MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 12

$\cos^ {2}\theta \left(1+\tan^ {2}\theta\right)=1$

0%
True
0%
False

Explanation

As we know that

$\text{sec}^2 A-\tan^2 A=1$

$\cos^2 \theta(1+\tan^2 \theta)=\cos^2 \theta\times \text{sec}^2 \theta$

$=\cos^2 \theta\times \dfrac{1}{\cos^2 \theta}$

$=1$

So the relation is

$\text{True}$

$\csc^{6}\theta -\cot^{6}\theta =1+3\csc^{2}\theta \cot^{2}\theta$

0%
True
0%
False

Explanation

$\text{csc}^6 \theta-\cot^6 \theta=(\text{csc}^2 \theta)^3-(\cot^2 \theta)^3$

$=(\text{csc}^2 \theta-\cot^2 \theta)(\text{csc}^4 \theta+\cot^4 \theta+\text{csc}^2 \theta\cot^2 \theta)$

$=(1)((\text{csc}^2 \theta)^2+(\cot^2 \theta)^2-2\text{csc}^2 \theta\cot^2 \theta+3\text{csc}^2 \theta\cot^2 \theta)$

$=(\text{csc}^2-\cot^2 \theta)^2+3\text{csc}^2 \theta\cot^2 \theta$

$=1+3\text{csc}^2 \theta\cot^2 \theta$

So the relation is

$\text{True}$

$x=a\cos\theta+b\sin\theta$ and

$y=a\sin\theta-b\cos\theta$ , then

$a^{2}+b^{2}=x^{2}+y^{2}$ .

0%
True
0%
False

Explanation

Given,

$x=a\cos\theta+b\sin\theta$ ......(1) and

$y=a\sin\theta-b\cos\theta$ ......(2).

Now squaring (1) and (2) and then adding we get,

$x^2+y^2=a^2(\cos^2 \theta+\sin^2 \theta)+2ab\sin \theta.\cos \theta-2ab\sin \theta.\cos \theta+b^2(\sin^2 \theta+\cos^2 \theta)$

or,

$x^2+y^2=a^2+b^2$ .

$(\cos A-\csc A)^{2}+(\sin A-\sec A)^{2}=(1-\sec A.\csc A)^{2}$

0%
True
0%
False

Explanation

Given

$\left( \cos A -\csc A\right)^2 + \left(\sin A-\sec A\right)^2=\left( 1-\sec A \csc A\right)^2$

Taking LHS

$\left( \cos A -\csc A\right)^2 + \left(\sin A-\sec A\right)^2$

$\implies \cos ^2 A +\csc ^2 A -2\cos A\csc A +\sin ^2 A -2\sin A\sec A +\sec ^2 A$

$\implies \left(\sin ^2A+\cos ^2A\right)+\csc ^2 A+\sec ^2A-2 \left(\dfrac {\cos A}{\sin A} \right)-2 \left(\dfrac {\sin A}{\cos A}\right)$

$\implies 1+\dfrac 1{\sin ^2A} +\dfrac 1{\cos ^2 A}-2\left( \dfrac {\cos ^2 A+\sin ^2}{\sin A\cos A}\right)$

$\implies 1+\left( \dfrac {\cos ^2 A+\sin ^2}{\sin^2 A\cos^2 A}\right)-2\left( \dfrac {1}{\sin A\cos A}\right)$

$\implies 1+\left( \dfrac {1}{\sin^2 A\cos^2 A}\right)-2\sec A\csc A$

$\implies 1+\sec ^2A\csc ^2 A-2\sec A\csc A$

$\implies \left(1-\sec A\csc A \right)^2$

$\therefore$ LHS

$=$ RHS

Hence Proved.

$\sin \theta +\sin^{2}\theta +\sin^{3}\theta =1$ , then

$\cos^{6}\theta -4\cos^{4}\theta +8 \cos^{2}\theta =4$ .

