Trigonometric Functions Test 17

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Trigonometric Functions Test 17

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Weekly Quiz Competition

Question 1
1 / -0

If $$cos \alpha + cos \beta = a , \, sin \alpha + sin \beta = b \, and \, \alpha - \beta = 2 \theta \, then \, \dfrac{cos \, 3 \theta }{cos \, \theta}$$ =

A
$$a^2 + b^2 - 2=0$$

B
$$a^2 + b^2 - 3 =0$$

C
$$3 - a^2 - b^2=0$$

D
$$2 - a^2 - b^2=0$$

Solution

we have
$$\cos \alpha+\cos \beta=a$$
$$\sin \alpha+\sin \beta=b$$

$$\alpha-\beta=2\theta$$

so, $$(\cos \alpha+\cos \beta)^2=a^2$$

$$\cos^2\alpha+\cos^2\beta+2\cos \alpha\cos \beta=a^2$$-----(i)

and $$(\sin \alpha+\sin \beta)^2=b^2$$

$$\sin^2\alpha+\sin^2\beta+2\sin \alpha\sin \beta=b^2$$-----(ii)

adding (i) and (ii)

$$2+\cos (\alpha-\beta)=a^2+b^2$$ $$\left[\sin^2\alpha+\cos^2\alpha=1\\\sin^2\beta+\cos^2\beta=1\\ \cos\alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta)\right.$$

$$a^2+b^2=2+2\cos (\alpha-\beta)$$

$$=2+2\cos^2\theta$$

$$\dfrac{\cos 3\theta}{\cos \theta} =\dfrac{4\cos^3\theta.3\cos \theta}{\cos \theta}$$

$$=4\cos^2\theta-3\rightarrow 4\cos^2\theta=3$$------(iii)

$$a^2+b^2=2+2(2\cos^2\theta-1)$$

$$=2+4\cos^2\theta-2$$

$$\therefore 4\cos^2\theta=a^2+b^2$$ from (iii)

$$\therefore 3=a^2+b^2$$

$$a^2+b^2-3=0$$
Question 2
1 / -0

If $$a$$, $$b$$, $$c$$, $$t$$ are the solution of the equation $$\tan\left(\theta+\dfrac{\pi}{4}\right)=3\ tan 3\theta$$, no two of which have equal tangents . Then, the value of $$\tan a+\tan b+\tan c+\tan t=$$

A
$$1/3$$

B
$$8/3$$

C
$$-8/3$$

D
$$0$$

Solution

Consider the given equation,

$$\tan \left( \dfrac{\pi }{4}+\theta \right)=3\tan 3\theta $$

Using identity ,

$$ \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\,\,\,\And \tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A} $$

$$ \Rightarrow \dfrac{\tan \left( \dfrac{\pi }{4} \right)+\tan \theta }{1-\tan \left( \dfrac{\pi }{4} \right)\tan \theta }=3\left( \dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right) $$

$$ \Rightarrow \dfrac{1+\tan \theta }{1-\tan \theta }=\dfrac{9\tan \theta-3{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } $$

$$ \Rightarrow \left( 1+\tan \theta \right)\left( 1-3{{\tan }^{2}}\theta \right)=\left( 1-\tan \theta \right)\left( 9\tan \theta -3{{\tan }^{3}}\theta \right) $$

$$ \Rightarrow 3{{\tan }^{4}}\theta -6{{\tan }^{2}}\theta +8\tan \theta -1=0 $$

As $$a,b,c,t,=$$ are roots of this equation ,

Hence, sum of roots ,

$$\tan \left( a \right)+\tan \left( b\right)+\tan \left( c\right)+\tan \left(t\right)=$$ $$-\dfrac{coeficient\,of\,{{\tan }^{3}}\theta }{coeficient\,of\,{{\tan }^{4}}\theta }$$ =$$-\dfrac{0}{1}$$

$$\tan \left( a\right)+\tan \left(b\right)+\tan \left( c\right)+\tan \left( t \right)=0$$

Hence, this is the answer.
Question 3
1 / -0

If $$\alpha$$ is a root of $$25\cos^2\theta +5\cos\theta -12=0, \dfrac{\pi}{2} < \alpha < \pi$$, then $$\sin 2θ $$ is equal to?

