \pi$$.

$$-\pi, -\pi /3, \pi /2, \pi/3, \pi$$.

None of these

Solution

Question 4

$$\theta =2n\pi -\dfrac{\pi}{6}$$.

$$ \theta =n\pi +\dfrac{\pi}{6}$$.

$$ \theta =2n\pi +\dfrac{5\pi}{6}$$.

$$ \theta =2n\pi +\dfrac{\pi}{6}$$.

Solution

Question 5

$$50 \sqrt{3}$$m

$$50 $$m

$$150 \sqrt{3}$$m

$$100 \sqrt{3}$$m

Solution

Question 6

$$7$$

$$23$$

$$16$$

$$9$$

Solution

Question 7

$$a \in \left( { - 1,1} \right)$$

$$a \in \left( { - 1, - {1 \over 2}} \right)$$

$$\eqalign{ & a \in \left( { - {1 \over 2},{1 \over 2}} \right) \cr & a \in \left( {{1 \over 2},1} \right) \cr} $$

$$a \in \left( {{1 \over 2},1} \right)$$

Solution

Question 8

$$0$$

$$\pm 1$$

$$\pm 2$$

$$\pm 3$$

Solution

Question 9

$$2n\pi + \frac{\pi }{6}$$

$$\left( b \right)2n\pi - \frac{{7\pi }}{6}$$

$$\left( c \right)2n\pi + \frac{{5\pi }}{6}$$

None of these

Solution

Question 10

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

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

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

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

Solution

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$$\eqalign{
& a \in \left( { - {1 \over 2},{1 \over 2}} \right) \cr
& a \in \left( {{1 \over 2},1} \right) \cr} $$