0%
True
0%
False

Explanation

Given

$\sin \theta+\sin^2 \theta+\sin^3 \theta=1$

$\implies \sin \theta(1+\sin^2 \theta)=1-\sin^2 \theta$

$\implies \sin \theta(2-\cos^2 \theta)=\cos^2 \theta$

Squaring on both sides

$\implies \sin^2 \theta(2-\cos^2 \theta)^2=\cos^4 \theta$

$\implies (1-\cos^2 \theta)(4+\cos^4 \theta-4\cos^2 \theta)=\cos^4 \theta$

$\implies 4+\cos^4 \theta-4\cos^2 \theta-4\cos^2 \theta-\cos^6 \theta+4\cos^4 \theta=\cos^4 \theta$

$\implies \cos^6 \theta-4\cos^4 \theta+8\cos^2 \theta=4$

So the relation is

$\text{True}$

$\sqrt{\dfrac{\sec A-1}{\sec A+1}}+\sqrt{\dfrac{\sec A+1}{\sec A-1}}=2\csc A$

0%
True
0%
False

Explanation

Given

$\sqrt{\dfrac {\sec A-1}{\sec A+1}} +\sqrt{\dfrac {\sec A+1}{\sec A-1}}$

Mulitiplying and Diving first term by

$\sec A-1$

Mulitiplying and Diving second term by

$\sec A+1$

$\sqrt{\dfrac {\left(\sec A-1\right)^2}{\left(\sec A+1\right)\left(\sec A-1\right)}} +\sqrt{\dfrac {\left(\sec A+1\right)^2}{\left(\sec A-1\right)\left(\sec A+1\right)}}$

$\sqrt{\dfrac {\left(\sec A-1\right)^2}{\left(\sec^2 A-1\right)}} +\sqrt{\dfrac {\left(\sec A+1\right)^2}{\left(\sec^2 A-1\right)}}$

$\bigg( (a+b)(a-b)=a^2-b^2\bigg)$

$\sqrt {\left(\dfrac {\sec A-1}{\tan A}\right)^2}+\sqrt {\left(\dfrac {\sec A+1}{\tan A}\right)^2}$

$\bigg( \sec ^2 A-1=\tan ^2 A\bigg)$

$\dfrac {\sec A-1}{\tan A}+\dfrac {\sec A+1}{\tan A}$

$\dfrac {2\sec A}{\tan A}$

$\dfrac {2\left(\dfrac 1{\cos A}\right)}{\left(\dfrac {\sin A}{\cos A}\right)}$

$\dfrac 2{\sin A}$

$2\csc A$

State true or false.

$\dfrac{\tan^{2}\theta }{\sec^{2}\theta}+\dfrac{\cot^{2}\theta }{cosec^{2}\theta}=1$

0%
True
0%
False

Explanation

Given,

$\dfrac{tan^2\theta }{sec^2\theta }+\dfrac{cot^2\theta }{csc^2\theta }$

$=\dfrac{\sin ^2\theta }{\cos ^2\theta } \times \cos ^2\theta + \dfrac{\cos ^2\theta }{\sin ^2\theta } \times \sin ^2\theta$

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

$=1$

State whether the following statement is true or false.

$\dfrac { \sin A }{ \sec A+\tan A-1 } +\dfrac { \cos A }{ \text{cosec}A+\cot A-1 } =1$

0%
True
0%
False

Explanation

$\Rightarrow$

$L.H.S=\dfrac{\sin A}{\sec A+\tan A-1}+\dfrac{cot A}{cosec A+\cot A-1}$

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

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

$=\dfrac{\sin A\times \cos A}{1+\sin A-\cos A}+\dfrac{\sin A\times \cos A}{1+\cos A-\sin A}$

$=\sin A\cos A\left[\dfrac{1}{1+\sin A-\cos A}+\dfrac{1}{1+\cos A-\sin A}\right]$

$=\sin A\cos A\left[\dfrac{1+\cos A-\sin A+1+\sin A -\cos A}{(1+\sin A-\cos A)(1+\cos A-\sin A)}\right]$

$=\sin A\cos A\left[\dfrac{2}{1+\cos A-\sin A+\sin A+\sin A\cos A-\sin^2A-\cos A-\cos^2 A+\cos A\sin A}\right]$

$=\sin A\cos A\left[\dfrac{2}{1-\sin^2A-\cos^2A+2\sin A\cos A}\right]$

$=\sin A\cos A\left[\dfrac{2}{1-(\sin^2 A+\cos^2 A)+2\sin A\cos A}\right]$

$=\sin A\cos A\left[\dfrac{2}{1-1+2\sin A\cos A}\right]$ [ Since,

$\sin^2 \theta+\cos^2\theta=1$ ]

$=\sin A\cos A\times\dfrac{2}{2\sin A\cos A}$

$=1$

$=R.H.S$

$\left( \csc { A-\sin { A } } \right) \left( \sec { A-\cos { A } } \right) =\dfrac { 1 }{ \tan { A+\cot { A } } }$

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True
0%
False

Check whether following statement is true or false.

