If $$\cos { \left( A-B \right) } =3/5$$ and $$\tan { A } \tan { B } =2$$, then

$$\cos { \left( A+B \right) } =-1/5$$

If $$\sin \theta = n \sin(\theta + 2 \alpha) $$ then $$\tan (\theta + \alpha)$$ =

$$\dfrac{1 + n}{1 - n} \tan \, \alpha$$

$$\dfrac{1 - n}{1 + n} \tan \, \alpha$$

MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 8

$\sin { \alpha } =12/13\left( 0<\alpha <\pi /2 \right)$ and

$\cos { \beta } =-\dfrac { 3 }{ 5 } \left( \pi <\beta <\dfrac { 3 }{ 2 } \pi \right)$ , the value of

$\sin { \left( \alpha +\beta \right) }$ is

0%
$\dfrac{-56}{65}$
0%
$\dfrac{16}{65}$
0%
$\dfrac{56}{65}$
0%
$\dfrac{-16}{65}$

Explanation

Ans.

$(a)$
Since

$0<\alpha <\pi$ we have

$\sin { \alpha }$ =

$\dfrac{12}{13}$

$\Rightarrow \cos { \alpha } =\surd \left[ 1-{ \left( 12/13 \right) }^{ 2 } \right]$ =

$\dfrac{5}{13}$
And since

$\pi <\beta <3\pi /2$ , we have

$\cos { \beta } =-3/5\Rightarrow \sin { \beta } =-4/5$
Hence

$\sin { \left( \alpha +\beta \right) } =\sin { \alpha } \cos { \beta } +\cos { \alpha } \sin { \beta }$

$=\dfrac { 12 }{ 13 } \left( -\dfrac { 3 }{ 5 } \right) +\left( \dfrac { 5 }{ 13 } \right) \left( -\dfrac { 4 }{ 5 } \right) =-\dfrac { 56 }{ 65 }$

The maximum value of the expression

$\dfrac { 1 }{ \sin ^{ 2 }{ \theta } +3\sin { \theta } \cos { \theta } +5\cos ^{ 2 }{ \theta } }$ is

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

Explanation

Ans. Ans.

$(a)$ .

$f\left( \theta \right) =\dfrac { 1 }{ \sin ^{ 2 }{ \theta } +3\sin \theta \cos \theta +5\cos ^{ 2 }{ \theta } }$

$=\dfrac { 1 }{ \dfrac { 1-\cos { 2\theta } }{ 2 } +\dfrac { 3 }{ 2 } \sin { 2\theta } +\dfrac { 5\left( 1+\cos { 2\theta } \right) }{ 2 } }$

$=\dfrac{ 2 }{ 6+3\sin 2\theta+4 \cos 2\theta }$
Hence,

$\left[ f\left( \theta \right) \right] _{ maximum }=\dfrac { 2 }{ 6-5 } =2$

$\cos { \left( A-B \right) } =3/5$ and

$\tan { A } \tan { B } =2$ , then

0%
$\cos { A } \cos { B } =1/5$
0%
$\sin { A } \sin { B } =-2/5$
0%
$\cos { \left( A+B \right) } =-1/5$
0%
$\sin { A } \cos { B } =4/5$

Explanation

Ans.

$(a)$ and

$(c)$

$\cos { \left( A-B \right) = } 3/5$

$\therefore\quad 5\cos { A } \cos { B } +5\sin { A } \sin { B } =3$
But from

$2$ nd relation

$\sin { A } \sin { B } =2\cos { A } \cos { B }$

$\therefore\quad \left( 5+10 \right) \cos { A } \cos { B } =3$

$\therefore\quad \cos { A } \cos { B } =1/5$ i.e.

$(a)$
Or

$5\left( \dfrac { 1 }{ 2 } +1 \right) \sin { A } \sin { B } =3$

$\therefore\quad \sin { A } \sin { B } =2/5$

$\quad\therefore\quad$

$(b)$ is not correct.

