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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 10
In $$\Delta ABC$$, a $$sin(B-C)+b{\,}sin (C-A)+c{\,}sin(A-B)=$$
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$$0$$
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$$a+b+c$$
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$$a^2+b^2+c^2$$
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$$2(a^2+b^2+c^2)$$
Explanation
In $$\triangle ABC,a=2 R\sin A=2 R\sin (180^{\circ}-(B+C))=2 R\sin (B+C)$$
$$a\sin(B-C)+b\sin (C-A)+c\sin(A-B)=\displaystyle\sum a\sin (B-C)$$
$$=2 R\sum \sin (B+C)\sin (B-C)$$
$$=2 R\sum (\sin ^2 B-\sin^2 C)$$ $$(\because \sin(A+B)\sin (A-B)=\sin^2 A-\sin^2 B)$$
$$=2 R(\sin^2 B-\sin^2 C+\sin^2 C-\sin^2 A+\sin^2 A-\sin^2 B)$$
$$=0$$
$$\sin^2 20+\sin^270$$ is equal to_____
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$$1$$
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$$-1$$
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$$0$$
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$$2$$
Explanation
$$\sin^2 20^o+\sin^270^o=\sin^2 20^o+\sin^2(90^o-20^o)=\sin^220^o+\cos^220^o=1$$
Let $$x$$ and $$y$$ be $$2$$ real numbers which satisfy the equations $$(\tan^{2} x - \sec^{2}y) = \dfrac {5a}{6} - 3$$ and $$(-\sec^{2}x + \tan^{2}y) = a^{2}$$, then the value of a can be equal to
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$$\dfrac {2}{3}$$
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$$\dfrac {-2}{3}$$
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$$\dfrac {3}{2}$$
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$$\dfrac {-3}{2}$$
$$ \sin (A+B) . \sin (A-B) = $$
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$$ \sin^2 A - \cos^2 B $$
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$$ \cos^2 A - \sin^2 B $$
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$$ \sin^2 A - \sin^2 B $$
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$$ \cos^2 A - \cos^2 B $$
Explanation
$$\begin{array}{l} \sin \left( { A+B } \right) .\sin \left( { A-B } \right) \\ =\left[ { \sin \left( A \right) \cos \left( B \right) +\cos \left( A \right) \sin \left( B \right) } \right] \left[ { \sin \left( A \right) \cos \left( B \right) -\cos \left( A \right) \sin \left( B \right) } \right] \\ ={ \sin ^{ 2 } }A{ \cos ^{ 2 } }A-{ \cos ^{ 2 } }A{ \sin ^{ 2 } }A \\ ={ \sin ^{ 2 } }A\left[ { 1-{ { \sin }^{ 2 } }B } \right] -\left[ { 1-{ { \sin }^{ 2 } }A } \right] { \sin ^{ 2 } }B \\ ={ \sin ^{ 2 } }A-{ \sin ^{ 2 } }A{ \sin ^{ 2 } }B-{ \sin ^{ 2 } }B+{ \sin ^{ 2 } }A{ \sin ^{ 2 } }B \\ ={ \sin ^{ 2 } }A-{ \sin ^{ 2 } }B \\ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$$
If $$4\ \sin^{2}x-1=0$$ and $$0 < x < 2 \pi$$, then positive values of $$x$$ are
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$$30^{o},120^{o},210^{o},300^{o}$$
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$$30^{o},150^{o},210^{o},330^{o}$$
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$$30^{o},120^{o},150^{o},210^{o}$$
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$$30^{o},160^{o},210^{o},320^{o}$$
Solution set of the equation $$\sin ^{ 2 }{ x } +\cos ^{ 2 }{ 3x }$$ =1$$ is given by
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$$\left\{ \frac { n\pi }{ 4 } ,n\epsilon I \right\}$$
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$$\left\{ n\pi ,n\epsilon I \right\}$$
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$$\left\{ \frac { n\pi }{ 2 } ,n\epsilon I \right\}$$
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$$none of these$$
Explanation
According to question,
$$\sin^2x+\cos^23x=1$$
$$\implies cos^23x= 1-\sin^2x$$
$$\implies \cos^23x=\cos^2x$$
Solving through trigonometric eq,
$$3x=2n\pi+x$$
$$\implies 2x=2n\pi$$
=> $$x=n\pi$$ , $$n\epsilon I$$
The no of solution of the equation: $$1+\sin { x } \cos { x } =2\sin { x } \cos { x } $$ in $$x$$, $$\left[ 0,40 \right] $$ are
