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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 8 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 8
If $$\sin { \alpha } =12/13\left( 0<\alpha <\pi /2 \right)$$ and
$$\cos { \beta } =-\dfrac { 3 }{ 5 } \left( \pi <\beta <\dfrac { 3 }{ 2 } \pi \right) $$, the value of $$\sin { \left( \alpha +\beta \right) }$$ is
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$$\dfrac{-56}{65}$$
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$$\dfrac{16}{65}$$
0%
$$\dfrac{56}{65}$$
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$$\dfrac{-16}{65}$$
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
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$$2$$
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$$3$$
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$$\dfrac { 1 }{ 2 } $$
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$$\dfrac { 1 }{ 3 } $$
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, $$\left[ f\left( \theta \right) \right] _{ maximum }=\dfrac { 2 }{ 6-5 } =2$$
If $$\cos { \left( A-B \right) } =3/5$$ and $$\tan { A } \tan { B } =2$$, then
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$$\cos { A } \cos { B } =1/5$$
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$$\sin { A } \sin { B } =-2/5$$
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$$\cos { \left( A+B \right) } =-1/5$$
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$$\sin { A } \cos { B } =4/5$$
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?
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Zero
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One
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Two
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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]
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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 $$\sin (x/3)$$.
If $$2\sin^2\theta -5\sin \theta +2 > 0, \theta \in (0, 2\pi)$$, then $$\theta \in$$
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$$\left(\dfrac{5\pi}{6}, 2\pi\right)$$
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$$\left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$$
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$$\left(0, \dfrac{\pi}{6}\right)$$
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$$\left(\dfrac{\pi}{80}, \dfrac{\pi}{6}\right)$$
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 $$\theta \in \left(\dfrac{5\pi}{6}, 2\pi\right)$$
$$\therefore \theta \in \left(0, \dfrac{\pi}{6}\right)\cup \left(\dfrac{5\pi}{6}, 2\pi\right)$$.
If $$\alpha, \beta, \gamma, \delta$$ are the smallest $$+$$ive angles in ascending order of magnitude which have their sines equal to a $$+$$ive quantity $$\lambda$$ then the value of $$4\sin \dfrac{\alpha}{2}+3\sin \dfrac{\beta}{2}+2\sin \dfrac{\gamma}{2}+\sin \dfrac{\delta}{2}=$$.
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$$2\sqrt{1-\lambda}$$
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$$2\sqrt{1+\lambda}$$
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$$2\sqrt{\lambda}$$
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$$2\sqrt{\lambda +2}$$
Explanation
Ans. (b)
$$\sin\alpha =\lambda$$
$$\theta =n\pi +(-1)^n\alpha$$
For $$+$$ive values, $$n=0, 1, 2, 3$$
$$\therefore \alpha =\alpha, \beta =\pi -\alpha, \gamma =2\pi +\alpha , \delta =3\pi -\alpha$$
$$\therefore E=4\sin\dfrac{\alpha}{2}+3\sin\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)+2\sin\left(\pi +\dfrac{\alpha}{2}\right)+\sin\left(\dfrac{3\pi}{2}-\dfrac{\alpha}{2}\right)$$
$$=4\sin\dfrac{\alpha}{2}+3\cos\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}-\cos\dfrac{\alpha}{2}$$
$$=2\left(\cos\dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)=2\sqrt{\left(\cos \dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)^2}$$
$$=2\sqrt{1+\sin\alpha}=2\sqrt{1+\lambda}$$.
If $$0\leq x\leq \pi$$ and $$81^{\sin^2x}+81^{\cos^2x}=30$$, then x is equal to.
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$$\pi /6$$
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$$\pi /3$$
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$$5\pi /6$$
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$$2\pi /3$$
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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: $$2(\cos x+\cos 2x)+\sin 2x(1+2\cos x)=2\sin x, -\pi \leq x \leq \pi$$.
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$$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$$.
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$$-\pi, -\pi /2, \pi /3, \pi$$.
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$$-\pi, -\pi /3, \pi /2, \pi/3, \pi$$.
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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 $$(2+\sqrt{3})\cos\theta =1-\sin \theta$$.
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$$\theta =-2n\pi -\dfrac{2\pi}{3}$$
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$$\theta =2r\pi -\dfrac{2\pi}{3}$$.