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

0%
True
0%
False

State whether the following statement is true or false.

$\dfrac { 1 }{ \sec { A } +\tan { A } } -\dfrac { 1 }{ \cos { A } } =\dfrac { 1 }{ \cos { A } } -\dfrac { 1 }{ \sec { A } -\tan { A } }$

0%
True
0%
False

Explanation

Consider

$RHS=\dfrac{ 1}{secA-tanA} - \dfrac{1}{cosA}$

Multiplying by

$secA + tanA$ in the numerator and denominator of

first term,

we get

$\dfrac{secA + tanA}{(secA +tanA)(secA - tanA)} -\dfrac{ 1}{cosA}$

$= secA + tanA - secA........................ (Since,\ sec²A - tan²A = 1)$

$= tanA$

Adding and subtracting

$secA$ , we get

$=secA + tanA - secA$

$= \dfrac{1}{cosA} - (secA - tanA)$

Now multiplying and dividing

$(secA - tanA)$ by

$(secA + tanA)$ ,

we get

$=\dfrac{1}{cosA} - \dfrac{(sec²A - tan²A)}{(secA + tanA)}$

$=\dfrac{ 1}{ cosA} - \dfrac{1}{secA + tanA}$

= L.H.S

$\dfrac{\cos\theta-\sin \theta+1}{\cos\theta+\sin \theta-1}=\text{cosec} \theta+\cot \theta$

0%
True
0%
False

Explanation

Given

LHS

$\dfrac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$

Dividing Numerator and Denominator by

$\sin \theta$

$\implies \dfrac{\cot \theta +\text{cosec} \theta-1}{\cot \theta-\text{cosec} \theta+1}$

$\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} ^2\theta-\cot^2\theta)}{\cot \theta-\text{cosec} \theta+1}$

$\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} \theta-\cot\theta)(\text{cosec} \theta+\cot \theta)}{\cot \theta-\text{cosec} \theta+1}$

$\implies \dfrac{(\cot \theta +\text{cosec} \theta)(\cot \theta-\text{cosec} \theta+1)}{\cot \theta-\text{cosec} \theta+1}$

$\implies \text{cosec} \theta+\cot \theta$

State whether the following statement is true or false.

$\dfrac { \cos A-\sin A+1 }{ \cos A+\sin A-1 }= \text{cosec} A+\cot A$ . (by using the identity

$\text{cosec}^{ 2 }A=1+{ \cot }^{ 2 }A.)\quad$

0%
True
0%
False

Explanation

$LHS=\dfrac{(cosA-sinA)+1}{(cosA+sinA)-1}$

taking conjugate

$=\dfrac{(cosA-sinA)+1}{(cosA+sinA)-1 } \times \dfrac{(cosA+sinA)+1}{(cosA+sinA)+1}$

multiplying we get

$=\dfrac{cos^2A-sin^2A+2cosA+1}{2sinAcosA}$

$=\dfrac{cos^2A-sin^2A+2cosA+sin^2A+cos^2A}{2sinAcosA}$

on solving we get

$=\dfrac{2cos^2A+2cosA}{2sinAcosA}$

taking

$2$ common and divided, we get

$=\dfrac{cosA+cos^2A}{sinAcosA}$

we can also write in the form of

$=\dfrac{cosA}{sinAcosA} +\dfrac{cos^2A}{sinAcosA}$

on solving we get

$cosecA+cotA =R.H.S$

The value of

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

0%
$0$
0%
$\cos^{2}\theta$
0%
$\tan^{2}\theta$
0%
$\dfrac{1}{\sin^{2}\theta}$

Explanation

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

$=\cot^2 \theta+\text{sec}^2 \theta-\text{cosec}^{2}\theta$

$=\text{sec}^2 \theta-(\text{cosec}^2 \theta-\cot^2 \theta)$

$=\text{sec}^2 \theta-1$

$=\tan^2 \theta$

State whether the following statement is true or false.