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$$4$$
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$$5$$
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$$7$$
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$$None\ of\ these$$
Explanation
$$ 1+sinx cosx = 2sinx cosx $$
$$ x\epsilon [ 0,40]$$
$$ 1 = \dfrac{2}{2}sinx cosx $$
$$ [\because 2 sinx cosx - sin2x]$$
$$ 2 = sin 2x.$$
$$ 2x = \dfrac{1}{2}sin^{-1}(2)$$
$$ x = \dfrac{1}{2}sin^{-1}(2)$$
If $$\sin { x } sin{ \left( { 60 }^{ O }+x \right) .\left( { 60 }^{ O }-x \right) }=\frac { 1 }{ 8 } $$ then$$x=$$
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$$\pi+(-1)^\alpha\frac{\pi}{6}$$
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$$\frac{\pi}{3}+(-1)^\alpha\frac{\pi}{18}$$
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$$n\pi \left( -1 \right) ^{ \alpha }\frac { \pi }{ 3 }$$
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$$\frac { n\pi }{ 3 } +\left( -1 \right) ^{ \alpha }\frac { \pi }{ 9 }$$
Explanation
Given
$$\sin x\sin (60^{\circ} +x) \sin (60^{\circ}-x)=\dfrac 18$$
$$\sin x\left( \sin ^2 60^{\circ} -\sin ^2 x\right)$$
$$\bigg( \sin (A+B)\sin (A-B) =\sin ^2 A-\sin ^2B\bigg)$$
$$\sin x\left(\dfrac 34 -\sin ^2 x\right)=\dfrac 18$$
$$\left(\dfrac {3\sin x-4\sin ^3 x}{4}\right)=\dfrac 18$$
$$\sin 3x =\dfrac 12$$
$$\bigg(\sin 3x=3\sin x-4\sin ^3 x\bigg)$$
$$\implies 3x=n\pi + (-1)^n \dfrac \pi 6$$
$$\implies x=\dfrac {n \pi}3 + (-1)^n \dfrac \pi {18}$$
Find number of solutions to the equation:
$$\left[ \sin { x+\cos { x } } \right] =3+\left[ -\sin { x } \right] +\left[ -\cos { x } \right]$$
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$$0$$
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$$1$$
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$$2$$
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infinite
The number of solutions of the equation $$|\cot x|=\cot x+\dfrac{1}{\sin x}(0\le x\le 2\pi)$$ is :
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
$$(\cot x)=\cot x+\dfrac{1}{\sin x}$$ $$0\le x\le 2\pi$$
$$0\pi\le x < \dfrac{3\pi}{2}$$ and $$0\le x\le \dfrac{\pi}{2}$$
$$\cot x=\cot x+\dfrac{1}{\sin x}$$
$$\sin x=0$$
$$x=0$$
$$2\cot x=\dfrac{1}{\sin x}$$
$$\dfrac{2}{\sin x}\cos x=\dfrac{1}{\sin x}$$
$$\sin x=1\Rightarrow x=\dfrac{\pi}{2}$$
$$0\cos x=\dfrac{1}{2}$$
$$x=\dfrac{\pi}{3}$$
If $$\cos 3x=-1$$, where $$0^{o} \ge x \ge 360^{o}$$, then $$x$$
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$$60^{o},180^{o},300^{o}$$
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$$180^{o}$$
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$$60^{o},180^{o}$$
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$$180^{o},300^{o}$$
Explanation
Given $$\cos 3 x=-1$$
$$\implies 3 x=2 n \pi\pm \pi=(2 n \pm 1)\pi$$ where $$n$$ is an integer
$$\implies x=\dfrac{(2 n\pm 1)\pi}{3}$$
$$\implies x=(2 n\pm 1)60^{\circ}=\pm 60^{\circ},\pm 180^{\circ},\pm 300^{\circ},\pm 420^{\circ},\cdots$$
As $$0^{\circ}\le x\le 360^{\circ},x=60^{\circ},180^{\circ},300^{\circ} $$
If $$\theta$$ is an acute angle such that $$\sec^2 \theta = 3$$. then the value of $$\dfrac{\tan^2 \theta - \text{cosec}^2 \theta}{\tan^2 \theta - \text{cosec}^2 \theta}$$ is
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$$\dfrac{4}{7}$$
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$$\dfrac{3}{7}$$
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$$1$$
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$$\dfrac{1}{7}$$
Explanation
Given $$\text{sec}^2 \theta=3$$
for any value of $$\theta$$,$$\dfrac{\tan^2 \theta-\text{cosec}^2 \theta}{\tan^2 \theta-\text{cosec}^2 \theta}=1$$ $$(\because \text{the numerator and denominator is same})$$
If $$3\sec\theta+2\sec\dfrac{\pi}{4}=2\cos\theta$$, where $$\theta\in (0, 2\pi)$$, then which of the following can be correct?
