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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 12
$$\cos^ {2}\theta \left(1+\tan^ {2}\theta\right)=1$$
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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 $$\text{True}$$
$$\csc^{6}\theta -\cot^{6}\theta =1+3\csc^{2}\theta \cot^{2}\theta$$
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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 $$\text{True}$$
If $$x=a\cos\theta+b\sin\theta$$ and $$y=a\sin\theta-b\cos\theta$$, then $$a^{2}+b^{2}=x^{2}+y^{2}$$.
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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, $$x^2+y^2=a^2+b^2$$.
$$(\cos A-\csc A)^{2}+(\sin A-\sec A)^{2}=(1-\sec A.\csc A)^{2}$$
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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.
If $$\sin \theta +\sin^{2}\theta +\sin^{3}\theta =1$$, then $$ \cos^{6}\theta -4\cos^{4}\theta +8 \cos^{2}\theta =4$$.
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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 $$\text{True}$$
$$\sqrt{\dfrac{\sec A-1}{\sec A+1}}+\sqrt{\dfrac{\sec A+1}{\sec A-1}}=2\csc A$$
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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.
$$ \dfrac{\tan^{2}\theta }{\sec^{2}\theta}+\dfrac{\cot^{2}\theta }{cosec^{2}\theta}=1 $$
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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.
$$\dfrac { \sin A }{ \sec A+\tan A-1 } +\dfrac { \cos A }{ \text{cosec}A+\cot A-1 } =1$$
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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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Check whether following statement is true or false.
$${ \cos }^{ 4 } \theta -{ \sin }^{ 4 }\theta = 1+2{ \sin }^{ 2 } \theta $$
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State whether the following statement is true or false.
$$\dfrac { 1 }{ \sec { A } +\tan { A } } -\dfrac { 1 }{ \cos { A } } =\dfrac { 1 }{ \cos { A } } -\dfrac { 1 }{ \sec { A } -\tan { A } } $$
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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
$$\dfrac{\cos\theta-\sin \theta+1}{\cos\theta+\sin \theta-1}=\text{cosec} \theta+\cot \theta$$
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Explanation
Given
LHS
$$\dfrac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$$
Dividing Numerator and Denominator by $$\sin \theta $$
$$\implies \dfrac{\cot \theta +\text{cosec} \theta-1}{\cot \theta-\text{cosec} \theta+1}$$
$$\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} ^2\theta-\cot^2\theta)}{\cot \theta-\text{cosec} \theta+1}$$
$$\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} \theta-\cot\theta)(\text{cosec} \theta+\cot \theta)}{\cot \theta-\text{cosec} \theta+1}$$
$$\implies \dfrac{\cot \theta +\text{cosec} \theta-(\text{cosec} \theta-\cot\theta)(\text{cosec} \theta+\cot \theta)}{\cot \theta-\text{cosec} \theta+1}$$
$$\implies \dfrac{(\cot \theta +\text{cosec} \theta)(\cot \theta-\text{cosec} \theta+1)}{\cot \theta-\text{cosec} \theta+1}$$
$$\implies \text{cosec} \theta+\cot \theta$$
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 $$ \text{cosec}^{ 2 }A=1+{ \cot }^{ 2 }A.)\quad $$
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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
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$$0$$
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$$\cos^{2}\theta$$
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$$\tan^{2}\theta$$
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$$\dfrac{1}{\sin^{2}\theta}$$
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.
$$\dfrac { \cot A-\cos A }{ \cot A+\cos A } =\dfrac { \text{cosec}A-1 }{ \text{cosec}A+1 } $$
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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.
$$\dfrac{1}{secA-1}+\dfrac{1}{secA+1}=2cosecAcotA$$
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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.
$$\sec { ^{ 4 } } A\left( 1-\sec { ^{ 4 } } A \right) -2\tan { ^{ 2 } } A=1$$
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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.
