MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 9

The number of solution of the equation

$\sin 5x \cos 3x=\sin 6x \cos 2x$ in the interval

$[0,\pi]$ is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\sin 5x\cos 3x=\sin 6x \cos 2x$

$\sin \theta\cos \phi =\cfrac{1}{2}[\sin(\theta+\phi)+\sin(\theta-\phi)]$

$\cfrac{1}{2}[\sin 8x+\sin 2x]=\cfrac{1}{2}[\sin 8x+\sin 4x]$

$\cfrac{1}{2}\sin 2x=\cfrac{1}{2}\sin 4x$

$\sin 4x-\sin 2x=0$

$2\sin 2x\cos 2x-\sin 2x=0$

$\sin 2x[2\cos 2x-1]=0$

$\sin 2x=0$

$2\cos 2x-1=0$

$2x=0$

$\cos 2x=\cfrac{1}{2}$

$x=0$

$2x=\cfrac{\pi}{3}$

$x=\cfrac{\pi}{6}$

$two$ solutions.

The general solution of the equation

$\tan^{2}\theta+2\sqrt {3} \tan \theta=1$

0%
$\theta=\dfrac {\pi}{2}$
0%
$\left(n+\dfrac {1}{2}\right)\pi$
0%
$(6n+1)\dfrac {\pi}{12}$
0%
$\dfrac {n \pi}{12}$

Explanation

$\tan^2\theta +2\sqrt{3}\tan \theta =1$

$\tan^2\theta +2\sqrt{3}\tan\theta -1=0$

Add

$+4$ on both sides

$\tan^2\theta +2\sqrt{3}\tan \theta +3=4$

$(\tan\theta +\sqrt{3})^2=4$

$\tan\theta+\sqrt{3}=\pm 2$

$\tan\theta =2-\sqrt{3}$ ,

$\tan\theta =-(2+\sqrt{3})$

$\tan\left(\dfrac{\pi}{12}\right)$

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

$x=n\pi\pm \dfrac{\pi}{12}$ for

$n=0, 1, 2$ …..

Find

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

0%
$\cos (A+B)$
0%
$\tan (A-B)$
0%
$\cot (A+B)$
0%
$\tan (A+B)$

Explanation

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

$=\cfrac { 1-\cos 2 A-1+\cos 2 B }{ \sin 2 A-\sin 2 B }$

$=\cfrac {- (\cos 2 A-\cos 2 B) }{ \sin 2 B-\sin 2 A }$

$=\cfrac { -\{ -2\sin(\cfrac { 2A+2B }{ 2 } )\sin(\cfrac { 2A-2B }{ 2 } )\} }{ \sin 2B-\sin 2A }$

$==\cfrac { -\{ -2\sin(\cfrac { 2A+2B }{ 2 } )\sin(\cfrac { 2A-2B }{ 2 } )\} }{ 2\sin(\cfrac { 2A-2B }{ 2 } )\cos(\cfrac { 2A+2B }{ 2 } ) }$

$=\cfrac { \sin(A+B) }{ \cos(A+B) }$

$=\tan(A+B)$

$$2\sin 2\beta + 4\cos (\alpha + \beta)\sin \alpha \sin \beta + \cos 2(\alpha + \beta)$$

0%
$\sin 2\alpha$
0%
$\cos \, 2 \beta$
0%
$\cos \, 2 \alpha$
0%
$\sin \, 2 \beta$

Explanation

$2 \sin^{2}\beta +4 \cos (\alpha +\beta ) \sin \alpha \sin\beta +\cos 2 (\alpha +\beta )$

$= 2 \sin^{2}\beta +4 (\cos \alpha \cos \beta -\sin\alpha \sin\beta ) \sin \alpha \sin\beta +\cos 2 \alpha \cos 2 \beta -\sin 2\alpha \sin 2\beta$

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

$=2\sin^{2}\beta +\sin^{2}\alpha \sin^{2}\beta -4\sin^{2}\alpha \sin^{2}\beta +\cos 2\alpha \cos^{2}\alpha -\sin^{2}\alpha \sin^{2}\beta$

$=(-\cos^{2}\beta )-(2 \sin^{2}\alpha )(2 \sin^{2}\beta )+\cos^{2}\alpha \cos^{2}\beta$