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$$\cos 2\theta=0$$
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$$\sin 2\theta=1$$
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$$\tan \theta=1$$
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$$\tan \theta=-1$$
Explanation
$$ 3\sec \theta +2\sec \frac{\pi }{4}=2\cos \theta $$
$$ \Rightarrow \frac{3}{\cos \theta }+2\sqrt{2}=2\cos \theta $$
$$ \Rightarrow \frac{3+2\sqrt{2}\cos \theta }{\cos \theta }=2\cos \theta $$
$$ \Rightarrow 3+2\sqrt{2}\cos \theta =2co{{s}^{2}}\theta $$
$$ \Rightarrow 2{{\cos }^{2}}\theta -2\sqrt{2}\cos \theta -3=0 $$
$$ \cos \theta =\frac{2\sqrt{2}\pm \sqrt{{{\left( 2\sqrt{2} \right)}^{2}}-4\left( 3 \right)\left( 2 \right)}}{4} $$
$$ \,\,\,\,\,\,\,\,\,=\frac{2\sqrt{2}\pm 4\sqrt{2}}{4}=\frac{\sqrt{2}\pm 2\sqrt{2}}{2} $$
$$ \,\,\,\,\,\,\,=\frac{3\sqrt{2}}{2},\frac{-1}{\sqrt{2}} $$
$$ \sin ce\,\,1\le \cos \theta \le -1 $$
$$ Hence\,\cos \theta =\frac{-1}{\sqrt{2}} $$
$$ \theta =\frac{\pi }{4},\frac{7\pi }{4}\,\,\,\,\,\,\because \theta \in \left( 0,2\pi \right) $$
$$ a)\,\cos 2\theta =1 $$
$$ \,\,\,\cos \left( 2\frac{\pi }{4} \right) $$
$$ =\cos \left( \frac{\pi }{2} \right) $$
$$ =0 $$
$$ b)\,sin2\theta =1 $$
$$ \,\,\,\sin \left( 2\frac{\pi }{4} \right) $$
$$ =\sin \left( \frac{\pi }{2} \right) $$
$$ =1 $$
$$ c)\,\tan \theta =1 $$
$$ \,\,\,\tan \left( \frac{\pi }{4} \right) $$
$$ =1 $$
$$ d)\tan \theta =-1 $$
$$ \,\,\,\tan \left( \frac{7\pi }{4} \right)=\tan \left( 2\pi -\frac{\pi }{4} \right) $$
$$ =-\tan \left( \frac{\pi }{4} \right) $$
$$ =-1 $$
For $$x\ \in\ (0,\pi)$$, the equation $$\sin x+2\sin 2x-\sin 3x=3$$ has
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infinitely solutions
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three solution
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one solution
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no solution
Explanation
$$\sin x+2\sin 2x-\sin 3x=3$$
$$\sin x+4\sin x\cos x=3+\sin 3x$$
$$\sin x (1+4\cos x)=3+\sin 3x$$
$$\sin x(1+4\cos x)=3+\sin x(3-4\sin^2 x)$$
$$\sin x(4\cos^2x-4\cos x-2)=-2$$
$$4\cos^2x-4\cos x-2=9$$
$$(2\cos x-1)^2-3 \ge -3$$
and on $$(0, \pi)$$ we have $$0 < \sin x \le 1$$ they
The $$LHS$$ of your last equation. should be $$\ge -3$$ on $$(0,\ \pi)$$
No Solution in $$(0,\ \pi)$$
The number of value of $$x\in (0,2\pi)$$ satisfying $$\log_{\tan x}(2+4\cos^2x)=2$$
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{6}$$
The sum $$s=sin\theta+sin2\theta+.......+sin n \theta$$, equals
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$$sin\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ sin \dfrac{\theta}{2}$$
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$$cos\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ sin \dfrac{\theta}{2}$$
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$$sin\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ cos \dfrac{\theta}{2}$$
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$$cos\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ cos\dfrac{\theta}{2}$$