A
$$\dfrac{24}{25}$$

B
$$-\dfrac{24}{25}$$

C
$$\dfrac{13}{18}$$

D
$$-\dfrac{13}{18}$$

Solution
Question 4
1 / -0

The value of $$\sin {27^ \circ } - \cos {27^ \circ }$$ is equal to

A
$$ - \dfrac{{\sqrt {3 - \sqrt 5 } }}{2}$$

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

C
$$ - \dfrac{{\sqrt 5 - 1}}{{2\sqrt 2 }}$$

D
$$ \dfrac{{\sqrt {3 - \sqrt 5 } }}{2}$$

Solution

$$ \sin {{27}^{0}}-\cos {{27}^{0}} $$

$$ \Rightarrow \sin {{27}^{0}}-\cos \left( {{90}^{0}}-{{63}^{0}} \right) $$

$$ \Rightarrow \sin {{27}^{0}}-\sin {{63}^{0}} $$

$$ \Rightarrow 2\cos \left( \dfrac{{{27}^{0}}+{{63}^{0}}}{2} \right)\sin \left( \dfrac{{{27}^{0}}-{{63}^{0}}}{2} \right) $$

$$ \Rightarrow 2\cos {{45}^{0}}\sin \left( -{{36}^{0}} \right) $$

$$ \Rightarrow 2\times \dfrac{1}{\sqrt{2}}\times -\left[ \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right] $$

$$ \Rightarrow -\dfrac{1}{\sqrt{2}}\times \sqrt{2}\left( \dfrac{\sqrt{5-\sqrt{5}}}{2} \right) $$

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

Hence, this is the answer
option $$(B)$$ is correct.
Question 5
1 / -0

The value of $${\cos ^2}{73^ \circ } + {\cos ^2}{47^ \circ } - {\sin ^2}{43^ \circ } + {\sin ^2}{107^ \circ }$$ is equal to :

A
$$1$$

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

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

D
$${\sin ^2}{73^ \circ } + {\sin ^2}{47^ \circ }$$

Solution

Consider the given equation.
$$ cos^2{{73}^{0}}+co{{s}^{2}}{{47}^{0}}-si{{n}^{2}}{{43}^{0}}+si{{n}^{2}}{{107}^{0}} $$

$$ \Rightarrow co{{s}^{2}}{{73}^{0}}+co{{s}^{2}}{{47}^{0}}-si{{n}^{2}}\left( {{90}^{0}}-{{47}^{0}} \right)+si{{n}^{2}}\left( {{180}^{0}}-{{73}^{0}} \right) $$

$$ \Rightarrow \left( co{{s}^{2}}{{73}^{0}}+{{\sin }^{2}}{{73}^{0}} \right)+co{{s}^{2}}{{47}^{0}}-{{\cos }^{2}}{{47}^{0}} $$

$$ \Rightarrow 1 $$

Hence, this is the answer.
Question 6
1 / -0

If $$\tan \theta + \tan 4\theta + \tan 7 = \tan \theta \tan 4\theta \tan 7\theta ,\,\,then\,\,\theta = $$

A
$$\dfrac{{n\pi }}{4}$$

B
$$\dfrac{{n\pi }}{7}$$

C
$$\dfrac{{n\pi }}{{12}}$$

D
$$n\pi $$

Solution

We know,
$$\tan\theta+\tan 4\theta+\tan 7\theta=\tan\theta\tan 4\theta\tan 7\theta$$
occurs only if $$\theta,4\theta,7\theta$$ are angles of a triangle.
Therefore,
sum of all angles of a triangle is $$\pi.$$
$$=>\theta+4\theta+7\theta=\pi$$
$$=>\theta=\dfrac{n\pi}{12}$$
Question 7
1 / -0

If $$\tan \theta + \sec \theta = \sqrt{3}$$, then the principal value of $$\left(\theta + \dfrac{\pi}{6}\right)$$ is

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

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

C
$$\dfrac{2\pi}{3}$$

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

Solution
Question 8
1 / -0

$$\cot{\dfrac{\pi}{20}}\cot{\dfrac{3\pi}{20}}\cot{\dfrac{5\pi}{20}}\cot{\dfrac{7\pi}{20}}\cot{\dfrac{9\pi}{20}}$$ is equal to

A
$$-1$$

B
$$0$$

C
$$1$$

D
None of these

Solution

We know that $$\pi={180}^{\circ}$$
$$\dfrac{\pi}{20}={9}^{\circ}$$, $$\dfrac{3\pi}{20}={27}^{\circ}$$, $$\dfrac{5\pi}{20}={45}^{\circ}$$, $$\dfrac{7\pi}{20}={63}^{\circ}$$, $$\dfrac{9\pi}{20}={81}^{\circ}$$

$$\cot{\dfrac{\pi}{20}}\cot{\dfrac{3\pi}{20}}\cot{\dfrac{5\pi}{20}}\cot{\dfrac{7\pi}{20}}\cot{\dfrac{9\pi}{20}}$$

$$=\cot{{9}^{\circ}}\cot{{27}^{\circ}}\cot{{45}^{\circ}}\cot{{63}^{\circ}}\cot{{81}^{\circ}}$$

