Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 28

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Question 1
1 / -0

If $$a\sin^{3}x+b\cos^{3}x=\sin x\cos x$$ and $$a\sin x=b\cos x$$ . If $$a,b,\sin x$$ and $$\cos x $$ all are non zero real numbers, then $$a^{2}+b^{2}=$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Given,
$$a\ { \sin }^{ 3 }x+b\ { \cos }^{ 3 }x=\sin x \ \cos x ........... (1)$$
$$a\sin x=b\cos x......... (2)$$
Consider equation $$(1)$$,
$$a\ { \sin }^{ 3}x+b \ { \cos }^{ 3 }x=\sin x \ \cos x$$
$$a\ {\sin }x\ { \sin }^{ 2 }x+b \ { \cos }^{ 3 }x=\sin x \ \cos x\\ b \ \cos x \ { \sin }^{ 2 }x+b{ \cos }^{ 3 }x=\sin x \ \cos x\\ b\cos x \ \left( { \sin }^{ 2 }x+{ \cos }^{ 2 }x \right) =\sin x \ \cos x$$ (using $$(2)$$)
Solving this, we get
$$b=\sin x$$
Then $$a =\dfrac{b \cos x}{\sin x} = \cos x$$
Hence,
$${ a }^{ 2 }+{ b }^{ 2 }={ \sin }^{ 2 }x+{ \cos }^{ 2 }x=1$$
Question 2
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If $$m\cos^{2}A+n\sin^{2}A=p,$$ then $$\cot^{2}A=$$

A
$$\dfrac{p+n}{m+p}$$

B
$$\dfrac{p-n}{m+p}$$

C
$$\dfrac{p-n}{m-p}$$

D
None of these

Solution

$$m{ \cos }^{ 2 }A+n{ \sin }^{ 2 }A=p\\ m{ \cos }^{ 2 }A+n\left( 1-{ \cos }^{ 2 }A \right) =p\\ { \cos }^{ 2 }A=\cfrac { p-n }{ m-n } \\ m{ \cos }^{ 2 }A+n{ \sin }^{ 2 }A=p\\ m\left( 1-{ \sin }^{ 2 }A \right) +n{ \sin }^{ 2 }A=p\\ { \sin }^{ 2 }A=\cfrac { p-m }{ n-m } =\cfrac { m-p }{ m-n } \\ { \cot }^{ 2 }A=\cfrac { { \cos }^{ 2 }A }{ { \sin }^{ 2 }A } =\cfrac { p-n }{ m-p } $$

Question 3

1 / -0

Match the following columns with the values obtained for the solution.

$$I.$$ $$x\cos \theta+ y\sin \theta=a$$, $$x\sin \theta- y\cos \theta=b$$	$$a)$$ $$(x^{2}-y^{2})^{2}=16xy$$
$$II.$$ $$x= \sec \theta+\tan\theta$$, $$y=\sec\theta-\tan\theta$$	$$b)$$ $$xy = 1$$
$$III.$$ $$x\sec \theta+ y\tan \theta=a$$, $$x\tan \theta+ y\sec \theta=b$$	$$c)$$ $$x^{2}-y^{2}=a^{2}-b^{2}$$
$$IV.$$ $$x=\cot\theta+\cos\theta$$, $$y=\cot\theta-\cos\theta$$	$$d)$$ $$x^{2}+y^{2}=a^{2}+b^{2}$$

$$I-c, II-d, III-a, IV-b$$

$$I-d, II-b, III-c, IV-a$$

$$I-a, II-c, III-d, IV-b$$

$$I-b, II-c, III-a, IV-d$$

Solution

I. $$x\cos { \theta } +y\sin { \theta } =a.....(i)$$
$$x\sin { \theta } -y\cos { \theta } =b.....(ii)$$
Squaring and adding equations $$(i)$$ and $$(ii)$$, we get
$${ \left[ x\cos { \theta } +y\sin { \theta } \right] }^{ 2 }+{ \left[ x\sin { \theta } -y\cos { \theta } \right] }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$$
$${ x }^{ 2 }\cos ^{ 2 }{ \theta } +{ y }^{ 2 }\sin ^{ 2 }{ \theta } +2xy\sin { \theta } \cos { \theta } +{ x }^{ 2 }\sin ^{ 2 }{ \theta } +{ y }^{ 2 }\cos ^{ 2 }{ \theta } -2xy\sin { \theta } \cos { \theta } ={ a }^{ 2 }+{ b }^{ 2 }$$
$${ x }^{ 2 }\left[ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right] +{ y }^{ 2 }\left[ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right] ={ a }^{ 2 }+{ b }^{ 2 }$$
$${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$$
which is (d).