$\dfrac { \cot A-\cos A }{ \cot A+\cos A } =\dfrac { \text{cosec}A-1 }{ \text{cosec}A+1 }$

0%
True
0%
False

Explanation

$LHS = \dfrac{(cotA - cosA)}{(cotA+cosA)}$

$=\dfrac{(\dfrac{cosA}{sinA} - cosA)}{(\dfrac{cosA}{sinA} + cosA)}$

$=\dfrac{cosA(\dfrac{1}{sinA} - 1)}{cosA(1/sinA + 1)}$

$=\dfrac{ (\dfrac{1}{sinA} -1)}{(\dfrac{1}{sinA}+1)}$

$=\dfrac{(cosecA-1)}{(cosecA+1)}$ ..............[ since

$1/sinA= cosecA$ ]

$=RHS$

State whether the following statement is true or false.

$\dfrac{1}{secA-1}+\dfrac{1}{secA+1}=2cosecAcotA$

0%
True
0%
False

Explanation

$LHS=\dfrac{1}{secA-1}+\dfrac{1}{secA+1}$

$=\dfrac{secA+1+secA-1}{sec^2A-1}$

$=\dfrac{2secA}{tan^2A}=\dfrac{2cos^2A}{cosAsin^2A}=2\dfrac{1}{sinA}\dfrac{cosA}{sinA}=2cosecAcotA=RHS$

State whether the following statement is true or false.

$\sec { ^{ 4 } } A\left( 1-\sec { ^{ 4 } } A \right) -2\tan { ^{ 2 } } A=1$

0%
True
0%
False

Explanation

$sec^4A(1-sin^4A)-2tan^2A$

$=sec^4A-sec^4Asin^4A-2tan^2A$

$=sec^4A-sin^4A(1/cos^4A)-2tan^2A$

$=sec^4A-tan^4A-2tan^2A$

$={(sec^2A)^2-(tan^2A)^2}-2tan^2A$

$=(sec^2A+tan^2A)(sec^2A-tan^2A)-2tan^2A$

$=sec^2A+tan^2A-2tan^2A \quad [∵ sec^2A-tan^2A=1]$

$=sec^2A-tan^2A$

$=1$

State whether the following statement is true or false.

$\cos A+{ \cos }^{ 2 }A=1,$ then

${ \sin }^{ 2 }A+{ \sin }^{ 4 }A=1.$

0%
True
0%
False

Explanation

Given:

$cosA+cos^2A=1$

$\implies cosA=1-cos^2A=sin^2A$ ..........[1]

Now,

$sin^2A+sin^4A=cosA+cos^2A=1$ .....................[from

$1$ ]

${ cosec }\theta -cot\theta =p$ , then

$cosec\theta +cot\theta =$

0%
1/p
0%
-1/p
0%
-p
0%
${ p }^{ 2 }$

Explanation

$\csc{\theta}-\cot{\theta}=p$

$\Rightarrow \left(\csc{\theta}-\cot{\theta}\right)\left(\csc{\theta}+\cot{\theta}\right)=p\left(\csc{\theta}+\cot{\theta}\right)$

$\Rightarrow {\csc}^{2}{\theta}-{\cot}^{2}{\theta}=p\left(\csc{\theta}+\cot{\theta}\right)$

$\Rightarrow 1=p\left(\csc{\theta}+\cot{\theta}\right)$

$\therefore \csc{\theta}+\cot{\theta}=\dfrac{1}{p}$

State whether the following statement is true or false.

$\dfrac { 1+\cos\theta -\sin^{ 2 }\theta }{ \sin { \theta \left( 1+\cos { \theta } \right) } } =\cot { \theta }$

0%
True
0%
False

Explanation

$\dfrac { 1+\cos\theta -\sin^{ 2 }\theta }{ \sin { \theta \left( 1+\cos { \theta } \right) } } =\cot { \theta }$

$\Rightarrow$

$L.H.S=\dfrac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}$

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

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

$=\dfrac{\cos\theta+\cos^2\theta}{\sin\theta(1+\cos\theta}$

$=\dfrac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)}$

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

$=\cot\theta$

$=R.H.S$

$\dfrac{{\tan\theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta .\text{cosec}\theta$