$\cos { \left( A+B \right) } =\cos { A } \cos { B } -\sin { A } \sin { B } =\cfrac { 1 }{ 5 } -\cfrac { 2 }{ 5 } =-\cfrac { 1 }{ 5 }$ i.e.

$(c)$

The number of solutions of the pair of equations.

$2\sin^2\theta -\cos 2\theta =0$

$2\cos^2\theta -3\sin \theta =0$
in the interval

$[0, 2\pi]$ is?

0%
Zero
0%
One
0%
Two
0%
Four

Explanation

From

$1$ st,

$2\sin^2\theta -(1-2\sin^2\theta)=0$

$\therefore 4\sin^2\theta =1$ or

$\sin\theta =\pm 1/2$

$\therefore \theta =\pi/6, \pi +\pi/6$ in

$[0, 2\pi]$
Also from

$2$ nd equation, we have

$2(1-\sin^2\theta)-3\sin\theta =0$
or

$2\sin^2\theta +3\sin\theta -2=0$
or

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

$\therefore \sin\theta =1/2$ as

$-2$ is rejected

$\therefore \sin\theta =1/2$ which in already included in

$1$ st. Hence there are only two solutions

$\Rightarrow$ (c).

State true or false
If sin x = sin

$\lambda$ , then the values of sin(x/3) are sin (

$\lambda$ /3), sin [

$(\pi- \lambda)$ /3] and - sin [

$(\pi+\lambda)$ /3]

0%
True
0%
False

Explanation

We have

$\sin x - \sin\lambda$

$\Rightarrow x = n\pi + (-1)^{n} \lambda,n\epsilon l$

$\therefore$

$\sin (x/3) = \sin [n\pi/3 + (- 1)^{2} (X/3)]$
Hence for n = 0,3, we have

$\sin(x/3) = \sin(\lambda/3)$
and for n = 1,2, we have

$\sin (x /3)$ =

$\sin{\pi/3-\lambda/3}$
[Note that for n = 2, sin(x /3)
=

$\sin (2\pi /3 + \lambda/3)$ =

$\sin [\pi - [\pi /3 - \lambda/3]]$
=

$\sin(\pi/3-\lambda/3)$
and for n = 4,5, we get
sin (x/3) = -

$\sin(\pi/3 + \lambda/3)$ .
It is easy to see that all other values of n repeat only these values of

$\sin (x/3)$ .

$2\sin^2\theta -5\sin \theta +2 > 0, \theta \in (0, 2\pi)$ , then

$\theta \in$

0%
$\left(\dfrac{5\pi}{6}, 2\pi\right)$
0%
$\left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$
0%
$\left(0, \dfrac{\pi}{6}\right)$
0%
$\left(\dfrac{\pi}{80}, \dfrac{\pi}{6}\right)$

Explanation

$(2\sin \theta -1)(\sin \theta -2) > 0$

$\therefore \sin \theta < \dfrac{1}{2}=\sin \dfrac{\pi}{6}=\sin \left(\pi -\dfrac{\pi}{6}\right)=\sin \dfrac{5\pi}{6}$

$\therefore \theta \in \left(0, \dfrac{\pi}{6}\right)$ or

$\theta \in \left(\dfrac{5\pi}{6}, 2\pi\right)$

$\therefore \theta \in \left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$ .

$\alpha, \beta, \gamma, \delta$ are the smallest

$+$ ive angles in ascending order of magnitude which have their sines equal to a

$+$ ive quantity

$\lambda$ then the value of

$4\sin \dfrac{\alpha}{2}+3\sin \dfrac{\beta}{2}+2\sin \dfrac{\gamma}{2}+\sin \dfrac{\delta}{2}=$ .

