MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 12

$\cos 9\theta=\sin \dfrac{\pi}{4}$ , then the value of

$\tan 6\theta$ is

0%
$\dfrac {1}{\sqrt {3}}$
0%
$\sqrt {3}$
0%
$1$
0%
$0$

Explanation

We have,

$\cos 9\theta=\sin \dfrac{\pi}{4}$

$\cos 9\theta =\cos \left( {{90}^{o}}-45^0 \right)$

$9\theta =45^0$

$10\theta ={45}^{o}$

$\theta =5^0$

Value of $\tan 6\theta$

$=\tan 6\theta$

$=\tan 6\times {{5}^{o}}$

$=\tan {{30}^{o}}$

$=\dfrac{1}{\sqrt 3}$

Hence, this is the answer.

$\tan { x } =3\cot { x } ,$ then x =?

0%
${ 45 }^{ \circ }$
0%
${ 60 }^{ \circ }$
0%
${ 30 }^{ \circ }$
0%
${ 15 }^{ \circ }$

Explanation

$\tan x = 3 \cot x$

$\tan x = 3 \times 1/ \tan x . (\cot x = t/tanx)$

$\tan^2 x = 3$

$\tan x = \sqrt{3}$

So the value of x will be

$60^o$ because the value of

$\tan \, 60$ is

$\sqrt{3}$

$\cos 35 ^ { \circ } + \cos 85 ^ { \circ } + \cos 155 ^ { \circ }$ is

0%
0
0%
$\frac { 1 } { \sqrt { 3 } }$
0%
$\frac { 1 } { \sqrt {2 } }$
0%
$\cos 275 ^ { \circ }$

${cos}^{2}0+{cos}^{2}10+{cos}^{2}20+{cos}^{2}30+{cos}^{2}40+{cos}^{2}50+{cos}^{2}60+{cos}^{2}70+{cos}^{2}80+{cos}^{2}90=$

0%
$0$
0%
$1$
0%
$5$
0%
$10$

$\sin \theta = \cos \theta$ , then

$\theta = ?$

0%
$45^{o}$
0%
$90^{o}$
0%
$0^{o}$
0%
$30^{o}$

Explanation

$\sin\theta =\cos \theta$

$\Rightarrow \dfrac {\sin\theta}{\cos\theta}=1$

$\Rightarrow \tan\theta=1$

$\Rightarrow \theta=\tan^{-1}1$

$\therefore \theta= 45^{\circ}$

$\cot{1^0}\cot{2^0}\cot{3^0}\cot{4^0}...\cot{89^0}$ .

0%
1
0%
0
0%
-1
0%
NONE OF THE GIVEN

Explanation

$\cot{1^0}\cot{2^0}\cot{3^0}\cot{4^0}...\cot{89^0}$

$=\cot{\left(90^0-89^0\right)}\cot{\left(90^0-88^0\right)}\cot{\left(90^0-87^0\right)}\cot{\left(90^0-86^0\right)}...\cot{45^0}.....\cot{86^0}\cot{87^0}\cot{88^0}\cot{89^0}$

There are

$\dfrac{89}{2}=44$ terms

$+1$ term

That

$1$ term is the middle term

$=\cot{45^0}=1$

and other terms will form the reciprocals

$=\tan{89^0}\tan{88^0}\tan{87^0}\tan{86^0}...\cot{45^0}.....\cot{86^0}\cot{87^0}\cot{88^0}\cot{89^0}$

$=\left(\tan{89^0}\cot{89^0}\right)\times \left(\tan{88^0}\cot{88^0}\right)\times \left(\tan{87^0}\cot{87^0}\right)\times \left(\tan{86^0}\cot{86^0}\right)\times ...\tan{45^0}$ by re-arranging

$=1\times 1\times 1\times 1...1=1$

Choose the correct alternative:

$\ cosec \theta = \dfrac { 1 } { \dots \ldots }$

0%
$\cos \theta$
0%
$\sin \theta$
0%
$\tan \theta$
0%
$\cot \theta$

Explanation

By using trigonometric formula,

$cosec{\theta}$

$=1/sin {\theta}$

Consider the following statements :

$\cos\theta+\sec\theta$ can never be equal to

$1.5$

$\tan\theta+\cot\theta$ can never be less than to

$2$
Which of the above statements is/are correct ?

