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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 6 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 6
$$\sin30^{\small\circ} + \cos60^{\small\circ}$$ equals
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$$\displaystyle\frac{1+\sqrt3}{2}$$
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$$\sqrt3$$
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$$1$$
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None of these
Explanation
$$\sin30^{0}+\cos60^{0}$$
$$=\sin30^{0}+\sin(90^{0}-60^{0})$$
$$=2\sin30^{0}$$
$$=2\times\dfrac{1}{2}$$
$$=1$$
In $$\triangle ABC$$, $$\angle B = 90^{\small\circ}$$. If $$AB = 14\space cm$$ and $$AC = 50\space cm$$ then $$\tan A$$ equals:
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$$\displaystyle\frac{24}{25}$$
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$$\displaystyle\frac{24}{7}$$
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$$\displaystyle\frac{7}{24}$$
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$$\displaystyle\frac{25}{24}$$
Explanation
In triangle ABC
$$B=90^{0}$$
Hence
$$AC=50cm$$ will be the hypotenuse.
And $$AB=14cm$$
Therefore
$$BC=\sqrt{AC^{2}-AB^{2}}cm$$
$$=48cm$$
Hence
$$tanA$$
$$=\dfrac{perpendicular}{base}$$
$$=\dfrac{BC}{AB}$$
$$=\dfrac{48}{14}$$
$$=\dfrac{24}{7}$$
The value of the expression $$[\text{cosec(}75^{\small\circ}+\theta) - \sec(15^{\small\circ}- \theta) - \tan(55^{\small\circ} + \theta) + \cot(35^{\small\circ} - \theta)]$$ is
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$$-1$$
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$$0$$
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$$1$$
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$$\displaystyle\frac{3}{2}$$
Explanation
$$cosec(75^{0}+\theta)-\sec(15^{0}-\theta)-\tan(55^{0}+\theta)+\cot(35^{0}-\theta)$$
$$=\sec(90^{0}-75^{0}-\theta)-\sec(15^{0}-\theta)-\cot(90^{0}-55^{0}-\theta)+\cot(35^{0}-\theta)$$
$$=\sec(15^{0}-\theta)-\sec(15^{0}-\theta)-\cot(35^{0}+\theta)+\cot(35^{0}+\theta)$$
$$=0$$
If $$\sec\theta = \displaystyle\frac{\sqrt{p^2 + q^2}}{q}$$, then the value of $$\displaystyle\frac{p\sin\theta - q\cos\theta}{p\sin\theta + q\cos\theta}$$
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$$\displaystyle\frac{p}{q}$$
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$$\displaystyle\frac{p^2}{q^2}$$
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$$\displaystyle\frac{p^2 - q^2}{p^2+q^2}$$
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$$\displaystyle\frac{p^2 + q^2}{p^2-q^2}$$
Explanation
$$\sec\theta=\cfrac{hypotenuse}{base}=\cfrac{\sqrt{p^{2}+q^{2}}}{q}$$
Hence,
$$hypotenuse=\sqrt{p^{2}+q^{2}}$$
$$base=q$$
Therefore by applying, Pythagoras theorem, we get the altitude as
$$p$$.
Hence
$$\sin\theta=\cfrac{p}{\sqrt{p^{2}+q^{2}}}$$
$$\cos\theta=\cfrac{q}{\sqrt{p^{2}+q^{2}}}$$
Hence substituting, the values in the above question we get
$$\cfrac{p\sin\theta-q\cos\theta}{p\sin\theta+q\cos\theta}$$
$$=\cfrac{p^{2}-q^{2}}{p^{2}+q^{2}}$$
Is LHS=RHS?