0%
$2\sqrt{1-\lambda}$
0%
$2\sqrt{1+\lambda}$
0%
$2\sqrt{\lambda}$
0%
$2\sqrt{\lambda +2}$

Explanation

Ans. (b)

$\sin\alpha =\lambda$

$\theta =n\pi +(-1)^n\alpha$
For

$+$ ive values,

$n=0, 1, 2, 3$

$\therefore \alpha =\alpha, \beta =\pi -\alpha, \gamma =2\pi +\alpha , \delta =3\pi -\alpha$

$\therefore E=4\sin\dfrac{\alpha}{2}+3\sin\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)+2\sin\left(\pi +\dfrac{\alpha}{2}\right)+\sin\left(\dfrac{3\pi}{2}-\dfrac{\alpha}{2}\right)$

$=4\sin\dfrac{\alpha}{2}+3\cos\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}-\cos\dfrac{\alpha}{2}$

$=2\left(\cos\dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)=2\sqrt{\left(\cos \dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)^2}$

$=2\sqrt{1+\sin\alpha}=2\sqrt{1+\lambda}$ .

$0\leq x\leq \pi$ and

$81^{\sin^2x}+81^{\cos^2x}=30$ , then x is equal to.

0%
$\pi /6$
0%
$\pi /3$
0%
$5\pi /6$
0%
$2\pi /3$
0%
All correct

Explanation

Put

$\cos^2x=1-\sin^2x$
If

$y=81^{\sin^2x}$ then

$y+\dfrac{81}{y}=30$

$y^2-30y+81=0$ ,

$y=27, 3$

$3^{4\sin^2x}=3^3, 3^1$

$\therefore \sin^2x=\dfrac{3}{4}, \dfrac{1}{4}$

$\therefore \sin x=\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}$
as

$\sin x$ is

$+$ ive in the given interval

$0 < x < \pi$

$\therefore x=\dfrac{\pi}{3}, \pi -\dfrac{\pi}{3}=\dfrac{2\pi}{3}, \dfrac{\pi}{6}, \pi -\dfrac{\pi}{6}=\dfrac{5\pi}{6}$ .

Solve:

$2(\cos x+\cos 2x)+\sin 2x(1+2\cos x)=2\sin x, -\pi \leq x \leq \pi$ .

0%
$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$ .
0%
$-\pi, -\pi /2, \pi /3, \pi$ .
0%
$-\pi, -\pi /3, \pi /2, \pi/3, \pi$ .
0%
None of these

Explanation

$2(\cos x+2\cos^2 x-1)+2\sin x\cos x. (1+2\cos x)-2\sin x=0$
or

$2(2\cos^2x+\cos x-1)+2\sin x(2\cos^2x+\cos x-1)=0$

$2(1+\sin x)(\cos x+1)(2\cos x-1)=0$
We have to determine values of x s.t.

$-\pi \leq x\leq \pi$

$1+\sin x=0$

$\therefore \sin x=-1$

$\therefore x=2n\pi +\dfrac{3\pi}{2}$

$\therefore x=-\dfrac{\pi}{2}$ ,
for

$n=-1$

$\because -\pi \leq x < \pi$

$\cos x=-1=\cos \pi$

$\therefore \cos x=1/2=\cos (\pi/3)$

$\therefore x=2n\pi \pm \pi/3$

$\therefore x=\pi /3$ ,

$-\pi /3$
Hence the values of x s.t.

$-\pi \leq x \leq \pi$ are

$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$ .

Solve

$(2+\sqrt{3})\cos\theta =1-\sin \theta$ .

0%
$\theta =-2n\pi -\dfrac{2\pi}{3}$
0%
$\theta =2r\pi -\dfrac{2\pi}{3}$ .
0%
$\theta =2r\pi +\dfrac{2\pi}{3}$ .
0%
$\theta =n\pi -\dfrac{2\pi}{3}$

Explanation

$(2+\sqrt{3})\cos\theta =1-\sin\theta$

$\dfrac{1-\sin\theta}{\cos\theta}=2+\sqrt{3}$

$\dfrac{\left(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}\right)^2}{\cos^2\dfrac{\theta}{2}-\sin^2\dfrac{\theta}{2}}=2+\sqrt{3}$
or

$\dfrac{1-t}{1+t}=\tan 75^o, t=\tan \dfrac{\theta}{2}$
or

$\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)=\tan\dfrac{5\pi}{12}$

$\therefore \dfrac{\pi}{4}-\dfrac{\theta}{2}=n\pi +\dfrac{5\pi}{12}$
or

$\theta =-2n\pi -\dfrac{2\pi}{3}$
or

$\theta =2r\pi -\dfrac{2\pi}{3}, r\in I$ .