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$$\theta =2r\pi +\dfrac{2\pi}{3}$$.
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$$\theta =n\pi -\dfrac{2\pi}{3}$$
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.
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$$50 \sqrt{3}$$m
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$$50 $$m
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$$150 \sqrt{3}$$m
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$$100 \sqrt{3}$$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 $$100 \sqrt{3} $$
Solve $$\tan \theta +\sec \theta =\sqrt{3}; 0\leq \theta \leq 2\pi$$.
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$$\theta =2n\pi -\dfrac{\pi}{6}$$.
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$$ \theta =n\pi +\dfrac{\pi}{6}$$.
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$$ \theta =2n\pi +\dfrac{5\pi}{6}$$.
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$$ \theta =2n\pi +\dfrac{\pi}{6}$$.
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 $$\tan\left(\dfrac{\pi}{4} +\dfrac{\theta}{2}\right)=\tan\dfrac{\pi}{3}$$
$$\therefore \dfrac{\theta}{2}+\dfrac{\pi}{4}=n\pi +\dfrac{\pi}{3}$$
$$\therefore \theta =2n\pi +\dfrac{\pi}{6}$$.
If $$P\left( 4 \right) = 3$$ 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
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$$7$$
0%
$$23$$
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$$16$$
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$$9$$
The equation $${\sin ^6}x + {\cos ^6}x = {a^2}$$ has real solution if
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$$a \in \left( { - 1,1} \right)$$
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$$a \in \left( { - 1, - {1 \over 2}} \right)$$
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$$\eqalign{
& a \in \left( { - {1 \over 2},{1 \over 2}} \right) \cr
& a \in \left( {{1 \over 2},1} \right) \cr} $$
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($$\sin^{2}x+\cos^{2}x=1)$$
$$\Rightarrow a^{2}=(1-3\sin^{2}x\cos^{2}x)$$
$$\Rightarrow a^{2}=(1-\cfrac{3}{4}\sin^{2}2x)$$
$$\sin^{2}2x \in [0,1]$$
$$\therefore 1-\cfrac{3}{4}\sin^{2}(2x) \in [\cfrac{1}{4},1]$$
$$\cfrac{1}{4}\le a^{2} \le 1$$
$$a \in [(-\infty,\cfrac{-1}{2}]\cup[\cfrac{1}{2}, \infty)]\cap[-1,1]$$
$$ \therefore a\in[-1,\cfrac{-1}{2}]\cup[\cfrac{1}{2},1]$$
If $$\sec A + \tan A = m $$ and $$ \sec A - \tan A = n$$, find the value of $$\sqrt{mn}$$.
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$$0$$
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, $$\sqrt{mn}=\pm 1$$
If $$8\sin(p+2q)= 5\sin p$$ , then $$3(\tan p+\tan q)= \dfrac{2\tan p}{\cos^2q}$$.
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0%
True
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False
The most general value of $$\theta $$ satisfying both the equations $$\sin \theta = \frac{1}{2},\tan \theta = \frac{1}{{\sqrt 3 }}\;is\;\left( {n \in I} \right)$$
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$$2n\pi + \frac{\pi }{6}$$
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$$\left( b \right)2n\pi - \frac{{7\pi }}{6}$$
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$$\left( c \right)2n\pi + \frac{{5\pi }}{6}$$
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None of these
The expression $$\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$$ reduces to :
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$$\dfrac {1+\sin A}{\cos A}$$
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$$\dfrac {1-\sin A}{\cos A}$$
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$$\dfrac {1+\cos A}{\sin A}$$
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$$\dfrac {1+\cos A}{\cos A}$$
Explanation
$$\sec^2\theta=1+\tan^2\theta$$
$$\Rightarrow \cfrac {\tan A+\sec A-1}{\tan A-\sec A+1}$$
$$\Rightarrow \cfrac {\tan A+\sec A-(\sec^2A-\tan^2A)}{\tan A-\sec A+1}$$
$$\Rightarrow \cfrac {(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$$
$$\Rightarrow \tan A+\sec A$$
$$\Rightarrow \cfrac {\sin A}{\cos A}+\cfrac {1}{\cos A}$$
$$\Rightarrow \cfrac {1+\sin A}{\cos A}$$
If $$\sin^2 \theta +\cos^2\theta =1$$ then
$$\sin^{12}\theta +3 \sin^{10}\theta +3 \sin^8\theta +\sin^6\theta +2\sin^4\theta +2\sin^2\theta -4=1$$
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0%
True
0%
False