If $$(1+\tan{A}).(1+\tan{B})=2$$, then $$A+B$$ is
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$$\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{6}$$
Explanation
$$(1+\tan A)(1+\tan B)=2$$
$$\therefore \tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$$ ........... $$(1)$$
Now $$1+\tan A+\tan B+\tan A\tan B=2$$
$$\therefore \tan A+\tan B=1-\tan A\tan B$$
$$\therefore\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1$$ ........ $$(2)$$
$$\therefore$$ From $$(1)$$ & $$(2)$$
$$\tan (A+B)=1$$
$$\therefore A+B=\dfrac{\pi}{4}$$
If $$3\sin { 2\theta } =2\sin { 3\theta }$$ and $$0<\theta <\pi$$, then value of $$\sin { \theta }$$ is
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$$\dfrac{\sqrt{2}}{3}$$
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`$$\dfrac{\sqrt{3}}{\sqrt{5}}$$
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$$\dfrac{\sqrt{15}}{4}$$
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$$\dfrac{\sqrt{2}}{\sqrt{5}}$$
Explanation
$$3 \sin 2 \theta = 2 \sin 3 \theta$$
$$\rightarrow 3 \times 2 \sin \cos \theta = 2(3 \sin \theta - 4 \sin{3} \theta )$$
$$\rightarrow 6 \sin \theta \cos \theta = 6 \sin \theta - 8 \sin{3} \theta$$
$$\rightarrow \sin \theta (6 \cos \theta - 6+8 \sin^{2} \theta )=0$$
$$\therefore \sin \theta =0$$ or $$3 \cos \theta+ 4 \sin{2} \theta =3$$
$$\therefore \sin \theta =0$$ or $$ 3 \cos \theta + 4 (1- \cos{2} \theta)=3$$
$$\therefore \sin \theta =0$$ or $$3 \cos \theta -4 \cos{2} \theta +1 =0$$
$$\therefore 4 \cos^{2} \theta- 3 \cos \theta -1 =0$$
$$\therefore 4 \cos^{2} \theta - 4 \cos \theta + \cos \theta -1 =0$$
$$\therefore 4 \cos \theta (\cos \theta -1) +1 (\cos \theta -1)=0$$
$$\therefore (4 \cos \theta +1) (\cos \theta -1)=0$$
$$\therefore \cos \theta = \dfrac{-1}{4}$$ or $$ \cos \theta =1$$
$$\therefore \sin \theta = \sqrt{1-\left( \dfrac{-1}{4} \right)^{2}}$$
$$\therefore \sin \theta =\sqrt{1- \dfrac{1}{16}}$$
$$\therefore \sin \theta = \dfrac{\sqrt{15}}{4}$$
If $$\cos2x\ +\ 2\cos\ x\ =\ 1$$, then $$\sin^{2}x(2-\cos^{2}x)$$ is
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$$1$$
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$$2$$
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$$4$$
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none of these
Explanation
Solution:- (A) $$1$$
$$\cos{2x} + 2 \cos{x} = 1$$
$$2 \cos^{2}{x} - 1 + 2 \cos{x} = 1 \; \left( \because \cos{2x} = 2 \cos^{2}{x} - 1 \right)$$
$$\Rightarrow 2 \cos^{2}{x} + 2 \cos{x} - 2 = 0$$
$$\Rightarrow \cos^{2}{x} + \cos{x} = 1 ..... \left( 1 \right)$$
$$\Rightarrow 1 - \cos^{2}{x} = \cos{x}$$
$$\Rightarrow \sin^{2}{x} = \cos{x} ..... \left( 2 \right)$$
Now,
$$\sin^{2} \left( 2 - \cos^{2}{x} \right)$$
$$= \sin^{2}{x} \left( 1 + \sin^{2}{x} \right)$$
$$= \cos{x} \left( 1 + \cos{x} \right) \; \left[ \text{From} \left( 2 \right) \right]$$
$$= \cos{x} + \cos^{2}{x}$$
$$= 1 \; \left[ \text{From} \left( 1 \right) \right]$$
Hence $$\sin^{2} \left( 2 - \cos^{2}{x} \right) = 1$$.