If $$\cos A+{ \cos }^{ 2 }A=1,$$ then $${ \sin }^{ 2 }A+{ \sin }^{ 4 }A=1.$$
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Explanation
Given:
$$cosA+cos^2A=1$$
$$\implies cosA=1-cos^2A=sin^2A$$..........[1]
Now,
$$sin^2A+sin^4A=cosA+cos^2A=1$$.....................[from $$1$$]
If $${ cosec }\theta -cot\theta =p$$, then $$cosec\theta +cot\theta =$$
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1/p
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-1/p
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-p
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$${ p }^{ 2 }$$
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.
$$\dfrac { 1+\cos\theta -\sin^{ 2 }\theta }{ \sin { \theta \left( 1+\cos { \theta } \right) } } =\cot { \theta }$$
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Explanation
$$\dfrac { 1+\cos\theta -\sin^{ 2 }\theta }{ \sin { \theta \left( 1+\cos { \theta } \right) } } =\cot { \theta }$$
$$\Rightarrow$$ $$L.H.S=\dfrac{1+\cos\theta-\sin^2\theta}{\sin\theta(1+\cos\theta)}$$
$$=\dfrac{1+\cos\theta-(1-\cos^2\theta)}{\sin\theta(1+\cos\theta)}$$
$$=\dfrac{1+\cos\theta-1+\cos^2\theta}{\sin\theta(1+\cos\theta)}$$
$$=\dfrac{\cos\theta+\cos^2\theta}{\sin\theta(1+\cos\theta}$$
$$=\dfrac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)}$$
$$=\dfrac{\cos\theta}{\sin\theta}$$
$$=\cot\theta$$
$$=R.H.S$$
$$\dfrac{{\tan\theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta .\text{cosec}\theta $$
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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
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$$\pm \frac{1}{2}$$
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$$\frac{1}{\sqrt 2}$$
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$$\pm \frac{1}{3}$$
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$$2$$
The maximum value of $$2 \sin x-4 \cos x$$ is
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$$-\sqrt { 12 } $$
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$$2$$
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$$\sqrt { 20 } $$
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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
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$$\pm\dfrac {1}{2}$$
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$$\dfrac {1}{\sqrt {2}}$$
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$$\pm\dfrac {1}{3}$$
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$$2$$
If $$x_{1}$$ 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
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$$a/b$$
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$$b/a$$
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$$c/a$$
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$$a/c$$
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.
$$\dfrac{1-\sin A}{1+\sin A}=(\sec A -\tan A)^2$$
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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.
If $$sin{(2cos^{-1}(\dfrac{1}{\sqrt{5}})} + cos{(2tan^{-1}(\dfrac{1}{3}))}=\dfrac{p}{q}$$, where p & q are relatively prime then digit at units place of $${(p-q)}^{2k+1}, k\epsilon{N}$$, can be ________.
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1
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3
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7
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9
Given $$A={ sin }^{ 2 }\theta +{ cos }^{ 4 }\theta $$, then for all real $$\theta $$
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$$1\le A\le 2$$
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$$\cfrac { 3 }{ 4 } \le A\le 1$$
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$$\cfrac { 13 }{ 16 } \le A\le 1$$
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$$\cfrac { 3 }{ 4 } \le A\le \cfrac { 13 }{ 16 } $$
The values of x in $$\left(0, \dfrac{\pi}{2}\right)$$ satisfying the equation $$\sin x\cos x=\dfrac{1}{4}$$ are ________.
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$$\dfrac{\pi}{6}, \dfrac{\pi}{12}$$
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$$\dfrac{\pi}{12}, \dfrac{5\pi}{12}$$
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$$\dfrac{\pi}{8}, \dfrac{3\pi}{8}$$
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$$\dfrac{\pi}{8}, \dfrac{\pi}{4}$$
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 ?
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$$\tan A$$
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$$\cot A$$
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$$2\tan A$$
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$$2\cot A$$
Explanation
$$\cot \left(\dfrac{A}{2}\right)-\tan\left(\dfrac{A}{2}\right)$$
$$\Rightarrow$$ $$\dfrac{\cos \dfrac{A}{2}}{\sin\dfrac{A}{2}}-\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$$
$$\Rightarrow$$ $$\dfrac{\cos^2\dfrac{A}{2}-\sin^2\dfrac{A}{2}}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$
$$\Rightarrow$$ $$\dfrac{\cos 2\left(\dfrac{A}{2}\right)}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$
$$\Rightarrow$$ $$\dfrac{\cos A}{\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$
$$\Rightarrow$$ $$\dfrac{2\cos A}{2\sin\dfrac{A}{2}.\cos\dfrac{A}{2}}$$
$$\Rightarrow$$ $$\dfrac{2\cos A}{\sin 2\left(\dfrac{A}{2}\right)}$$
$$\Rightarrow$$ $$\dfrac{2\cos A}{\sin A}$$
$$\Rightarrow$$ $$2\cot A$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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