0%
$1$ only
0%
$2$ only
0%
Both $1$ and $2$
0%
Neither $1$ nor $2$

Explanation

$\cos\theta=\dfrac{1}{\sec\theta}$

$\tan\theta=\dfrac{1}{cot\theta}$

$\Rightarrow A.M\ge G.M$

$\Rightarrow x+\dfrac{1}{x}\ge 2, x>0$
From above inequality, we can write

$x+\dfrac{1}{x}\le -2, x<0$

$\tan\theta+\dfrac{1}{\tan\theta}\ge 2$ (when

$\tan \theta > 0$ )

and

$\cos\theta+\dfrac{1}{\cos\theta}\ge 2$ (when

$\cos \theta > 0$ )

Similarly

$\tan\theta+\dfrac{1}{\tan\theta}\le -2$ (when

$\tan \theta < 0$ )

and

$\cos\theta+\dfrac{1}{\cos\theta}\le -2$ (when

$\cos \theta < 0$ )

Hence

$\cos \theta +\sec \theta$ can never be

$1.5$ .

But

$\tan \theta + \cot \theta$ will always be less than

$2$ when they are negative.

Hence only (1) is correct.

$\sin A+\cos A= {\sqrt2}$ and

$\tan A+\cot A=2$ , then the value of

$\sec A\cdot \csc A$ is equal to

0%
$2$
0%
$3$
0%
$4$
0%
$\dfrac{1}{2}$

If x is acute, then

$\displaystyle \sqrt{\frac{1 + sin x}{1 - sin x}} = ?$

0%
$\displaystyle sec x + cosec x$
0%
$\displaystyle sec x + tan x$
0%
$\displaystyle cosec x + cot x$
0%
$\displaystyle tan x + cot x$

Explanation

$\displaystyle \sqrt{\frac{1 + sin x}{1 - sin x}} \\= \dfrac{\sqrt{1 + sin x}}{\sqrt{1 - sin x}} \times \dfrac{\sqrt{1 + sin x}}{\sqrt{1 + sin x}}\\= \dfrac{(1 + sin x)}{\sqrt{1 - sin^2 x}} \\ = \dfrac{(1 + sin x)}{cos x}$

$=\displaystyle \left(\frac{1}{cos x} + \frac{sin x}{cos x}\right) \\= (sec x + tan x)$ .

$\displaystyle \tan \theta + \cot \theta = 2$ , then

$\displaystyle \sin \theta =$ ?

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{\sqrt{2}}$
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$\displaystyle \frac{\sqrt{3}}{2}$
0%
$\displaystyle \frac{1}{\sqrt{3}}$

Explanation

$\displaystyle \tan \theta + \cot \theta = 2 \Rightarrow \tan \theta + \frac{1}{\tan \theta} = 2$

$\displaystyle \Rightarrow \tan^2 \theta + 1 = 2 \tan \theta \Rightarrow 1 + \tan^2 \theta - 2\tan \theta = 0$

$\displaystyle \Rightarrow ( 1 - \tan \theta )^2 = 0 \Rightarrow 1 - \tan \theta = 0 \Rightarrow \tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4}$ .

$\displaystyle \therefore \ \ \ \sin \theta = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\displaystyle (sec \theta - cos \theta) (cosec \theta - sin \theta) (cot \theta + tan \theta) = ?$

0%
$\displaystyle 1$
0%
$\displaystyle -1$
0%
$\displaystyle 0$
0%
none of these

Explanation

Given exp.

$\displaystyle \Big( \frac{1}{cos\theta} - cos\theta\Big)\Big(\frac{1}{sin\theta} - sin\theta\Big) \Big(\frac{cos\theta}{sin\theta} + \frac{sin\theta}{cos\theta}\Big) \\= \dfrac{sin^2\theta cos^2 \theta}{sin^2\theta cos^2\theta} = 1$

$\displaystyle \Bigg( \frac{\sec \theta + \tan \theta - 1}{\tan \theta - \sec \theta + 1} \Bigg)^2 =$ ?