State whether the following statement is true or false:
$$cosec{20^ \circ }\sec {20^ \circ } = $$
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$$2$$
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$$2 cosec {40^ \circ}$$
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$$4$$
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$$2\sin{45^ \circ} . cosec {40^ \circ}$$
Explanation
$$csc 20 \sec 20$$
$$\implies \dfrac 1{\sin 20\cos 20}$$
$$\implies \dfrac 2{2\sin 20\cos 20}$$
$$\implies \dfrac 2{\sin 2(20)}$$
$$\implies \dfrac 2{\sin 40}$$
$$\implies 2{\csc 40}$$
$${ cos }^{ 2 }2x+2{ cos }^{ 2 }x=1,x\epsilon \left( -\pi ,\pi \right)$$, then $$x$$ can take the values
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$$\pm \dfrac { \pi }{ 2 }$$
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$$\pm \dfrac { \pi }{ 4 }$$
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$$\pm \dfrac { 3\pi }{ 8 }$$
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Non of these
Explanation
Given $$x\in (-\pi,\pi)\implies 2 x\in (-2\pi,2\pi)$$
$$\cos^2 2 x+2\cos^2 x=1$$
$$\implies \cos^2 2 x+2\cos^2 x-1=0$$
$$\implies \cos^2 2 x+(2\cos^2 x-1)=0$$
$$\implies \cos^2 2 x+\cos 2 x=0$$
$$\implies \cos 2 x(\cos 2 x+1)=0$$
So $$\cos 2 x=0$$ or $$-1$$
$$\implies 2 x=\pm \dfrac{\pi}{2},\pm \dfrac{3\pi}{2}$$ or $$\pm \pi$$
$$\implies x=\pm \dfrac{\pi}{4},\pm \dfrac{3\pi}{4}$$ or $$\pm\dfrac{\pi}{2}$$
The number of solutions to the equation $$sin\ 5x+\ sin\ 3x\ +\ sin\ x=0$$ for $$\ 0\leq x\leq \pi$$ is
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1
0%
2
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3
0%
None of these
Explanation
$$ sin x+sin3x+sin5x = 0 x\in [0,\pi ]$$
$$ (sinx+sin5x)+sin3x = 0$$
$$ 2sin 3x cos2x + sin3x = 0$$ $$\displaystyle [\because sinx+siny = 2\left | sin \dfrac{x+y}{2} \right | \left | cos\dfrac{x-y}{2} \right |]$$
$$ sin 3x (2cos2x+1) = 0$$
$$ \Rightarrow sin 3x = 0 , 2cos2x +1 = 0$$
$$ 3x = n\pi$$ $$\quad|$$ , $$cos 2x = \frac{-1}{2}$$
$$ x = \dfrac{n\pi }{3}$$$$\quad|$$ $$ 2x =2n\pi \pm \dfrac{2\pi }{3} [\because cos2\dfrac{2\pi }{3} = \dfrac{-1}{2}]$$
$$ \quad \quad \quad \quad \, | x = n\pi \pm \dfrac{\pi }{3}$$
For $$ x\in [ 0,\pi ], x = 0, \dfrac{\pi }{3}, \dfrac{2\pi }{3}, \pi $$ [ 4 solutions]
$$ \therefore $$ None of these is correct option
State true or false.