A balloon is observed simultaneously from three points A B and C, on a straight road directly under it. The angular elevation at B is twice of what it is at A and the angular elevation at C is thrice of what it is at A. If the distance between A and B is 200 meters and the distance between B and C is 100 meters, then find the height of the balloon.

0%
$50 \sqrt{3}$ m
0%
$50$ m
0%
$150 \sqrt{3}$ m
0%
$100 \sqrt{3}$ m

Explanation

$x = h \cot 3 \alpha$ ...............(i)

$(x + 100) = h \cot 2 \alpha$ ............ (ii)

$(x + 300) = h \cot \alpha$ ............(iii)

From (i) and (ii)

$-100 = h (\cot 3 \alpha - \cot 2 \alpha) = h\dfrac{(sin2\alpha \ cos3\alpha - cos2\alpha \ sin3\alpha)}{sin3\alpha \ sin2\alpha} = \dfrac{sin(3\alpha - 2\alpha)}{sin3\alpha \ sin2\alpha}$

Similarly,

From (ii) and (iii)

$-200 = h (\cot 2 \alpha - \cot \alpha) = \dfrac{sin(2\alpha - \alpha)}{sin2\alpha \ sin\alpha}$

$100 = h \left (\dfrac{\sin \alpha}{\sin 3 \alpha \sin 2 \alpha} \right )$ on solving

$200 = h \left (\dfrac{\sin \alpha}{\sin 2 \alpha \sin \alpha} \right )$ on solving

On dividing the above equations we get,

$\dfrac{\sin 3 \alpha}{\sin \alpha} = \dfrac{200}{100} \Rightarrow \dfrac{\sin 3 \alpha}{\sin \alpha} = 2$ ......... (3)

We know that:

$sin3\alpha = 3sin\alpha - 4sin^3\alpha$

From eq (3) we get

$\Rightarrow 3 \sin \alpha - 4 \sin^{3} \alpha - 2 \sin \alpha = 0$

$\Rightarrow 4 \sin^{3} \alpha - \sin \alpha = 0 \Rightarrow \sin \alpha = 0$ or

$\sin^{2} \alpha = \dfrac{1}{4}$

$\sin^{2} \alpha = \dfrac{1}{4} = \sin^{2} \left (\dfrac{\pi}{6} \right ) \Rightarrow \alpha = \dfrac{\pi}{6}$

Hence

$h = 200 \ sin2\alpha= 200 \sin \dfrac{\pi}{3} = 200 \dfrac{\sqrt{3}}{2} = 100 \sqrt{3}$

So the height of the balloon

$100 \sqrt{3}$

2095090_1007556_ans_e10129e3bc234ce3aaf92c46d7f6000d.png

Solve

$\tan \theta +\sec \theta =\sqrt{3}; 0\leq \theta \leq 2\pi$ .

0%
$\theta =2n\pi -\dfrac{\pi}{6}$ .
0%
$\theta =n\pi +\dfrac{\pi}{6}$ .
0%
$\theta =2n\pi +\dfrac{5\pi}{6}$ .
0%
$\theta =2n\pi +\dfrac{\pi}{6}$ .

Explanation

$\tan\theta +sec\theta =\sqrt{3}$

$\dfrac{1+\sin\theta}{\cos\theta}=\sqrt{3}$

$\dfrac{\left(\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}\right)^2}{\cos^2\dfrac{\theta}{2}-\sin^2\dfrac{\theta}{2}}=\sqrt{3}$
or

$\dfrac{\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}}{\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}}=\sqrt{3}$ or

$\dfrac{1+t}{1-t}=\sqrt{3}$
or

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

$\therefore \dfrac{\theta}{2}+\dfrac{\pi}{4}=n\pi +\dfrac{\pi}{3}$

$\therefore \theta =2n\pi +\dfrac{\pi}{6}$ .