0%
$\displaystyle \frac{ ( 1 + \sin \theta)}{1-\sin\theta)}$
0%
$\displaystyle \frac{ ( 1 - \sin \theta)}{1+\sin\theta)}$
0%
$\displaystyle 1$
0%
none of these

Explanation

Given exp. =

$\displaystyle \Bigg\{ \frac{(sec\theta + tan \theta) -(sec^2 \theta - tan^2 \theta )}{(tan \theta -sec\theta + 1)} \Bigg\}^2 = (sec \theta + tan \theta)^2 = \frac{(1 + sin \theta)}{( 1 - sin \theta)}$

$\displaystyle \cos^2 \theta + sec^2 \theta = p$ , then

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$\displaystyle p < 1$
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$\displaystyle p = 1$
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$\displaystyle 1 < p < 2$
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$\displaystyle p \geq 2$

Explanation

$\displaystyle p = ( \cos^2 \theta + sec^2 \theta )= (cos \theta - sec \theta)^2 + 2 cos \theta sec \theta = (cos\theta - sec \theta )^2 + 2 \geq 2$ .

$\displaystyle \sin \theta + \sin^2 \theta = 1$ , then

$\displaystyle (\cos^2 \theta + \cos^4 \theta) =$ ?

0%
$\displaystyle 0$
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$\displaystyle 1$
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$\displaystyle 2$
0%
none of these

Explanation

$\displaystyle \sin \theta + \sin^2 \theta = 1\\ \Rightarrow \sin \theta = (1 - \sin^2 \theta) = \cos^2 \theta \\ \Rightarrow \sin^2 \theta = \cos^4\theta$

$\displaystyle \therefore \ \ \ 1 - \cos^2 \theta = \cos^4 \theta$

$\displaystyle \Rightarrow \cos^2 \theta + \cos^4 \theta = 1$

$\displaystyle sin \theta \left(\frac{sin\theta}{1 + cos \theta} + \frac{1 + cos \theta}{sin \theta}\right) = ?$

0%
$\displaystyle 1$
0%
$\displaystyle 2$
0%
$\displaystyle 3$
0%
$\displaystyle 4$

Explanation

Given exp.

$=\displaystyle \frac{sin \theta [sin^2 \theta + (1 + cos \theta)^2]}{sin \theta ( 1 + cos\theta)}\\ = \dfrac{2(1 + cos\theta)}{( 1 + cos \theta)} = 2$

$\theta =30^o$ , then

$\dfrac{1-\sin^22\theta}{\cos 2\theta}$ is

0%
$1$
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$-1$
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$2$
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$-2$

Explanation

$\dfrac{1-sin^22\theta}{\cos 2\theta}=\dfrac{1-\sin^22\times 30^o}{\cos 2\times 30^o}$

$=\dfrac{1-\sin^2 60^o}{\cos 60^o}=\dfrac{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}{\dfrac{1}{2}}$

$\dfrac{1-\dfrac{3}{4}}{\dfrac{1}{2}}=\dfrac{\dfrac{4-3}{4}}{\dfrac{1}{2}}\rightarrow \dfrac{1}{4}\times \dfrac{2}{1}=\dfrac{1}{2}$

$\dfrac{1}{cosec \theta - \cot \theta}$ is equal to:

0%
$cosec \theta + \cot \theta$
0%
$\dfrac{1}{cosec \theta + \cot \theta}$
0%
$cosec \theta - \cot \theta$
0%
$\tan \theta$

$a \sin^{2}x+b\cos^{2}x=c, b\sin^{2}y+a\cos^{2}y=d$ and

$a \tan x=b\tan y,$ then

$\displaystyle \frac{a^{2}}{b^{2}}$ equals to

0%
$^{\displaystyle \frac{(a-d)(c-a)}{(b-c)(d-b)}}$
0%
$^{\displaystyle \frac{(b-c)(b-d)}{(a-c)(a-d)}}$
0%
$\displaystyle \frac{(b-c)(d-b)}{(a-d)(c-a)}$
0%
$^{\displaystyle \frac{(d-a)(c-a)}{(b-c)(d-b)}}$