$${(\cos x + {\mathop{\rm cosy}\nolimits} )^2} + {(\sin x - \sin y)^2} = 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right)$$
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True
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False
Explanation
$$(\cos x+\cos y)^2+(\sin x-\sin y)^2$$ $$=\left[\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right)\right]^2+\left[2\cos\left(\dfrac{x+y}{2}\right)\sin\left(\dfrac{x-y}{2}\right)\right]^2$$
$$=\left[(2)^2\cos^2\left(\dfrac{x+y}{2}\right)\cos^2\left(\dfrac{x-y}{2}\right)\right]+\left[(2)^2\cos^2\left(\dfrac{x+y}{2}\right)\sin^2\left(\dfrac{x-y}{2}\right)\right]$$
$$=4\cos^2\left(\dfrac{x+y}{2}\right).\cos^2\left(\dfrac{x-y}{2}\right)+4\cos^2\left(\dfrac{x+y}{2}\right)\sin^2\left(\dfrac{x-y}{2}\right)$$
$$=4\cos^2\left(\dfrac{x+y}{2}\right)\left[\cos^2\left(\dfrac{x-y}{2}\right)+\sin^2\left(\dfrac{x-y}{2}\right)\right]$$ [ $$\sin^2\theta+\cos^2\theta=1$$ ]
$$=4\cos^2\left(\dfrac{x+y}{2}\right)\times 1$$
$$=4\cos^2\left(\dfrac{x+y}{2}\right)$$
$$=R.H.S$$
$$\therefore$$
$$(\cos x+\cos y)^2+(\sin x-\sin y)^2$$
$$=4\cos^2\left(\dfrac{x+y}{2}\right)$$
If $$\cos^{2}x=\sin x$$, then the value of $$\cos^{2}x(1+\cos^{2}x)$$ is equal to
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$$-1$$
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$$1$$
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$$0$$
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$$\dfrac{1}{2}$$
Explanation
Given $$\cos^2 x=\sin x$$
Squaring on both sides
$$\implies (\cos^2 x)^2=\sin^2 x$$
$$\implies \cos^4 x=1-\cos^2 x$$
$$\implies \cos^4 x+\cos^2 x=1$$
$$\implies \cos^2 x(1+\cos^2 x)=1$$
$$\dfrac{{\sec 8{\text{A}} - 1}}{{\sec 4{\text{A}} - 1}} = $$
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$$\dfrac{{\tan 2{\text{A}}}}{{\tan 8{\text{A}}}}$$
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$$\dfrac{{\tan 8{\text{A}}}}{{\tan 2{\text{A}}}}$$
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$$\dfrac{{{\text{cot8A}}}}{{\cot 2{\text{A}}}}$$
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$$\dfrac{{{\text{tan6A}}}}{{\tan 2{\text{A}}}}$$
Explanation
Solution :-
$$\displaystyle \frac{sec8A-1}{sec4A-1} = \frac{1/cos8A-1}{1/cos4A-1} $$
$$\displaystyle \Rightarrow \frac{\frac{1-cos8A}{cos8A}}{\frac{1-cos4A}{cos4A}} = \frac{cos4A(1-cos8A)}{cos8A(1-cos4A)} $$
$$\displaystyle \Rightarrow \frac{cos4A(1-(1-2sin^{2}4A))}{cos8A(1-(1-2sin^{2}2A))} = \frac{cos4A.sin^{2}2A}{cos8A.sin^{2}2A} $$
$$\displaystyle \Rightarrow \frac{(2cos4Asin4A)sin4A}{(2cos8A\,sin^{2}2A)} = \frac{sin8Asin4A}{2cos8A.sin^{2}2A} $$
$$\displaystyle \Rightarrow tan8A(\frac{sin4A}{2sin^{2}A}) = tan8A.(\frac{2sin2A.cos2A}{2sin^{2}2A}) $$
$$\displaystyle \Rightarrow \frac{tan8A}{tan2A} $$
If $$\tan \theta + \, \cot \theta = 4$$, then the value of $$\tan^3 \theta + \cot^3 \theta$$ is
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$$52$$
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$$16$$