$=(1-\cos 2\beta )-(1-\cos 2\alpha )(1-\cos 2\beta )+\cos 2\alpha \cos 2\beta$

$=\cos 2\alpha$

$\tan\theta=\dfrac{a}{b}$ then

$\dfrac{a\sin\theta-b\cos\theta}{a\sin\theta +b\cos\theta}=$

0%
$\dfrac{a^2b^2}{a^2+b^2}$
0%
$\dfrac{ab}{a^2+b^2}$
0%
$\dfrac{a^2b^2}{a+b}$
0%
None of these

Explanation

$\tan\theta =\cfrac { a }{ b } ,$ then

$\cfrac { a \sin\theta -b \cos\theta }{ a \sin\theta +b \cos }=\cfrac { \cfrac { a \sin\theta }{ b \cos\theta } -1 }{ \cfrac { a \sin\theta }{ b \cos\theta } +1 }$

$=\cfrac { \cfrac { a }{ b } \tan\theta -1 }{ \cfrac { a }{ b } \tan\theta +1 }$

$=\cfrac { \cfrac { { a }^{ 2 } }{ { b }^{ 2 } } -1 }{ \cfrac { { a }^{ 2 } }{ { b }^{ 2 } } +1 }$

$=\cfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } }$

The expression

$3\left[ \sin ^{ 4 }{ \left\{ \dfrac { 3 }{ 2 } \pi -\alpha \right\} } +\sin ^{ 4 }{ \left( 3\pi +\alpha \right) } \right] -2\left[ \sin ^{ 6 }{ \left\{ \dfrac { 1 }{ 2 } \pi +\alpha \right\} } +\sin ^{ 6 }{ \left( 5\pi -\alpha \right) } \right]$ is equal to

0%
$0$
0%
$1$
0%
$3$
0%
$\sin { 4\alpha } +\cos { 6\alpha }$

Explanation

$3[cos^{4}\alpha+sin^{4}\alpha]-2[cos^{6}\alpha+sin^{6}\alpha]$

$= 3[(cos^{2}\alpha)^{2}+(sin^{2}\alpha^{2})]-2[(cos^{2}\alpha)^{3}+(sin^{2}\alpha)^{3}]$

$= 3[1-\dfrac{1}{2}sin^{2}2\alpha] -2[1-\dfrac{3}{4}sin^{2}2\alpha]$

$=3-2$

$=1$

$\sin { x } +\cos { x } =a,$ then

$\left| \sin { x } -\cos { x } \right|$ equals

0%
$\sqrt { 2-{ a }^{ 2 } }$
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$\sqrt { 2+{ a }^{ 2 } }$
0%
$\sqrt { { a }^{ 2 }-2 }$
0%
$\sqrt { { a }^{ 2 }-4 }$

Explanation

$\left|\sin{x}-\cos{x}\right|=\sqrt{{\left(\sin{x}-\cos{x}\right)}^{2}}$

$=\sqrt{{\sin}^{2}{x}+{\cos}^{2}{x}-2{\sin}{x}{\cos}{x}}$

$=\sqrt{1-2\sin{x}\cos{x}}$

Given

$\sin{x}+\cos{x}=a$

squaring both sides we get

${\left(\sin{x}+\cos{x}\right)}^{2}={a}^{2}$

$\Rightarrow {\sin}^{2}{x}+{\cos}^{2}{x}+2\sin{x}\cos{x}={a}^{2}$

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

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

$\therefore \left|\sin{x}-\cos{x}\right|=\sqrt{1-\left({a}^{2}-1\right)}=\sqrt{2-{a}^{2}}$

$\tan x+\tan 2x+\tan 8x=\tan x.\tan 2x.\tan 8x$ then the general solution of

$x=$

0%
$n \pi,\forall n \epsilon Z$
0%
$n \pi,\dfrac {\pi}{4},\forall n \epsilon Z$
0%
$\dfrac {n \pi}{11},\forall n \epsilon Z$
0%
$n \pi,\dfrac {\pi}{3},\forall n \epsilon Z$