If $$\sin \left( {\pi \cos x} \right) = \cos \left( {\pi \sin x} \right)$$, then $$\sin 2x = $$
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$$ -\dfrac{3}{4}$$
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$$ -\dfrac{4}{3}$$
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$$ \dfrac{1}{3}$$
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-
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$$-2$$
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$$2$$
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$$-1$$
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$$1$$
Explanation
$$\begin{array}{l} f\left( x \right) =a\sin x+\frac { 1 }{ 3 } \sin 3x \\ { f^{ ' } }\left( x \right) =a\cos x+\dfrac { 1 }{ 3 } \cos 3x\times 3 \\ =a\cos x+\cos 3x \\ Since\, f\left( x \right) has\, \max imum\, at\, x=\dfrac { \pi }{ 3 } \\ so,\, f'\left( { \dfrac { \pi }{ 3 } } \right) =0 \\ \Rightarrow a\cos \dfrac { \pi }{ 3 } +\cos 3\times \dfrac { \pi }{ 3 } =0 \\ \Rightarrow a\times \dfrac { 1 }{ 2 } +\left( { -1 } \right) =0 \\ \Rightarrow \dfrac { a }{ 2 } =1 \\ a=2 \end{array}$$
If $$\alpha cos^23\theta +\beta cos^4\theta= 16 cos^6\theta + 9 cos^2\theta$$ is an identity then-
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$$\alpha = 1, \beta = 18$$
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$$\alpha = 1, \beta = 24$$
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$$\alpha = 3, \beta = 24$$
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$$\alpha = 4, \beta = 2$$
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.
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$$20\ m$$
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$$30\ m$$
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
$$\tan 45= \dfrac{x +80}{100}$$
$$\therefore 1 =\dfrac{x+80}{100} $$
$$\therefore x= 100-80$$
$$\therefore x= 20$$
If $$x=\dfrac{{2\left( {\sin {1^0} + \sin {2^0} + \sin {3^0} + ....... + \sin {{89}^0}} \right)}}{{2\left( {\cos {1^0} + \cos {2^0} + .............\cos {{44}^0}} \right) + 1}}$$ , then the value of $${\log_x}2$$ is equal
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$$0$$
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$$\dfrac { 1 }{ 2 } $$
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$$1$$
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$$2$$
If $$A + B + C = \pi $$, then $${\sin ^4}A + {\sin ^4}B + {\sin ^4}C = \cfrac{3}{2} + 2\cos A\cos B\cos C + \cfrac{1}{2}\cos 2A\cos 2B\cos 2C$$
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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.
If $$x \in (\pi, 2\pi)$$ 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 $$x\in (\pi, 2\pi)$$
$$\therefore \tan x=\dfrac{-4+\sqrt{7}}{3}$$.
If $$\dfrac{{\left( {1 - \cos A} \right)}}{2} = x$$ then find the value of
x
is
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$${\cos ^2}\left( {\dfrac{A}{2}} \right)$$
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$$\sqrt {\sin \left( {\dfrac{A}{2}} \right)} $$
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$$\sqrt {\cos \left( {\dfrac{A}{2}} \right)} $$
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$${\sin ^2}\left( {\dfrac{A}{2}} \right)$$
Explanation
$$\implies\cfrac { 1-\cos A }{ 2 } =x$$
$$\implies\cfrac { 2{ \sin }^{ 2 }\cfrac { A }{ 2 } }{ 2 } =x$$
$$\implies x={ \sin }^{ 2 }\cfrac { A }{ 2 } $$
If $$\sin \theta = n \sin(\theta + 2 \alpha) $$ then $$\tan (\theta + \alpha)$$ =
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$$\dfrac{1 + n}{1 - n} \tan \, \alpha$$
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$$\dfrac{1 - n}{1 + n} \tan \, \alpha$$
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$$\tan \, \alpha$$
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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
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$$1$$
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$$2$$
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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
$$='3'$$
$$\boxed{Number\, of \, solutions=3}$$
$$\sin^{-1}x+\sin^{-1}\dfrac{1}{x}+\cos^{-1}x+\cos^{-1}\dfrac{1}{x}, x\notin \pm 1$$ 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
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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