$P\left( 4 \right) = 3$ and

$\displaystyle {\sin ^6}x + {\cos ^6}x = {a \over b}\left( {a,b \in N} \right)$ and

$a,b$ are relatively prime, then

$a + b$ is equal to

0%
$7$
0%
$23$
0%
$16$
0%
$9$

The equation

${\sin ^6}x + {\cos ^6}x = {a^2}$ has real solution if

0%
$a \in \left( { - 1,1} \right)$
0%
$a \in \left( { - 1, - {1 \over 2}} \right)$
0%
$\eqalign{ & a \in \left( { - {1 \over 2},{1 \over 2}} \right) \cr & a \in \left( {{1 \over 2},1} \right) \cr}$
0%
$a \in \left( {{1 \over 2},1} \right)$

Explanation

$a^{2}=\sin^{6}x+\cos^{6}x$

$\Rightarrow a^{2}=(\sin^{2}x+\cos^{2}x)(\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x)$

$\Rightarrow a^{2}=((\sin^{2}x+\cos^{2}x)^{2}-3\sin^{2}x \cos^{2}x)$ as(

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

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

$\Rightarrow a^{2}=(1-\cfrac{3}{4}\sin^{2}2x)$

$\sin^{2}2x \in [0,1]$

$\therefore 1-\cfrac{3}{4}\sin^{2}(2x) \in [\cfrac{1}{4},1]$

$\cfrac{1}{4}\le a^{2} \le 1$

$a \in [(-\infty,\cfrac{-1}{2}]\cup[\cfrac{1}{2}, \infty)]\cap[-1,1]$

$\therefore a\in[-1,\cfrac{-1}{2}]\cup[\cfrac{1}{2},1]$

$\sec A + \tan A = m$ and

$\sec A - \tan A = n$ , find the value of

$\sqrt{mn}$ .

0%
$0$
0%
$\pm 1$
0%
$\pm 2$
0%
$\pm 3$

Explanation

Use trigonometry formula:

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

$secA+tanA=m$ ......(1)

$secA-tanA=n$ .........(2)

multiply the equations (1) and (2),

$\left ( secA+tanA \right )\left ( secA-tanA \right )=mn$

$sec^{2}A-tan^{2}A=mn$

$1=mn$

Therefore,

$\sqrt{mn}=\pm 1$

$8\sin(p+2q)= 5\sin p$ , then

$3(\tan p+\tan q)= \dfrac{2\tan p}{\cos^2q}$ .

0%
True
0%
False

The most general value of

$\theta$ satisfying both the equations

$\sin \theta = \frac{1}{2},\tan \theta = \frac{1}{{\sqrt 3 }}\;is\;\left( {n \in I} \right)$

0%
$2n\pi + \frac{\pi }{6}$
0%
$\left( b \right)2n\pi - \frac{{7\pi }}{6}$
0%
$\left( c \right)2n\pi + \frac{{5\pi }}{6}$
0%
None of these

The expression

$\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$ reduces to :

0%
$\dfrac {1+\sin A}{\cos A}$
0%
$\dfrac {1-\sin A}{\cos A}$
0%
$\dfrac {1+\cos A}{\sin A}$
0%
$\dfrac {1+\cos A}{\cos A}$

Explanation

$\sec^2\theta=1+\tan^2\theta$

$\Rightarrow \cfrac {\tan A+\sec A-1}{\tan A-\sec A+1}$

$\Rightarrow \cfrac {\tan A+\sec A-(\sec^2A-\tan^2A)}{\tan A-\sec A+1}$

$\Rightarrow \cfrac {(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$

$\Rightarrow \tan A+\sec A$

$\Rightarrow \cfrac {\sin A}{\cos A}+\cfrac {1}{\cos A}$

$\Rightarrow \cfrac {1+\sin A}{\cos A}$

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

$\sin^{12}\theta +3 \sin^{10}\theta +3 \sin^8\theta +\sin^6\theta +2\sin^4\theta +2\sin^2\theta -4=1$