Explanation

Given

$\tan x+\tan 2x+\tan 8x =\tan x\tan 2x\tan 8x$

Consider

$\tan (x+2x+8x) =\dfrac{\tan x+\tan 2x+\tan 8x - \tan x\tan 2x\tan 8x}{1-\tan x\tan 2x-\tan x\tan 8x -\tan 2x\tan 8x}$

$\implies\tan (x+2x+8x) =\dfrac{\tan x\tan 2x\tan 8x- \tan x\tan 2x\tan 8x}{1-\tan x\tan 2x-\tan x\tan 8x -\tan 2x\tan 8x}$

$\implies \tan (x+2x+8x) =0$

$\implies\tan 11x =0$

$\implies 11x=n\pi$

$\implies x=\dfrac {n\pi }{11}$

The value of

$\theta$ satisfying

$\sin{7\theta}=\sin{4\theta}-\sin{\theta}$ and

$\theta<0<\pi/2$ are-

0%
$\dfrac{\pi}{9}, \dfrac{\pi}{4}$
0%
$\dfrac{\pi}{3}, \dfrac{\pi}{9}$
0%
$\dfrac{\pi}{6}, \dfrac{\pi}{9}$
0%
$\dfrac{\pi}{3}, \dfrac{\pi}{4}$

Evaluate

$\tan \left[ {\dfrac{\pi }{4}\, + \,\dfrac{1}{2}\,{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, + \,\tan \left[ {\dfrac{\pi }{4}\, - \,\dfrac{1}{2}{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, =$

0%
$\dfrac{{2a}}{b}$
0%
$\dfrac{{2b}}{a}$
0%
$\dfrac{a}{b}$
0%
$\dfrac{b}{a}$

Explanation

$\textbf{Given expression}$

$= \tan \left[ {\dfrac{\pi }{4}\, + \,\dfrac{1}{2}\,{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\, + \,\tan \left[ {\dfrac{\pi }{4}\, - \,\dfrac{1}{2}{{\cos }^{ - 1}}\,\dfrac{a}{b}} \right]\,$

$\textbf{Step : 1 : Do relevant substitution to simplify given equation}$

Let

$x=\cos ^{ -1 }{ \cfrac { a }{ b } } \Rightarrow \cos{x} = \dfrac{a}{b}$

$........(1)$

$= \tan\left[\cfrac { \theta }{ 4 } +\cfrac { x }{ 2 } \right]+\tan\left[\cfrac { \theta }{ 4 } -\cfrac { x }{ 2 } \right]$

$\textbf{Step : 2 Apply relevant identity }$ :

$=\cfrac { \tan \cfrac {\theta }{ 4 } +\tan\cfrac { x }{ 2 } }{ 1-\tan\cfrac { \theta }{ 4 } \times \tan \cfrac {x }{ 2 } } +\cfrac { \tan \cfrac {\theta }{ 4 } -\tan\cfrac { x }{ 2 } }{ 1+\tan\cfrac { \theta }{ 4 } \times \tan \cfrac {x }{ 2 } } ,$

$\left[\tan(A \pm B) = \dfrac{tan A \pm tan B}{1\mp tanA tanB}\right]$

$=\cfrac { 1+ \tan\cfrac { x }{ 2 } }{ 1- \tan \cfrac { x }{ 2 } } +\cfrac { 1- \tan\cfrac { x }{ 2 } }{ 1+ \tan \cfrac { x }{ 2 } }$

$=\cfrac { { \left(1+ \tan \cfrac{x }{ 2 } \right) }^{ 2 }+{ \left(1- \tan\cfrac{ x }{ 2 } \right) }^{ 2 } }{ 1-{ \tan }^{ 2 }\cfrac { x }{ 2 } }$

$=\cfrac { 2\left(1+{ \tan }^{ 2 }\cfrac { x }{ 2 } \right) }{ \left(1- { \tan }^{ 2 }\cfrac {x }{ 2 } \right) }$

$[\because(a\pm b)^2 = a^2 + b^2 \pm 2ab]$

$=\cfrac { 2 }{ \cos x }$

$[\because \cos{x} =\cfrac { \left(1-{ \tan }^{ 2 }\cfrac { x }{ 2 } \right) }{ \left(1+ { \tan }^{ 2 }\cfrac {x }{ 2 } \right) }$