Explanation

dividing equation

$asin^2x+bcos^2x=c$ by

$cos^2x$

$a\ tan^2x+b=c\ sec^2x\Rightarrow a\ tan^2x+b=c(1+tan^2x)\Rightarrow tan^2x(c-a)=b-c\Rightarrow tan^2x=\dfrac{b-c}{c-a}$

Similarly , dividing equation

$bsin^2y+acos^2y=d$ by

$cos^2y$

$b\ tan^2y+a=d\ sec^2y\Rightarrow b\ tan^2y+a=d(1+tan^2y)\Rightarrow tan^2y(d-b)=a-d\Rightarrow tan^2y=\dfrac{a-d}{d-b}$

Now,

$atanx=btany\Rightarrow a^2tan^2x=b^2tan^2y\Rightarrow \dfrac{a^2}{b^2}=\dfrac{tan^2y}{tan^2x}\Rightarrow \dfrac{a^2}{b^2}=\dfrac{\dfrac{a-d}{d-b}}{\dfrac{b-c}{c-a}}=\dfrac{(a-d)(c-a)}{(b-c)(d-b)}$

Therefore, Answer is

$\dfrac{(a-d)(c-a)}{(b-c)(d-b)}$

$\dfrac {\sin\alpha}{\sin\beta}=\dfrac {\sqrt 3}{2}$ and

$\dfrac {\cos\alpha}{\cos\beta}=\dfrac {\sqrt 5}{2}, 0 < \alpha < \beta < \dfrac {\pi}{2}$ , then

0%
$\tan\alpha=1$
0%
$\tan\alpha=\dfrac {\sqrt 3}{\sqrt 5}$
0%
$\tan\beta=\dfrac {\sqrt 3}{\sqrt 5}$
0%
$\tan\beta=1$

Explanation

$\cfrac { \sin\alpha }{ \sin\beta } =\cfrac { \sqrt { 3 } }{ 2 }$ ...(1)

$\cfrac { \cos\alpha }{ \cos\beta } =\cfrac { \sqrt { 5 } }{ 2 }$ ...(2)
Dividing (1) by (2), we get

$\cfrac { \sin\alpha }{ \sin\beta } .\dfrac { \cos\beta }{ \cos\alpha } =\dfrac { \sqrt {3} }{ \sqrt { 5 } }$

$\Rightarrow \cfrac { \tan\alpha }{ \tan\beta } =\cfrac { \sqrt { 3 } }{ \sqrt { 5 } }$

$\Rightarrow \tan\alpha =\cfrac { \sqrt { 3 } }{ \sqrt { 5 } }\tan\beta$
As

$\alpha <\beta$ and

$1>\cfrac { \sqrt { 3 } }{ \sqrt { 5 } }$
We get

$\tan\beta =1$ and

$\tan\alpha =\cfrac { \sqrt { 3 } }{ \sqrt { 5 } }$

The value of the expression

$(\tan1^{0} \tan2^{0} \tan 3^{0}...\tan89^{0})$ is equal to

0%
$0$
0%
$1$
0%
$2$
0%
$\dfrac{1}{2}$

Explanation

Given:

$A = \tan1 ^o\cdot\tan2^o\cdot\tan3^o\cdots\cdots\tan89^o$

We know that, the value of

$\tan89^o=\tan(90^o-1^o)=\cot 1^o$
In the same way,

$\tan 88^o=\cot 2^o$

$\tan 87^o=\cot 3^o$
And .... so on, up to

$\tan 46^o=\cot 44^o$
And the middle one is

$\tan 45^o=1$
So,

$A = \tan1^o\cdot\tan2^o\cdot\tan3^o\cdots\cdots\tan{44}^o\cdot\tan45^o\cdot\cot44^o\cdots\cdots\cot 2^o\cdot\cdot\cot1^o$

$\tan$ and

$\cot$ cancels out each other, then

$\tan45^o = 1$ is remaining

$\therefore$ The value of the expression

$V$ is

$1$ .