$$\left\{\displaystyle\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}\right\}^2 = \displaystyle\frac{1+\sin\theta}{1-\sin\theta},\quad 1-\cos\theta \ne 0$$
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Yes
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No
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Ambiguos
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Data insufficient
Explanation
Given
LHS
$$ \left\{\dfrac{1+\cos \theta+ \sin \theta}{1+\cos \theta- \sin \theta} \right\}^2$$
$$ \dfrac{1+\cos ^2 \theta +\sin ^2 \theta +2\cos \theta +2 \sin \theta + 2\cos \theta \sin \theta }{1+\cos ^2 \theta +\sin ^2 \theta +2\cos \theta -2 \sin \theta -2\cos \theta \sin \theta }$$
$$ \dfrac{2 +2\cos \theta +2 \sin \theta + 2\cos \theta \sin \theta }{2 +2\cos \theta -2 \sin \theta -2\cos \theta \sin \theta }$$
$$ \dfrac{(1+\cos \theta) + \sin \theta (1 + \cos \theta) }{(1+\cos \theta) - \sin \theta (1 + \cos \theta)}$$
$$ \dfrac{(1+\cos \theta )(1+\sin \theta)}{(1+\cos \theta )(1-\sin \theta)}$$
$$ \implies \dfrac {1+\sin \theta }{1-\sin \theta }$$
If $$\sec{\theta}+\tan{\theta}=P$$, then the value of $$\sin{\theta}$$ is
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$$\displaystyle\frac{{P}^{2}+1}{2P}$$
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$$\displaystyle\frac{{P}^{2}-1}{2P}$$
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$$\displaystyle\frac{{P}^{2}-1}{{P}^{2}+1}$$
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$$\displaystyle\frac{{P}^{2}+1}{{P}^{2}-1}$$
Explanation
$$\sec { \theta } +\tan { \theta } =P\\ \Rightarrow { \left( \sec { \theta } +\tan { \theta } \right) }^{ 2 }={ P }^{ 2 }\\ \Rightarrow \sec ^{ 2 }{ \theta } +\tan ^{ 2 }{ \theta } +2\tan { \theta } \sec { \theta } ={ P }^{ 2 }$$
$$\displaystyle \Rightarrow 2\tan ^{ 2 }{ \theta } +1+\frac { 2\sin { \theta } }{ \cos { \theta } } ={ P }^{ 2 }$$
$$\displaystyle \Rightarrow \frac { 2\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } }{ \cos ^{ 2 }{ \theta } } ={ P }^{ 2 }$$
Applying componendo and dividendo
$$\displaystyle \frac { 2\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } -\cos ^{ 2 }{ \theta } }{ 2\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } -\cos ^{ 2 }{ \theta } } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } $$
$$\displaystyle \Rightarrow \frac { 2\sin { \theta } \left( \sin { \theta } +1 \right) }{ 2\left( \sin { \theta } +1 \right) } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } \Rightarrow \sin { \theta } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } $$
Is LHS=RHS?
$$\sqrt{\displaystyle\frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta}} + \sqrt{\displaystyle\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta}} = 2\csc\theta$$
Say true or false.
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True
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False
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Ambiguous
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Data insufficient
Explanation
$$LHS =\sqrt{\displaystyle\frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta}} + \sqrt{\displaystyle\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta}}\\=\dfrac{\csc\theta - \cot\theta + \csc\theta + \cot\theta}{\sqrt{\csc^{2}\theta - \cot^{2}\theta}}$$
$$= \dfrac{2\csc\theta}{1}$$
$$= 2\csc\theta \\= RHS$$
If $$\sin\theta+co\sec\theta=2$$, then the value of $$\sin^8\theta+co\sec^8\theta$$ is equal to
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$$2$$
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$$2^8$$
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$$2^4$$
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none of these
Explanation
$$\sin\theta +cosec\theta =2\\ \Rightarrow \sin\theta +\cfrac { 1 }{ \sin\theta } =2$$
Squaring both sides
$${ \left( \sin\theta +\cfrac { 1 }{ \sin\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta } +2\sin\theta \cfrac { 1 }{ \sin\theta } =4\\ \Rightarrow { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta } =2$$
Again squaring bot sides