Put the value from eq.(1),

$=\cfrac { 2 }{ \cfrac { a }{ b } }$

$=\cfrac { 2b }{ a }$

$A+B+C=\pi$ then

$\sin^3A \cos(B-C)+ \sin^3\cos(C-A) + \sin^3C \cos (A-B)$ is equal to

0%
$3\sin A \sin B \sin C.$
0%
$6\sin A \sin B \sin C.$
0%
$4\sin A \sin B \sin C.$
0%
$0$

$(\cos\theta +\cos 2\theta)^3=\cos^3\theta +\cos^32\theta$ , then the least positive value of

$\theta$ is equal to?

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

Explanation

We are given

$(\cos\theta+\cos2\theta)^{3}=\cos^{3}\theta+\cos^{3}2\theta$

$\cos^{3}\theta+\cos^{3}2\theta+3\cos^{2}\theta\cos^{2}\theta+3\cos\theta\cos^{2}2\theta$

$=\cos^{3}\theta+\cos^{3}2\theta$

$(as\ (a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3b^{2}a)$

So,

$3\cos^{2}\theta\cos 2\theta+3\cos\theta\cos^{2}\theta=0$

$3\cos\theta\cos2\theta(\cos\theta+\cos2\theta)=0$

So,

$\cos\theta.\cos 2\theta.(\cos\theta+2\cos^{2}\theta-1)=0$

$\cos\theta.\cos 2\theta.(2\cos^{2}\theta+\cos\theta-1)=0$

We get

$\cos\theta=0, \cos 2\theta=0$ &

$2\cos^{2}\theta+\cos\theta-1=0$

$\cos\theta=0$ or

$\cos 2\theta=0$ or

$2\cos^{2}\theta+\cos\theta-1=0$

$\theta=\cos^{-1}0$ or

$2\theta=\cos^{-1}0$ or

$2\cos^{2}\theta+2\cos\theta-\cos\theta-1=0$

$2\cos\theta(\cos\theta+1)-1(\cos\theta+1)=0$

$(2\cos\theta-1)(\cos\theta+1)=0$

$2\cos\theta-1=0$ or

$\cos\theta+1=0$

$\cos\theta=\dfrac{1}{2}$ or

$\cos\theta+1=0$

$\cos\theta=\dfrac{1}{2}$ or

$\cos\theta=-1$

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

$\theta=\cos^{-1}(-1)$

By principal values we get

$\theta$ as

$\dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \pi$

$\Rightarrow$ then least positive value of

$\theta$ is equal to

$\left(\dfrac{\pi}{4}\right)$

General solution of

$\sin ^{ 3 }{ x } +\cos ^{ 3 }{ x } +\frac { 3 }{ 2 } \sin { 2x } =1$

0%
$x=n\pi$ where $n$ is even integer
0%
$x=n\pi +\dfrac{\pi}{2}$ where $n$ is odd integer
0%
$x=2n\pi$ where $n$ is odd integer
0%
$x=n\pi -\dfrac{\pi}{2}$ where $n$ is even integer

Explanation

$\sin ^3x+\cos ^3x+\dfrac{3}{2}\sin 2x=1$

$\Rightarrow (\sin x+\cos x)(\sin ^2x+\cos ^2x-\sin x\cos x)+\dfrac{3}{2}\times 2\sin x \cos x=1$

$\Rightarrow (\sin x+\cos x)(\sin ^2x+\cos ^2x-\sin x\cos x)+ 3\sin x \cos x=1$ ....