Hence, option B.

$\cos9 \alpha= \sin \alpha$ and

$9 \alpha < 90^{0}$ , then the value of

$\tan5 \alpha$ is

0%
$\dfrac{1}{\sqrt{3}}$
0%
$\sqrt{3}$
0%
$1$
0%
$0$

Explanation

$\cos(9\alpha)=\sin(\alpha)$

$\sin(90^0-9\alpha)=\sin(\alpha)$

For

$0<\alpha<90^0$

$90^0-9\alpha=\alpha$

$90^0=10\alpha$

$\alpha=9^0$
Now,

$\tan5\alpha$

$=\tan(5)(9^0)$

$=\tan45^0 = 1$

Hence, option C.

$x=\displaystyle \frac{\sin^{3}p}{\cos^{2}p}, y=\displaystyle \frac{\cos^{3}p}{\sin^{2}p}$ and

$\displaystyle \sin p+\cos p=\frac{1}{2}$ , then

$x+y=$

0%
$\dfrac {75}{18}$
0%
$\dfrac{44}9$
0%
$\dfrac {79}{18}$
0%
$\dfrac {48}9$

Explanation

$\sin p+\cos p=\cfrac { 1 }{ 2 } \Rightarrow { \left( \sin p+\cos p \right) }^{ 2 }=\cfrac { 1 }{ 4 } \\ 1+2\sin p.\cos p=\cfrac { 1 }{ 4 } \Rightarrow \sin p\cos p=-\cfrac { 3 }{ 8 }$

Therefore,

$x+y=\cfrac { { \sin }^{ 3 }p }{ { \cos }^{ 2 }p } +\cfrac { { \cos }^{ 3 }p }{ { \sin }^{ 2 }p }$

$=\cfrac { { \sin }^{ 5 }p+{ \cos }^{ 5 }p }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \sin }^{ 3 }p \, \left( 1-{ \cos }^{ 2 }p \right) +{ \cos }^{ 3 }p \, \left( 1-{ \sin }^{ 2 }p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \sin }^{ 3 }p-{ \sin }^{ 3 }p \, { \cos }^{ 2 }p+{ \cos }^{ 3 }p-{ \cos }^{ 3 }p \, { \sin }^{ 2 }p }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \sin }^{ 3 }p+{ \cos }^{ 3 }p-{ \sin }^{ 2 }p \, { \cos }^{ 2 }p \, \left( \sin p+\cos p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \left( \sin p+\cos p \right) }^{ 3 }-3\sin p \, \cos p\left( \sin p+\cos p \right) -{ \sin }^{ 2 }p \, { \cos }^{ 2 }p\left( \sin p+\cos p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \left( \cfrac { 1 }{ 2 } \right) }^{ 3 }-3\left( -\cfrac { 3 }{ 8 } \right) \left( \cfrac { 1 }{ 2 } \right) -\left( \cfrac { 9 }{ 64 } \right) \left( \cfrac { 1 }{ 2 } \right) }{ \left( \cfrac { 9 }{ 64 } \right) }$

$=\cfrac { \cfrac { 1 }{ 8 } +\cfrac { 9 }{ 16 } -\cfrac { 9 }{ 128 } }{ \cfrac { 9 }{ 64 } } =\cfrac { 79 }{ 18 }$

$\displaystyle {x}=\frac{\sin^{3}{p}}{\cos^{2}{p}}, \displaystyle {y}=\frac{\cos^{3}{p}}{\sin^{2}{p}}$ and

$\sin p$

$+$

$\cos p$

$= \dfrac 12$ then

${x}+{y}=$

0%
$\displaystyle \frac{75}{18}$
0%
$\displaystyle \frac{44}{9}$
0%
$\displaystyle \frac{79}{18}$
0%
$\displaystyle \frac{48}{9}$