0%
True
0%
False

Explanation

$\dfrac{\tan \theta}{1-\cot \theta}+\dfrac{\cot \theta}{1-\tan \theta}=\dfrac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\dfrac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$

$=\dfrac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)}-\dfrac{\cos^2 \theta}{\sin \theta(\sin \theta-\cos \theta)}$

$=\dfrac{1}{\sin \theta-\cos \theta}\bigg(\dfrac{\sin^2 \theta}{\cos \theta}-\dfrac{\cos^2 \theta}{\sin \theta}\bigg)$

$=\dfrac{\sin^3 \theta-\cos^3 \theta}{(\sin \theta-\cos \theta)\sin \theta\cos \theta}$

$=\dfrac{(\sin \theta-\cos \theta)(\sin^2 \theta+\cos^2 \theta+\sin \theta\cos \theta)}{(\sin \theta-\cos \theta)\sin \theta\cos \theta}$

$(\because a^3-b^3=(a-b)(a^2+b^2+a b))$

$=\dfrac{1+\sin \theta\cos \theta}{\sin \theta\cos \theta}=1+\text{sec}\theta\text{cosec}\theta$

So the relation is

$\text{True}$

The value of

$\sin \theta$ is not equal to

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

The maximum value of

$2 \sin x-4 \cos x$ is

0%
$-\sqrt { 12 }$
0%
$2$
0%
$\sqrt { 20 }$
0%
None of these

Explanation

$\textbf{Step – 1: Convert }\mathbf{2\sin x - 4\cos x}\textbf{ into single sine function.}$

$\text{Let }f(x) = 2\sin x - 4\cos x$

$\text{Here, }a = 2\ \text{and }b = -\ 4$

$\therefore$

$\sqrt {{a^2} + {b^2}} = \sqrt {{2^2} + {{( - 4)}^2}} \\$

$= \sqrt {4 + 16} \\$

$= \sqrt {20} \\$

${\text{Divide }}f(x){\text{ by }}\sqrt {{a^2} + {b^2}}$

$\dfrac{{f(x)}}{{\sqrt {20} }} = \dfrac{2}{{\sqrt {20} }}\sin x - \dfrac{4}{{\sqrt {20} }}\cos x$

${\text{By using triangular method,}}$

$\text{Let,}$

$\sin \theta = \dfrac{4}{{\sqrt {20} }}{\text{ and }}\cos \theta = \dfrac{2}{{\sqrt {20} }}$

$\therefore \dfrac{{f(x)}}{{\sqrt {20} }} = \cos \theta \sin x - \sin \theta \cos x$

$\Rightarrow \dfrac{{f(x)}}{{\sqrt {20} }} = \sin (x - \theta )$

$\mathbf{[\because \sin (a - b) = \sin a\cos b - \cos a\sin b]}$

$\Rightarrow f(x) = \sqrt {20} \sin (x - \theta )$

$\textbf{Step – 2: Sine function lies in the range}$

$\mathbf{[- 1,1]}$

$-1\leq \sin (x - \theta ) \leq 1$

$\Rightarrow - \sqrt {20} \leq \sqrt{20}\sin (x - \theta ) \leq \sqrt {20}$

$\textbf{ Hence, the maximum value of} \ \mathbf{2 \sin x - 4\cos x}{\textbf{ is equal to }}\mathbf{\sqrt {20}.}$

The value of

$\sin\theta$ is not equal to

0%
$\pm\dfrac {1}{2}$
0%
$\dfrac {1}{\sqrt {2}}$
0%
$\pm\dfrac {1}{3}$
0%
$2$

$x_{1}$ and

$x_{2}$ are two distinct roots of the equation

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

$\tan\dfrac {x_{1}+x_{2}}{2}$ is equal to

0%
$a/b$
0%
$b/a$
0%
$c/a$
0%
$a/c$

Explanation

$a \cos x + b \sin x =c$

$\dfrac{1-\tan^2\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}} +b \dfrac{2\tan \dfrac{x}{2}}{1+\tan^2 \dfrac{x}{2}}=c$

$(c+a) \tan^2 \dfrac{x}{2} -2b \tan \dfrac{x}{2} +(c-a) =0$

Since,

$x_1$ and

$x_2$ are roots,

$\tan \dfrac{x_1}{2}+\tan \dfrac{x_2}{2} =\dfrac{2b}{c+a}$

and

$\tan \dfrac{x_1}{2}.\tan \dfrac{x_2}{2} =\dfrac{c-a}{c+a}$

Hence,

$\tan (\dfrac{x_1+x_2}{2}) = \dfrac{\tan \dfrac{x_1}{2} +\tan \dfrac{x_2}{2}}{1-\tan \dfrac{x_1}{2} \tan \dfrac{x_2}{2}}$

$=\dfrac{\dfrac{2b}{c+a}}{1-\dfrac{c-a}{c+a}}=\dfrac{b}{a}$

Hence, option (b) is correct.