$(1)$

Take

$\Rightarrow \sin x +\cos x =t$

Squaring on both sides

$\Rightarrow (\sin x +\cos x )^2=t^2$

$\Rightarrow \sin^2+\cos^2+2\sin x\cos x=t^2$

$\Rightarrow 1+2\sin x\cos x=t^2$

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

Substituting these values in equation

$(1)$ we get

$\Rightarrow t(1+\dfrac{t^2}{2}-\dfrac{1}{2})+3(\dfrac{t^2}{2}-\dfrac{1}{2})$

$\Rightarrow t^3+3t^2+t-5=0$

$\Rightarrow (t-1)(t^2+4t+5)=0$

Therefore,

$\Rightarrow \sin x+\cos x=1$

Squaring on both sides

$\Rightarrow (\sin x+\cos x)^2=1$

$\Rightarrow 1+\sin 2x=1$

$\Rightarrow 2\sin x \cos x=0$

$\Rightarrow \sin x=0; \cos x=0$

$\Rightarrow x=2n\pi$

The general solution of the equation

${\sin ^{50}}x - {\cos ^{50}}x = 1$

0%
$2n\pi + \frac{\pi }{2}$
0%
$2n\pi + \frac{\pi }{3}$
0%
$n\pi + \frac{\pi }{2}$
0%
$n\pi + \frac{\pi }{3}$

Explanation

Consider the given equation.

${{\sin }^{50}}x-{{\cos }^{50}}x=1$

${{\sin }^{50}}x=1+{{\cos }^{50}}x$ …….. (1)

Here, ${{\cos }^{50}}x$ is always positive. So, the maximum value of ${{\sin }^{50}}x$ is $1$ .

So,

${{\cos }^{50}}x$ must be zero.

Hence, ${{\cos }^{50}}x=0$

$x=\left( 2n+1 \right)\dfrac{\pi }{2},$ where $n\in I$

$x=n\pi +\dfrac{\pi}{2}$

Hence, this is the answer.

$sin 2 \theta = cos 3 \theta \, and \, \theta$ is acute then

$sin 5 \theta$ = ...

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

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

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

Explanation

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)

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

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

General solution of equation

$\cot \theta + \text{cosec}\theta = \sqrt 3$ is

0%
$2n\pi + \frac{\pi }{4}$
0%
$\left( {2n - 1} \right)\pi$
0%
$2n\pi + \frac{\pi }{3}$
0%
$2n\pi + \frac{\pi }{6}$

Explanation

Given expression:

$\cot\theta+\text{cosec}\ \theta=\sqrt3$

$\dfrac{\cos \theta}{\sin \ \theta}+\dfrac{1}{\sin \ \theta}=\sqrt3$

Taking square both side,

$\cos ^2\theta+1+2\cos \theta=3-3\cos ^2\theta$ [\sin ce

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

$4\cos ^2\theta+2\cos \theta-2=0$

$4\cos ^2\theta+4\cos \theta-2\cos \theta-2=0$

We get

$(\cos \theta+1)(4\cos \theta -2)=0$

When,

$(\cos \theta+1)=0$

$\cos \theta =-1$

$\cos \pi=\cos (2n+1)\pi=-1$

Hence

$\theta =(2n+1)\pi.$

When,

$(4\cos \theta-2)=0$

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

$\cos \dfrac{\pi}{3}=\cos (2n\pi\pm \dfrac{\pi}{3})=\dfrac{1}{2}$

Hence,

$\theta=(2n\pi\pm \dfrac{\pi}{3})$

$a$ ,

$b$ ,

$c$ ,

$t$ 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

$\tan a+\tan b+\tan c+\tan t=$

0%
$1/3$
0%
$8/3$
0%
$-8/3$
0%
$0$

Explanation

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.

$\alpha$ is a root of

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

0%
$\dfrac{24}{25}$
0%
$-\dfrac{24}{25}$
0%
$\dfrac{13}{18}$
0%
$-\dfrac{13}{18}$

$\dfrac{x}{y} = \dfrac{{\cos \,A}}{{\cos \,B}}$ then

$\dfrac{{x\tan A + y\tan B}}{{x + y}} =$

0%
$\cot\dfrac{A+B}{2}$
0%
$\cot\dfrac{A-B}{2}$
0%
$\tan\dfrac{A-B}{2}$
0%
$\tan\dfrac{A+B}{2}$

Explanation

We have,

$\dfrac{x}{y}=\dfrac{\cos A}{\cos B}$

$x=\dfrac{y\cos A}{\cos B}$

Since,

$=\dfrac{x\tan A+y\tan B}{x+y}$ …….. (1)

On putting the value of $x$ in equation (1), we get

$=\dfrac{\dfrac{y\cos A}{\cos B}\times \dfrac{\sin A}{\cos A}+\dfrac{y\sin B}{\cos B}}{\dfrac{y\cos A}{\cos B}+y}$