II. $$x=\sec { \theta } +\tan { \theta } $$
$$y=\sec { \theta } -\tan { \theta } $$
$$xy=\left( \sec { \theta } +\tan { \theta } \right) \left( \sec { \theta } -\tan { \theta } \right) $$
$$xy=\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } $$
$$xy=1\left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] $$
which is (b).

III. $$x\sec { \theta } +y\tan { \theta } =a$$ ....... $$(iii)$$
$$x\tan { \theta } +y\sec { \theta } =b$$ ......... $$(iv)$$
Squaring and then subtracting equation $$(iii)$$ from $$(iv)$$, we get
$$\left[ { x }^{ 2 }\sec ^{ 2 }{ \theta } +{ y }^{ 2 }\tan ^{ 2 }{ \theta } +2xy\sec { \theta } \tan { \theta } \right] -\left[ { x }^{ 2 }\tan ^{ 2 }{ \theta } +{ y }^{ 2 }\sec ^{ 2 }{ \theta } +2xy\sec { \theta } \tan { \theta } \right] ={ a }^{ 2 }-{ b }^{ 2 }$$
$${ x }^{ 2 }\left[ \sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } \right] -{ y }^{ 2 }\left[ \sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } \right] ={ a }^{ 2 }-{ b }^{ 2 }$$
$${ x }^{ 2 }-{ y }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$$
which is (c).

IV. $$x=\cot { \theta } +\cos { \theta } $$
$$y=\cot { \theta } -\cos { \theta } $$
Squaring both equations, we get
$${ x }^{ 2 }=\cot ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\cot { \theta } \cos { \theta } $$ ......... $$(v)$$
$${ y }^{ 2 }=\cot ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } -2\cot { \theta } \cos { \theta } $$ ......... $$(vi)$$
Subtract equation $$(vi)$$ from $$(v)$$, we get
$${ x }^{ 2 }-{ y }^{ 2 }=4\cot { \theta } \cos { \theta } $$
Squaring both sides, we get
$${ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }=16\cot ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } $$ ......... $$(vii)$$
Now, $$xy=\left( \cot { \theta } +\cos { \theta } \right) \left( \cot { \theta } -\cos { \theta } \right) $$
$$xy=\cot ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } $$
$$xy=\cfrac { \cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } } \left[ 1-\sin ^{ 2 }{ \theta } \right] $$
$$xy=\cot ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } $$ ......... $$(viii)$$
Substitute $$(viii)$$ in equation $$(vii)$$, we get
$${ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }=16xy$$
which is (a).
Therefore, I-d, II-b, III-c, IV-a

Question 4
1 / -0

If $$\sin\theta+\cos\theta=a$$, then $$\sin^{4}\theta+\cos^{4}\theta=$$

A
$$1-\displaystyle \frac{1}{2}(a^{2}+1)^{2}$$

B
$$1-\displaystyle \frac{1}{2}(a^{2}-1)^{2}$$

C
$$1+\displaystyle \frac{1}{2}(a^{2}+1)^{2}$$

D
$$1+\displaystyle \frac{1}{2}(a^{2}-1)^{2}$$

Solution

$$ \sin \theta + \cos \theta = a $$
$$ {\sin}^{4} \theta + {\cos}^{4} \theta = {({\sin}^{2} \theta + {\cos}^{2} \theta)}^{2} - 2{\sin}^{2} \theta {\cos}^{2} \theta $$
= $$1$$ - $$ 2{\sin}^{2} \theta {\cos}^{2} \theta $$
Now, $$ {(\sin \theta + \cos \theta)}^{2} = {\sin}^{2} \theta + {\cos}^{2} \theta + 2sin \theta \cos \theta $$
$$ {a}^{2} = 1 + 2\sin \theta \cos \theta $$
$$ \sin \theta \cos \theta = \frac{1}{2} ({a}^{2} - 1) $$
Hence, substituting the value of $$ \sin \theta \cos \theta $$, we get,
$$ {\sin}^{4} \theta + {\cos}^{4} \theta = 1 - \frac{1}{2}{({a}^{2} - 1)}^{2} $$
Question 5
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If $$\displaystyle \sec\alpha=5x+\frac{1}{20x} $$, then $$\sec\alpha+\tan\alpha$$ is equal to