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$$7 \dfrac{9}{8}$$
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$$27 \dfrac{1}{27}$$
Explanation
$$\tan\theta +\cot\theta =4$$
$$(\tan\theta +\cot\theta )^{3}=4^{3}$$
$$\tan^{3}\theta +\cot^{3}\theta +(\tan\theta +\cot\theta )3\tan\theta \cot\theta =64 $$
$$\tan^{3}\theta +\cot^{3}\theta +(4)3=64 $$
$$\tan^{3}\theta +\cot^{3}\theta =64-12 $$
$$\tan^{3}\theta +\cot^{3}\theta =52$$
The positive integer value of $$n>3$$ satisfing the equation
$$\frac{1}{{\sin \left( {\frac{\pi }{n}} \right)}} = \frac{1}{{\sin \left( {\frac{{2\pi }}{n}} \right)}} + \frac{1}{{\sin \left( {\frac{{3\pi }}{n}} \right)}}$$
is
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$$7$$
0%
$$8$$
0%
$$9$$
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$$10$$
If $$cot\Theta =sin2\Theta (where\Theta \neq n\pi ,n\ is\ an\ integer)\Theta $$=
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45$$^{0}$$ and 60$$^{0}$$
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45 $$^{0}$$ and 90 $$^{0}$$
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only 45$$^{0}$$
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only 90$$^{0}$$
Explanation
$$\cot\theta =\sin 2\theta$$
$$\Rightarrow \dfrac{\cos\theta}{\sin\theta}=2\sin\theta \cos\theta$$
$$\Rightarrow \cos\theta =2\sin^2\theta \cos\theta$$
$$\Rightarrow \cos\theta -2\sin^2\theta \cos\theta =0$$
$$\Rightarrow \cos\theta(1-2\sin^2\theta)=0$$
$$\Rightarrow \cos\theta(\cos 2\theta)=0$$
$$\Rightarrow \cos\theta =0$$ or $$\cos 2\theta =0$$
$$\Rightarrow \theta =\dfrac{\pi}{2}$$ or $$2\theta =\dfrac{\pi}{2}$$
$$\Rightarrow \theta =\dfrac{\pi}{4}$$
$$\therefore \theta$$ can be $$45^o$$ or $$90^o$$.
$$(\sec\ A-\cos\ A)(\sec\ A+\cos\ A)=\sin^{2}\ A+\tan^{2}A$$.
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True
0%
False
Explanation
$$\left(\sec{A}-\cos{A}\right)\left(\sec{A}+\cos{A}\right)$$
$$={\sec}^{2}{A}-{\cos}^{2}{A}$$
$$=1+{\tan}^{2}{A}-{\cos}^{2}{A}$$
$$=1-{\cos}^{2}{A}+{\tan}^{2}{A}$$
$$={\sin}^{2}{A}+{\tan}^{2}{A}$$
In a triangle ABC,
$$b\,cos\,(C+\theta)+c\,cos(B-\theta)=$$
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$$a\,cos\,\theta$$
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$$a\,sin\,\theta$$
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$$a\,tan\,\theta$$
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$$a\,cot\,\theta$$
Explanation
In $$\triangle ABC,A+B+C=\pi$$ and $$a=2 R\sin A,b=2 R\sin B,c=2 R \sin C$$
$$b\cos (C+\theta)+c\cos (B-\theta)=R(2\sin B\cos (C+\theta)+2\sin C\cos (B-\theta))$$
$$=R(\sin (B+C+\theta)+\sin (B-C-\theta)+\sin (B+C-\theta)+\sin (C-B+\theta))$$
$$=R(\sin (\pi-(A-\theta))+\sin (\pi-(A+\theta))+\sin (B-C-\theta)-\sin (B-C-\theta))$$
$$=R(\sin (A-\theta)+\sin (A+\theta))$$
$$=R(2\sin A\cos \theta)$$
$$=(2 R\sin A)\cos \theta$$
$$=a\cos \theta$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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