$${ \left( { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta } \right) }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { \sin }^{ 4 }\theta +\cfrac { 1 }{ { \sin }^{ 4 }\theta } =2$$
Similarly
$${ \sin }^{ 8 }\theta +\cfrac { 1 }{ { \sin }^{ 8 }\theta } =2$$
$$\Rightarrow { \sin }^{ 8 }\theta +{ cosec }^{ 8 }\theta =2$$
If $$\cos 9\alpha = \sin \alpha$$, and $$9\alpha < 90^{\circ}$$, then $$\tan 5\alpha = .....$$
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$$\dfrac {1}{\sqrt {3}}$$
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$$\sqrt {3}$$
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$$1$$
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$$0$$
Explanation
$$\cos 9\alpha = \sin \alpha$$,
So, $$\cos 9\alpha = \cos (90 - \alpha)\\\therefore 9\alpha = 90 - \alpha\Rightarrow 9\alpha + \alpha = 90^{\circ}\Rightarrow 10\alpha=90^o\Rightarrow \alpha=9^o$$
Now, $$\tan5\alpha=\tan(5*9^o)=\tan45^o=1$$
Therefore, Answer is $$1$$
If $$ \sin{x}+\sin^{2}{x}=1$$, then the value of $$\cos^{2}{ x}+\cos^{4}{x}$$ is
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$$0$$
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$$2$$
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$$1$$
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None of these
Explanation
$$\textbf{Find the value of the given expression}$$
$$\text{We have given,}$$
$$ \sin{x}+\sin^{2}{x}=1$$
$$\Rightarrow\sin{x}=1-\sin^{2}{x}$$
$$\Rightarrow \sin x=\cos^{2}{x}$$
$$\boldsymbol{[\because\sin^2 x+ \cos^2 x=1]}$$
$$\text{Now value of required eq. is obtained by}$$
$$\text{Substituting the value of}\cos^2 x$$ $$\text{in given equation} $$
$$\cos^{2}{x}+\cos^{4}{x}$$
$$=\sin{x}+\sin^{2}{x}$$
$$=1$$
$$\Rightarrow \cos^{2}{x}+\cos^{4}{x}=1$$
$$\textbf{Hence value of the given expression is 1}$$
$$2(\sin^{6}{\theta}+\cos^{6}{\theta})-3(\sin^{4}{\theta}+\cos^{4}{\theta})+1$$ is equal to
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$$2$$
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$$0$$
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$$4$$
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$$6$$
Explanation
$$2\left( \sin ^{ 6 }{ \theta } +\cos ^{ 6 }{ \theta } \right) -3\left( \sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } \right) +1$$
Using $$(a^3+b^3=(a+b)^3-3\cdot a\cdot b(a+b)) \ and \ (a^2+b^2=(a+b)^2-2ab) $$, we get
$$ =2\left( { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 3 }-3\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) \right) -3\left( { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 2 }-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } \right) +1$$
$$ =2-6\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } -3+6\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } +1=0$$
If $$\sin \theta + \cos \theta = 1$$, then $$\sin \theta \cos \theta =$$.
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$$0$$
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$$1$$
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$$2$$
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$$\dfrac {1}{2}$$
Explanation
$$\textbf{Step 1: Apply relevant identity of a trigonometric function and simplify}$$
$$\text{We have,}$$
$$\sin\theta + \cos\theta = 1$$
$$\text{Squaring both sides, we get,}$$
$$\left(\sin\theta + \cos\theta\right)^2 = \left(1\right)^2$$
$$\Rightarrow \sin^2\theta + \cos^2\theta + 2\sin\theta.\cos\theta = 1$$ $$\boldsymbol{\left[\because\left(a + b\right)^2 = a^2 + b^2 + 2ab\right]}$$
$$\Rightarrow 1 + 2\sin\theta\cos\theta = 1$$
$$\Rightarrow \sin\theta\cos\theta = 0$$
$$\textbf{Hence, the answer is 0}$$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is incorrect but Reason is correct
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Both Assertion and Reason are incorrect
Explanation
Given, $$\sin\theta+\dfrac{1}{\sin\theta}=2$$
$$(\sin\theta+\dfrac{1}{\sin\theta})^{2}=4$$
$$\sin^{2}\theta+\dfrac{1}{\sin^{2}\theta}=2$$
$$(\sin^{2}\theta+\dfrac{1}{\sin^{2}\theta})^{2}=4$$
$$\sin^{4}\theta+\dfrac{1}{\sin^{4}\theta}=2$$
Therefore
$$\sin^{2k}\theta+\dfrac{1}{\sin^{2k}\theta}=2$$ where $$k=1,2,..N$$
Hence assertion is incorrect.
$$a+b=2$$ and $$ab=1$$
Then $$a=b=1$$ satisfies the above equation.