0%
True
0%
False

$\sin \left( {\pi \cos x} \right) = \cos \left( {\pi \sin x} \right)$ , then

$\sin 2x =$

0%
$-\dfrac{3}{4}$
0%
$-\dfrac{4}{3}$
0%
$\dfrac{1}{3}$
0%
none of these

Explanation

Given

$\sin(\pi\cos x)=\cos(\pi\sin x)$

$\sin(\pi cosx)=\sin(\dfrac{\pi}{2}-\pi\sin x)$

$\pi \cos x=\dfrac{\pi}{2}-\pi \sin x$

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

Squaring on both sides

$\sin^2x+\cos^2x+2\sin x \cos x= \dfrac{1}{4}$

$1+2\sin x\cos x=\dfrac{1}{4}$

$\sin 2x=-\dfrac{3}{4}$

The function

$f(x)=a \sin x+\dfrac {1}{3}\sin 3x$ has a maximum at

$x=\pi/3$ , then a equals-

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

Explanation

$\begin{array}{l} f\left( x \right) =a\sin x+\frac { 1 }{ 3 } \sin 3x \\ { f^{ ' } }\left( x \right) =a\cos x+\dfrac { 1 }{ 3 } \cos 3x\times 3 \\ =a\cos x+\cos 3x \\ Since\, f\left( x \right) has\, \max imum\, at\, x=\dfrac { \pi }{ 3 } \\ so,\, f'\left( { \dfrac { \pi }{ 3 } } \right) =0 \\ \Rightarrow a\cos \dfrac { \pi }{ 3 } +\cos 3\times \dfrac { \pi }{ 3 } =0 \\ \Rightarrow a\times \dfrac { 1 }{ 2 } +\left( { -1 } \right) =0 \\ \Rightarrow \dfrac { a }{ 2 } =1 \\ a=2 \end{array}$

$\alpha cos^23\theta +\beta cos^4\theta= 16 cos^6\theta + 9 cos^2\theta$ is an identity then-

0%
$\alpha = 1, \beta = 18$
0%
$\alpha = 1, \beta = 24$
0%
$\alpha = 3, \beta = 24$
0%
$\alpha = 4, \beta = 2$

Explanation

$\alpha\cos^2 3\theta+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$

$\Rightarrow$

$\alpha(4\cos^3\theta-3\cos\theta)^2+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$

$\Rightarrow$

$\alpha(16\cos^6\theta+9\cos^2\theta-24\cos^4\theta)+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$

$\Rightarrow$

$\alpha(16\cos^6\theta+9\cos^2\theta)-24\alpha\cos^4\theta+\beta\cos^4\theta=(16\cos^6\theta+9\cos^2\theta)$

$\Rightarrow$

$\alpha(16\cos^6\theta+9\cos^2\theta)-\cos^4\theta(24\alpha-\beta)=(16\cos^6\theta+9\cos^2\theta)$

$\Rightarrow$

$\alpha(16\cos^6\theta+9\cos^2\theta)-\cos^4\theta(24\alpha-\beta)=(16\cos^6\theta+9\cos^2\theta)+\cos^4\theta\times 0$

Now, we are going to compare,

$\alpha(16\cos^6+9\cos^2\theta)=16\cos^6+9\cos^2\theta$

$\therefore$

$\alpha=1$

Again,

$\Rightarrow$

$-\cos^4\theta(24\alpha-\beta)=\cos^4\theta\times 0$

$\Rightarrow$

$24(1)-\beta=0$ [ Since,

$\alpha=1$ ]

$\Rightarrow$

$\beta=24$

$\therefore$

$\alpha=1$ and

$\beta=24$

A flag staff on the top of the tower

$80\ meter$ high, subtends an angle

$\tan^{-1}\left(\dfrac{1}{9}\right)$ at point on the ground

$100\ meters$ away from the foot of the tower. Find the height of the flag-staff.