$=\dfrac{\dfrac{ysinA}{\cos B}+\dfrac{y\sin B}{\cos B}}{\dfrac{y\cos A+y\cos B}{\cos B}}$

$=\dfrac{\dfrac{y\left( sinA+\sin B \right)}{\cos B}}{\dfrac{y\left( \cos A+\cos B \right)}{\cos B}}$

$=\dfrac{\left( sinA+\sin B \right)}{\left( \cos A+\cos B \right)}$

We know that

$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)$

$\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)$

Therefore,

$=\dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)}{2\cos \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)}$

$=\tan \left( \dfrac{A+B}{2} \right)$

Hence, the value is $\tan \left( \dfrac{A+B}{2} \right)$ .

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

0%
$\dfrac{{n\pi }}{4}$
0%
$\dfrac{{n\pi }}{7}$
0%
$\dfrac{{n\pi }}{{12}}$
0%
$n\pi$

Explanation

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

$\pi.$

$=>\theta+4\theta+7\theta=\pi$

$=>\theta=\dfrac{n\pi}{12}$

the following is

$\frac{cos(A-B)}{cos(A+B)}= \frac{cosA. cotB +2}{costA. costB-1}$

0%
True
0%
False

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

0%
$-1$
0%
$0$
0%
$1$
0%
None of these

Explanation

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

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

$\tan \theta + \sec \theta = \sqrt{3}$ , then the principal value of

$\left(\theta + \dfrac{\pi}{6}\right)$ is

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

The value of

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

0%
$- \dfrac{{\sqrt {3 - \sqrt 5 } }}{2}$
0%
$- \dfrac{{\sqrt {5 - \sqrt 5 } }}{2}$
0%
$- \dfrac{{\sqrt 5 - 1}}{{2\sqrt 2 }}$
0%
$\dfrac{{\sqrt {3 - \sqrt 5 } }}{2}$

Explanation

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

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

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

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

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

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

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

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

Hence, this is the answer

option $(B)$ is correct.

$\sin\theta=\dfrac{1}{2}\implies$

$\theta =2n\pi +\dfrac \pi 6$

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True
0%
False

Explanation

$\sin \theta =\dfrac 12\\\sin \theta =\sin \dfrac \pi 6\\ \theta =2n\pi +(-1)^n\dfrac \pi 6$

The value of

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

0%
$1$
0%
$\dfrac{1}{2}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
${\sin ^2}{73^ \circ } + {\sin ^2}{47^ \circ }$

Explanation

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.

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

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

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

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True
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False

Explanation

$x^2+y^2=(asin\theta+ccos\theta)^2+(acos\theta-csin\theta)^2\\=a^2sin^2\theta+c^2cos^2\theta+2ac sin\theta cos\theta +a^2cos^2\theta+c^2sin^2\theta-2acsin\theta cos\theta\\=a^2(sin^2\theta+cos^2\theta)+c^2(sin^2\theta+cos^2\theta)\\=a^2+c^2$

The equation

$\sin x+2\sin 2x=3+\sin 3x$ has no solution in

$(0, \pi)$ .

0%
True
0%
False

Explanation

$\sin x + 2\sin 2x -\sin3x = 3$

$-2\sin x + 2\sin2x + 4\sin^3x = 3$

$-2\sin x(1-2\sin^2x)+2\sin2x= 3$

$-2\sin x \cos 2x + 2\sin2x = 3$

$2\sin x(2\cos x - \cos2x) = 3$

The maximum value attained by

$(2\cos x - 2\cos2x)$ is

$\cfrac{3}{2}$ at

$x = \cfrac{\pi}{3}$

$x = \cfrac{\pi}{3}$ is

$\cfrac{3\sqrt3}{2} < 3$ . Hence, it is not possible.

Maximum value of

$sin\alpha+sin\beta+2$ whwre

$\alpha+\beta=120^o$ &

$\alpha,\beta{\,}\in \left(0,\dfrac{\pi}{2}\right)$ is-

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$cot\dfrac{\pi}{12}$
0%
$cot\dfrac{\pi}{8}$
0%
$tan\dfrac{\pi}{12}$
0%
$tan\dfrac{\pi}{8}$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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