A
$$5x$$

B
$$\displaystyle \frac{1}{20x}$$

C
$$10x$$ or $$ -\displaystyle \frac{1}{10x}$$

D
$$10\displaystyle x$$ or $$\dfrac{1}{10x}$$

Solution

Let $$\sec\alpha + \tan\alpha = a$$ ..... $$(i)$$

It is known that, $$\sec^{2}\alpha - \tan^{2}\alpha=1$$
Or
$$(\sec\alpha+\tan\alpha)(\sec\alpha-\tan\alpha)=1$$

So we get $$\sec\alpha - \tan\alpha = \dfrac{1}{a}$$ ..... $$(ii)$$

Adding equations $$(i)$$ and $$(ii)$$, we get

$$2\sec\alpha = a + \dfrac{1}{a}$$

$$\Rightarrow \sec\alpha = \dfrac{1}{2}\left(a + \dfrac{1}{a}\right) $$

Put the value of $$\sec\alpha$$ in the given equation.

$$\Rightarrow 5x + \dfrac{1}{20x}=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)$$

$$\Rightarrow \dfrac{1}{2}\left(10x + \dfrac{1}{10x}\right)=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)$$
Solving for $$a$$ we get $$ a = 10x$$ or $$a = \dfrac{1}{10x}$$

Hence option (D) is correct
Question 6
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If $$k=(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)$$( $$\sec A$$$$-\tan A$$) ( $$\sec B$$$$-\tan B$$) ( $$\sec C$$$$-\tan C$$), then $$k$$ is equal to

A
$$0$$

B
$$ 1$$

C
$$\pm 3$$

D
$$\pm 4$$

Solution

$$k=(\text{sec}A+\tan A)(\text{sec}B+\tan B)(\text{sec}C+\tan C)(\text{sec}A-\tan A)(\text{sec}B-\tan B)(\text{sec}C-\tan C)$$
$$k=(\text{sec}A+\tan A)(\text{sec}A-\tan A)(\text{sec}B+\tan B)(\text{sec}B-\tan B)(\text{sec}C+\tan C)(\text{sec}C-\tan C)$$
$$k=(\text{sec}^2 A-\tan^2 A)(\text{sec}^2 B-\tan^2 B)(\text{sec}^2 C-\tan^2 C)$$
$$\implies k=1$$
Question 7
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If $$\tan { \theta } +\sin { \theta } =m, \tan { \theta - \sin { \theta =n } } $$, then $$(m^{2}-n^{2})^{2}=$$.

A
$$16mn$$

B
$$4mn$$

C
$$32mn$$

D
$$8mn$$

Solution

we solve to get
$$\tan\theta = \dfrac{m+n}{2} $$ and $$ \sin\theta = \dfrac{m-n}{2}$$
$$\therefore \cot\theta = \dfrac{2}{m+n}$$ and $$\mathrm{cosec}\theta = \dfrac{2}{m-n}$$
now we use $$\mathrm{cosec}^{2}\theta - \cot^{2}\theta = 1$$ to eliminate $$\theta$$
$$\dfrac{4}{(m-n)^2}-\dfrac{4}{(m+n)^2}=1$$
$$4[(m+n)^2-(m-n)^2]=(m^2-n^2)^2$$
$$4(4mn)=(m^2-n^2)^2$$
Question 8
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If $$\cos x + \sec x = - 2$$ for a positive odd integer $$n$$ then $$\cos^nx + \sec^nx$$ is

A
$$1$$

B
$$-1$$

C
$$-2$$

D
$$2$$

Solution

If $$\cos x + \sec x = - 2$$
$$\cos x + \displaystyle\frac{1}{\cos x} =-2 $$
$$\cos ^2x +1 =-2 \cos x$$
$$\cos ^2x +1 +2 \cos x =0$$
This is of the form $$a^2+b^2+2ab=(a+b)^2$$
$$\Rightarrow (\cos x +1)^2 = 0$$
$$\Rightarrow (\cos x +1) = 0$$
$$\Rightarrow \cos x =-1 $$
$$\implies \sec x=-1$$
If $$n$$ is odd,
$$ \cos ^nx + \sec ^nx =-1+(-1)= - 2$$
If $$n$$ is even,
$$\cos ^nx + \sec ^nx = 1+1=2$$
Question 9
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In the given figure, $$ \triangle ABC$$ is right angled at B and $$\cot A = \dfrac{3}{4}$$ . If $$AC =10$$cm, then the length of $$AB$$ is:

A
$$3$$cm

B
$$4$$cm

C
$$6$$cm

D
$$8$$cm

Solution

$$\cot { A } =\dfrac { 3 }{ 4 } =\dfrac { AB }{ BC } $$
$$\Rightarrow \dfrac { AB }{ BC } =\dfrac { 3 }{ 4 } $$
$$\Rightarrow AB:BC=3:4$$
$$\Rightarrow$$ Let $$ AB=3x\;\&\; BC=4x$$
In right angle $$\triangle ABC,$$
$$\Rightarrow { AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$$
$$\Rightarrow { \left( 10 \right) }^{ 2 }={ \left( 3x \right) }^{ 2 }+{ \left( 4x \right) }^{ 2 }$$
$$\Rightarrow 100= \left( 9x^2 \right) + \left( 16x^2 \right)$$
$$\Rightarrow 25{ x }^{ 2 }=100$$
$${ \Rightarrow x }^{ 2 }=4$$
$$\Rightarrow x=2$$
$$\Rightarrow AB=3x=6cm$$
Hence, the answer is $$6cm$$
Question 10
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If $$\cos x$$ $$+\cos^{2}{x}=1 $$ then $$ \sin^{12}{x}+3\sin^{10}{x}+$$$$3 \sin^{8}{x}+\sin^{6}{x}+\sin^{4}{x}+2\sin^{2}{x}-2$$ is equal to

A
$$0$$

B
$$\sin x$$

C
$$1 -\cos x$$

D
$$1$$

Solution

$$\cos x + \cos ^{2}x = 1$$

$$\cos x= 1 - \cos ^{2}x$$

$$\cos x= \sin ^{2}x$$

now in the given equation we substitute $$\sin ^{2}x = \cos x$$ everywhere this gives

$$= \cos ^{6}x + 3\cos ^{5}x + 3\cos ^{4}x + \cos ^{3}x + 2\cos ^{2}x + \cos x -2$$
$$ = \cos ^{3}x(\cos x+ 1)^{3} + 2\cos ^{2}x + \cos x -2$$
$$ = (\cos ^{2}x+ \cos x)^{3} + \cos x -2(1 - \cos ^{2}x)$$
$$ =(1)^{3} + \cos x -2\cos x$$
$$ =1-\cos x$$

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Introduction to Trigonometry Test - 28

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Introduction to Trigonometry Test - 28

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 2

$$\dfrac{p+n}{m+p}$$

$$\dfrac{p-n}{m+p}$$

$$\dfrac{p-n}{m-p}$$

None of these

Solution

Question 3

$$I-c, II-d, III-a, IV-b$$

$$I-d, II-b, III-c, IV-a$$

$$I-a, II-c, III-d, IV-b$$

$$I-b, II-c, III-a, IV-d$$

Solution

Question 4

$$1-\displaystyle \frac{1}{2}(a^{2}+1)^{2}$$

$$1-\displaystyle \frac{1}{2}(a^{2}-1)^{2}$$

$$1+\displaystyle \frac{1}{2}(a^{2}+1)^{2}$$

$$1+\displaystyle \frac{1}{2}(a^{2}-1)^{2}$$

Solution

Question 5

$$5x$$

$$\displaystyle \frac{1}{20x}$$

$$10x$$ or $$ -\displaystyle \frac{1}{10x}$$

$$10\displaystyle x$$ or $$\dfrac{1}{10x}$$

Solution

Question 6

$$0$$

$$ 1$$

$$\pm 3$$

$$\pm 4$$

Solution

Question 7

$$16mn$$

$$4mn$$

$$32mn$$

$$8mn$$

Solution

Question 8

$$1$$

$$-1$$

$$-2$$

$$2$$

Solution

Question 9

$$3$$cm

$$4$$cm

$$6$$cm

$$8$$cm

Solution

Question 10

$$0$$

$$\sin x$$

$$1 -\cos x$$

$$1$$

Solution

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