0%
$20\ m$
0%
$30\ m$
0%
$25\ m$
0%
$35\ m$

Explanation

$\rightarrow$ Let, height of the flag stuff be

$x$ .

$\rightarrow$ Angle subtended by the tower.

$\therefore \theta = \tan^{-1} \left( \dfrac{80}{100} \right)= 38.7$ degree.

$\rightarrow$ Angle subtended by the flag stuff

$\therefore \tan^{-1} (7/9)= 6.3$ degree.

$\rightarrow$ total angle subtended

$= 38.7 + 6.3$

$= 45$ degree.

$\rightarrow$ As shown in triangle

$\tan 45= \dfrac{x +80}{100}$

$\therefore 1 =\dfrac{x+80}{100}$

$\therefore x= 100-80$

$\therefore x= 20$

1425610_1058829_ans_4f17d2f44924471ab190bdd9494fbbd1.png

$x=\dfrac{{2\left( {\sin {1^0} + \sin {2^0} + \sin {3^0} + ....... + \sin {{89}^0}} \right)}}{{2\left( {\cos {1^0} + \cos {2^0} + .............\cos {{44}^0}} \right) + 1}}$ , then the value of

${\log_x}2$ is equal

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

$A + B + C = \pi$ , then

${\sin ^4}A + {\sin ^4}B + {\sin ^4}C = \cfrac{3}{2} + 2\cos A\cos B\cos C + \cfrac{1}{2}\cos 2A\cos 2B\cos 2C$

0%
True
0%
False

Explanation

We have,

$A + B + C = \pi$

We know that

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

$\sin^4 x=\left(\dfrac{1-\cos 2x}{2}\right)^2$

$\cos 2A+\cos 2B+\cos 2C=-1-4\cos A \cos B \cos C$

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

$\cos^2 2A+\cos^2 2B+\cos^2 2C=1+2\cos 2A \cos 2B \cos 2C$

$........(2)$

Now,

L.H.S

$={\sin ^4}A + {\sin ^4}B + {\sin ^4}C$

$=\left(\dfrac{1-\cos 2A}{2}\right)^2+\left(\dfrac{1-\cos 2B}{2}\right)^2+\left(\dfrac{1-\cos 2C}{2}\right)^2$

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

$=\dfrac{1}{2}\left[3+(\cos^2 2A+\cos^2 2B+\cos^2 2C)-2(\cos 2A+\cos 2B+\cos 2C)\right]$

From equations

$(1)$ and

$(2)$

$=\dfrac{1}{2}\left[3+(1+\cos 2A\cos 2B\cos 2C)-2(-1-4\cos A\cos B\cos C)\right]$

$=\dfrac{1}{2}\left[4+\cos 2A\cos 2B\cos 2C+2+8\cos A\cos B\cos C\right]$

$=\dfrac{1}{2}\left[6+\cos 2A\cos 2B\cos 2C+8\cos A\cos B\cos C\right]$

$= \cfrac{3}{2} + 2\cos A\cos B\cos C + \cfrac{1}{2}\cos 2A\cos 2B\cos 2C$

R.H.S

Hence, proved.

$x \in (\pi, 2\pi)$ and

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

$\tan x$ is

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$\dfrac{4 - \sqrt{7}}{3}$
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$\dfrac{\sqrt{7} - 4}{3}$
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$\dfrac{-4 + \sqrt{7}}{3}$
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$-\left(\dfrac{4 + \sqrt{7}}{3}\right)$

Explanation

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

$x\in (\pi, 2\pi)$

divide by

$\cos x$

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

$\Rightarrow 1+\tan x=\dfrac{1}{2}\sec x$ [we know

$\sec^2x-\tan^2x=1$ ]