$$=\cot{{9}^{\circ}}\cot{{27}^{\circ}}\times 1\times\cot{\left({90}^{\circ}-{27}^{\circ}\right)}\cot{\left({90}^{\circ}-{9}^{\circ}\right)}$$

$$=\cot{{9}^{\circ}}\cot{{27}^{\circ}}\tan{{9}^{\circ}}\tan{{27}^{\circ}}$$

$$=\cot{{9}^{\circ}}\tan{{9}^{\circ}}\cot{{27}^{\circ}}\tan{{27}^{\circ}}$$

$$=1$$ since $$\tan{\theta}\cot{\theta}=1$$
Question 9
1 / -0

If $$4\ \sin^{2}x-1=0$$ and $$0 < x < 2 \pi$$, then positive values of $$x$$ are

A
$$30^{o},120^{o},210^{o},300^{o}$$

B
$$30^{o},150^{o},210^{o},330^{o}$$

C
$$30^{o},120^{o},150^{o},210^{o}$$

D
$$30^{o},160^{o},210^{o},320^{o}$$

Solution
Question 10
1 / -0

If $$\sin { x } sin{ \left( { 60 }^{ O }+x \right) .\left( { 60 }^{ O }-x \right) }=\frac { 1 }{ 8 } $$ then$$x=$$

A
$$\pi+(-1)^\alpha\frac{\pi}{6}$$

B
$$\frac{\pi}{3}+(-1)^\alpha\frac{\pi}{18}$$

C
$$n\pi \left( -1 \right) ^{ \alpha }\frac { \pi }{ 3 }$$

D
$$\frac { n\pi }{ 3 } +\left( -1 \right) ^{ \alpha }\frac { \pi }{ 9 }$$

Solution

Given

$$\sin x\sin (60^{\circ} +x) \sin (60^{\circ}-x)=\dfrac 18$$

$$\sin x\left( \sin ^2 60^{\circ} -\sin ^2 x\right)$$

$$\bigg( \sin (A+B)\sin (A-B) =\sin ^2 A-\sin ^2B\bigg)$$

$$\sin x\left(\dfrac 34 -\sin ^2 x\right)=\dfrac 18$$

$$\left(\dfrac {3\sin x-4\sin ^3 x}{4}\right)=\dfrac 18$$

$$\sin 3x =\dfrac 12$$

$$\bigg(\sin 3x=3\sin x-4\sin ^3 x\bigg)$$

$$\implies 3x=n\pi + (-1)^n \dfrac \pi 6$$

$$\implies x=\dfrac {n \pi}3 + (-1)^n \dfrac \pi {18}$$

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Trigonometric Functions Test 17

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Trigonometric Functions Test 17

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SHARING IS CARING

Question 1

$$a^2 + b^2 - 2=0$$

$$a^2 + b^2 - 3 =0$$

$$3 - a^2 - b^2=0$$

$$2 - a^2 - b^2=0$$

Solution

Question 2

$$1/3$$

$$8/3$$

$$-8/3$$

$$0$$

Solution

Question 3

$$\dfrac{24}{25}$$

$$-\dfrac{24}{25}$$

$$\dfrac{13}{18}$$

$$-\dfrac{13}{18}$$

Solution

Question 4

$$ - \dfrac{{\sqrt {3 - \sqrt 5 } }}{2}$$

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

$$ - \dfrac{{\sqrt 5 - 1}}{{2\sqrt 2 }}$$

$$ \dfrac{{\sqrt {3 - \sqrt 5 } }}{2}$$

Solution

Question 5

$$1$$

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

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

$${\sin ^2}{73^ \circ } + {\sin ^2}{47^ \circ }$$

Solution

Question 6

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

$$\dfrac{{n\pi }}{7}$$

$$\dfrac{{n\pi }}{{12}}$$

$$n\pi $$

Solution

Question 7

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

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

$$\dfrac{2\pi}{3}$$

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

Solution

Question 8

$$-1$$

$$0$$

$$1$$

None of these

Solution

Question 9

$$30^{o},120^{o},210^{o},300^{o}$$

$$30^{o},150^{o},210^{o},330^{o}$$

$$30^{o},120^{o},150^{o},210^{o}$$

$$30^{o},160^{o},210^{o},320^{o}$$

Solution

Question 10

$$\pi+(-1)^\alpha\frac{\pi}{6}$$

$$\frac{\pi}{3}+(-1)^\alpha\frac{\pi}{18}$$

$$n\pi \left( -1 \right) ^{ \alpha }\frac { \pi }{ 3 }$$

$$\frac { n\pi }{ 3 } +\left( -1 \right) ^{ \alpha }\frac { \pi }{ 9 }$$

Solution

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