$\dfrac{1-\sin A}{1+\sin A}=(\sec A -\tan A)^2$

0%
True
0%
False

Explanation

L.H.S. :

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

$=\dfrac{(1-\sin A)^2}{(1-\sin A)(1+\sin A)}$

$=\dfrac{(1-\sin A)^2}{1-\sin^2A}$

$=\dfrac{(1-\sin A)^2}{\cos ^2A}$

$=[\dfrac{1-\sin A}{\cos A}]^2$

$=(\sec A -\tan A)^2$ = R.H.S.

$sin{(2cos^{-1}(\dfrac{1}{\sqrt{5}})} + cos{(2tan^{-1}(\dfrac{1}{3}))}=\dfrac{p}{q}$ , where p & q are relatively prime then digit at units place of

${(p-q)}^{2k+1}, k\epsilon{N}$ , can be ________.

0%
1
0%
3
0%
7
0%
9

Given

$A={ sin }^{ 2 }\theta +{ cos }^{ 4 }\theta$ , then for all real

$\theta$

0%
$1\le A\le 2$
0%
$\cfrac { 3 }{ 4 } \le A\le 1$
0%
$\cfrac { 13 }{ 16 } \le A\le 1$
0%
$\cfrac { 3 }{ 4 } \le A\le \cfrac { 13 }{ 16 }$

The values of x in

$\left(0, \dfrac{\pi}{2}\right)$ satisfying the equation

$\sin x\cos x=\dfrac{1}{4}$ are ________.

0%
$\dfrac{\pi}{6}, \dfrac{\pi}{12}$
0%
$\dfrac{\pi}{12}, \dfrac{5\pi}{12}$
0%
$\dfrac{\pi}{8}, \dfrac{3\pi}{8}$
0%
$\dfrac{\pi}{8}, \dfrac{\pi}{4}$

Explanation

We know that,

$\displaystyle \text{sin}(2x) = 2\text{ sin }x\text{ cos }x$ .

So, from

$\displaystyle \text{sin }x \text{ cos }x = \frac{1}{4}$ , we get that

$\text{ sin }(2x) = \frac{1}{2}$ .

So, we have to find the solutions of the equation

$\displaystyle\text{sin} (2x) = \frac{1}{2}$ , where

$\displaystyle x \in (0, \frac{\pi}{2})$ .

From this data, we get that

$\displaystyle 2x = \frac{\pi}{6}$ or

$\dfrac{5\pi}{6}$ .

That is,

$x = \dfrac{\pi}{12}$ or

$\dfrac{5\pi}{12}$ .

What is

$\cot(\dfrac{A}{2})-\tan(\dfrac{A}{2})$ equal to ?

0%
$\tan A$
0%
$\cot A$
0%
$2\tan A$
0%
$2\cot A$

Explanation

$\cot \left(\dfrac{A}{2}\right)-\tan\left(\dfrac{A}{2}\right)$

$\Rightarrow$

$\dfrac{\cos \dfrac{A}{2}}{\sin\dfrac{A}{2}}-\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$

$\Rightarrow$

$\dfrac{\cos^2\dfrac{A}{2}-\sin^2\dfrac{A}{2}}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$

$\Rightarrow$

$\dfrac{\cos 2\left(\dfrac{A}{2}\right)}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$

$\Rightarrow$

$\dfrac{\cos A}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$

$\Rightarrow$

$\dfrac{2\cos A}{2\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$

$\Rightarrow$

$\dfrac{2\cos A}{\sin 2\left(\dfrac{A}{2}\right)}$

$\Rightarrow$

$\dfrac{2\cos A}{\sin A}$

$\Rightarrow$

$2\cot A$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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