$\Rightarrow 1+\tan x=\dfrac{1}{2}\sqrt{1+\tan^2x}$

Let

$\tan x=t$

$1+t=\dfrac{\sqrt{1+t^2}}{2}$

$(2(1+t))^2=1+t^2$

$\Rightarrow 3t^2+8t+3=0$

$\therefore t=\dfrac{-8\pm \sqrt{8^2-4(3)(3)}}{2\times 3}$

$\Rightarrow \dfrac{-8\pm 2\sqrt{7}}{6}$

$\Rightarrow t=\dfrac{-4\pm \sqrt{7}}{3}$

$\therefore \tan x=\dfrac{-4\pm \sqrt{7}}{3}$ but

$x\in (\pi, 2\pi)$

$\therefore \tan x=\dfrac{-4+\sqrt{7}}{3}$ .

$\dfrac{{\left( {1 - \cos A} \right)}}{2} = x$ then find the value of x is

0%
${\cos ^2}\left( {\dfrac{A}{2}} \right)$
0%
$\sqrt {\sin \left( {\dfrac{A}{2}} \right)}$
0%
$\sqrt {\cos \left( {\dfrac{A}{2}} \right)}$
0%
${\sin ^2}\left( {\dfrac{A}{2}} \right)$

Explanation

$\implies\cfrac { 1-\cos A }{ 2 } =x$

$\implies\cfrac { 2{ \sin }^{ 2 }\cfrac { A }{ 2 } }{ 2 } =x$

$\implies x={ \sin }^{ 2 }\cfrac { A }{ 2 }$

$\sin \theta = n \sin(\theta + 2 \alpha)$ then

$\tan (\theta + \alpha)$ =

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$\dfrac{1 + n}{1 - n} \tan \, \alpha$
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$\dfrac{1 - n}{1 + n} \tan \, \alpha$
0%
$\tan \, \alpha$
0%
None

Explanation

$\sin\theta =n\sin(\theta +2\alpha )$

$\dfrac{\sin\theta }{\sin(\theta+2\alpha ) }=n$

apply componendo dividendo rule,

$\dfrac{\sin\theta +\sin(\theta +2\alpha )}{\sin\theta -\sin(\theta +2\alpha )}=\dfrac{n+1}{n-1}$

apply

$\sin C+\sin D$ and

$\sin C-\sin D$

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

$\dfrac{\tan(\theta +\alpha )}{\tan\alpha }=\dfrac{n+1}{n-1}$

$\tan(\theta +\alpha )=\dfrac{n+1}{n-1}\tan\alpha$

Total number of solution of the equation

$3x+2\tan x=\dfrac {5\pi}{2}$ in

$x\ \epsilon [0,2\pi]$ is equal to

0%
$1$
0%
$2$
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$3$
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$4$

Explanation

Given equation

$3x+2\tan x=\dfrac{5\pi}{2}$

$2\tan x =5\dfrac{\pi}{2}-3x$

$\tan x=\dfrac{5\pi}{4}-\dfrac{3x}{2}$

$y=\tan x=\dfrac{5\pi}{4}-\dfrac{3x}{2}$

Using graph we find point intersection of

$y=\tan x;y=\dfrac{5\pi}{4}-\dfrac{3x}{2}\left(m=-\dfrac{3}{2};C=\dfrac{5\pi}{4}\right)$

$x-intercept :y=0;$

$\dfrac{3x}{2}=\dfrac{5\pi}{4}\left(x=\dfrac{5\pi}{6},y=0\right)$

$y-intercept\, x=0;$

$y=\dfrac{5\pi}{4}\left(x=0;y=\dfrac{5\pi}{4}\right)$

Given region of graph

$[0,2\pi]$

Number of solution = Number of intersections

$='3'$

$\boxed{Number\, of \, solutions=3}$

2113947_1059257_ans_ff9fb51671dd41b3963fba9b18245df2.png

$\sin^{-1}x+\sin^{-1}\dfrac{1}{x}+\cos^{-1}x+\cos^{-1}\dfrac{1}{x}, x\notin \pm 1$ is equal to?

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$\pi$
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$\dfrac{\pi}{2}$
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$\dfrac{3\pi}{2}$
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None of these

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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