Explanation

$\sin p+\cos p=\cfrac { 1 }{ 2 } \Rightarrow { \left( \sin p+\cos p \right) }^{ 2 }=\cfrac { 1 }{ 4 } \\ 1+2\sin p.\cos p=\cfrac { 1 }{ 4 } \Rightarrow \sin p\cos p=-\cfrac { 3 }{ 8 }$

Therefore,

$x+y=\cfrac { { \sin }^{ 3 }p }{ { \cos }^{ 2 }p } +\cfrac { { \cos }^{ 3 }p }{ { \sin }^{ 2 }p }$

$=\cfrac { { \sin }^{ 5 }p+{ \cos }^{ 5 }p }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \sin }^{ 3 }p \, \left( 1-{ \cos }^{ 2 }p \right) +{ \cos }^{ 3 }p \, \left( 1-{ \sin }^{ 2 }p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \sin }^{ 3 }p-{ \sin }^{ 3 }p \, { \cos }^{ 2 }p+{ \cos }^{ 3 }p-{ \cos }^{ 3 }p \, { \sin }^{ 2 }p }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { { \sin }^{ 3 }p+{ \cos }^{ 3 }p-{ \sin }^{ 2 }p \, { \cos }^{ 2 }p \, \left( \sin p+\cos p \right) }{ { \sin }^{ 2 }p \, { \cos }^{ 2 }p }$

$=\cfrac { \cfrac { 1 }{ 8 } +\cfrac { 9 }{ 16 } -\cfrac { 9 }{ 128 } }{ \cfrac { 9 }{ 64 } } =\cfrac { 79 }{ 18 }$

$\sin x+\sin ^{2}x=1$ ,then the value of

$\cos ^{12}x+3\cos ^{10}x+3\cos ^{8}x+\cos ^{6}x-2$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

Explanation

We have

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

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

Now

$\cos ^{12}x+3\cos ^{10}x+3\cos ^{8}x+\cos ^{6}x-2=\sin ^{6}x+3\sin ^{5}x+3\sin ^{4}x+\sin ^{3}x-2$

$=\left ( \sin ^{2}x \right )^{3}+3 \left ( \sin ^{2}x \right )^{2}(\sin x)+3\left ( \sin ^{2}x \right )\left ( \sin x \right )^{2}+\left ( \sin x \right )^3-2$

$=\left ( \sin ^{2}x +\sin x\right )^{3}-2$ .............

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$

$=\left ( 1 \right )^{3}-2$

$=-1$

$\cos\theta +\cos^{2}\theta =1$ and

$a\sin^{12}\theta +b\sin^{10}\theta +c\sin^{8}\theta +d\sin^{6}\theta =1.$ Then

$\displaystyle \frac{b+c}{a+d}=$ ?

0%
$2$
0%
$3$
0%
$4$
0%
$6$

Explanation

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

$\Rightarrow \cos\theta +1-\sin^{ 2 }\theta =1$

$\Rightarrow \cos\theta = \sin^{ 2 }\theta$

$\Rightarrow { \sqrt { 1-\sin^{2}} \theta }={ \sin^{2}\theta}$

On squaring both sides:-

$\Rightarrow { \left( \sqrt { 1-\sin^{2}} \theta \right)}^{2}={\left( \sin^{2}\theta \right)}^{2}$

$\Rightarrow 1-\sin^{2}\theta = \sin^{4}\theta$

$\Rightarrow \sin^{4}\theta + \sin^{2}\theta =1$

On taking cube on both sides:-

$\Rightarrow { \left( \sin^{4}\theta +\sin^{2}\theta \right) }^{3}={\left( 1 \right) }^{3}$

$\Rightarrow \sin^{12}\theta +3\sin^{10}\theta +3\sin^{8}\theta +\sin^{6}\theta =1$

On comparing above obtained equation with

$a\sin^{12}\theta +b\sin^{10}\theta +c\sin^{8}\theta +d\sin^{6}\theta =1$ we get,

$a=1,b=3,c=3,d=1$

Hence,

$\ \dfrac {b+c}{a+d} =\dfrac { 3+3 }{ 1+1 }$

$=\dfrac { 6 }{ 2 }$

$=3$

Hence, option

$B$ is correct.

$a \sin^{2}\theta+b\cos^{2}\theta=a\cos^{2}\phi+b\sin^{2}\phi=1$ and

$a \tan\theta=b\tan\phi$ , then choose the correct option.

0%
${a+b}=2ab$
0%
${a-b}=2ab$
0%
${a-b}+2ab=0$
0%
${a+b}+2ab=0$

Explanation

Given:

$a\sin ^2\theta+b\cos ^2\theta=1$ ....

$(i)$ and

$a\cos ^2\phi+b\sin ^2\phi=1$ ....

$(ii)$

Dividing equation

$(i)$ by

$\cos ^2\theta$ , we get

$a\tan ^2\theta+b=\sec ^2\theta$

$\Rightarrow a\tan ^2\theta+b=1+\tan ^2\theta$

$\Rightarrow \tan ^2\theta=\dfrac{b-1}{1-a}=\tan \theta=\sqrt{\dfrac{b-1}{1-a}}$

Similarly in the equation

$(ii)$ , on dividing by

$\cos ^2\phi$ we get,

$b\tan ^2\phi+a=\sec ^2\phi$

$\Rightarrow b\tan ^2\phi+a=1+\tan ^2\phi$

$\Rightarrow \tan ^2\phi=\dfrac{1-a}{b-1}=\tan \phi=\sqrt{\dfrac{1-a}{b-1}}$

Now, it is given that

$a\tan \theta=b\tan \phi$

$\Rightarrow a\sqrt{\dfrac{b-1}{1-a}}=b\sqrt{\dfrac{1-a}{b-1}}$

$\Rightarrow \dfrac{a}{b}=\dfrac{1-a}{b-1}$

$\Rightarrow ab-a=b-ab$

$\Rightarrow a+b=2ab$

$a \sin \theta + b \cos \theta = c$ , then

$\dfrac {a - b \tan \theta}{b + a \tan \theta} =$

0%
$\dfrac {\sqrt {a^{2}+b^{2}-c^{2}}}{c}$
0%
$-\dfrac {\sqrt {a^{2}+b^{2}-c^{2}}}{c}$
0%
$\displaystyle \pm \dfrac {\sqrt {a^{2}+b^{2}-c^{2}}}{c}$
0%
$\displaystyle \pm\dfrac {\sqrt {a^{2}+b^{2}+c^{2}}}{c}$

Explanation

$a\sin {\theta +b\cos {\theta}} =\quad c$

$a\cos {\theta -b\sin {\theta}} = \quad k$

Squaring on both sides and adding gives

$a^{2} + b^{2} = c^{2} + k^{2}$

$k = \pm \sqrt {a^{2} + b^{2} - c^{2}}$

$\dfrac {a-b\tan {\theta}}{b+a\tan {\theta}} = \dfrac {a\cos {\theta - b\sin {\theta}}}{a\sin {\theta + b\cos {\theta}}} = \dfrac {k}{c} = \dfrac {\pm \sqrt {a^{2} + b^{2} - c^{2}}}{c}$

$\cfrac { \cos { \theta } }{ 1-\sin { \theta } } -\cfrac { \cos { \theta } }{ 1+\sin { \theta } } =2$ is satisfied by which one of the following values of

$\theta$ ?

0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$

$\tan { \theta } =\cfrac { p }{ q }$ , then what is

$\cfrac { p\sec { \theta } -qco\sec { \theta } }{ p\sec { \theta } +qco\sec { \theta } }$ equal to?

0%
$\cfrac { p-q }{ p+q }$
0%
$\cfrac { { q }^{ 2 }-{ p }^{ 2 } }{ { q }^{ 2 }+{ p }^{ 2 } }$
0%
$\cfrac { { p }^{ 2 }-{ q }^{ 2 } }{ { q }^{ 2 }+{ p }^{ 2 } }$
0%
$1$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 12 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 12 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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