Hence reason is correct.
Find $$\theta$$, if $$\displaystyle\frac{2\tan\displaystyle\frac{\theta}{2}}{1 + \tan^2\displaystyle\frac{\theta}{2}} = 1,\quad 0^{\small\circ} < \theta \le 90^{\small\circ}$$
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$$\theta = 60^{\small\circ}$$
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$$\theta = 40^{\small\circ}$$
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$$\theta = 90^{\small\circ}$$
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$$\theta = 70^{\small\circ}$$
Explanation
Put $$ \displaystyle\frac{\theta}{2} = A$$, then we must have $$\displaystyle\frac{2\tan A}{1+\tan^2A} = 1$$
$$ \Rightarrow 1+\tan^2A - 2\tan A = 0$$
$$ \Rightarrow(1-\tan A)^2 = 0$$
$$ \Rightarrow\tan A = 1 = \tan45^{\small\circ}$$
$$A = 45^{\small\circ}$$
We know $$\displaystyle\frac{\theta}{2} = A = 45^{\small\circ} $$
$$\Rightarrow \displaystyle\frac{\theta}{2} = 45^{\small\circ} $$
$$ \Rightarrow \theta = 90^{\small\circ}$$
If $$(\sec{A}-\tan{A})(\sec{B}-\tan{B})(\sec{C}-\tan{C})=(\sec{A}+\tan{A})(\sec{B}+\tan{B})(\sec{C}+\tan{C})$$ represents each side of a equilateral triangle, then each side is equal to -
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$$0$$
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$$1$$
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$$-1$$
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$$\pm 1$$
Explanation
Let $$\left ( \sec A+\tan A \right )\left ( \sec B+\tan B \right )\left ( \sec C+\tan C \right )=x$$............(i)
$$\therefore \left ( \sec A-\tan A \right )\left ( \sec B-\tan B \right )\left ( \sec C-\tan C \right )=x$$.............(ii)
Multiplying Eqs. (i) and (ii), we get
$$\left ( \sec ^{2}A-\tan ^{2}A \right )\left ( \sec ^{2}B-\tan ^{2}B \right )\left ( \sec ^{2}C-\tan ^{2}C \right )=x^{2}$$...........
$$\left ( \sec ^{2}A-\tan ^{2}A \right )=1$$
or $$x^{2}=1$$ $$\therefore $$ $$x=\pm 1$$
Hence, each side is equal to $$\pm 1$$.
If $$\displaystyle \sin \theta +\sin ^{2}\theta = 1,$$ then $$\displaystyle \cos ^{2}\theta +\cos ^{4}\theta =$$
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$$1$$
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$$\displaystyle \sqrt{2}$$
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$$0$$
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$$2$$
Explanation
Given: $$\displaystyle \sin\theta + \sin^{2}\theta=1$$
$$\therefore \displaystyle \sin\theta=1-\sin^{2}\theta$$
$$\therefore \sin \theta = \cos^{2}\theta$$
Now, $$ \cos^{2}\theta+\cos^{4}\theta$$ $$=\sin\theta+\sin^{2}\theta=1$$
If $$\displaystyle 4\sin \theta =3\cos \theta ,$$ Then $$\displaystyle \frac{\sec ^{2}\theta }{4(1-\tan ^{2}\theta )}$$ is
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$$\displaystyle \frac{25}{16}$$
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$$\displaystyle \frac{25}{28}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{16}{25}$$
Explanation
Given: $$4 \sin \theta = 3 \cos \theta$$
$$\implies \tan \theta = \dfrac{3}{4}$$
Now, $$\displaystyle \dfrac{\sec ^{2}\theta }{4(1-\tan ^{2}\theta )}$$
$$=\displaystyle \dfrac{1 + \tan ^{2}\theta }{4(1-\tan ^{2}\theta )}$$
$$=\displaystyle \dfrac{1 + \left(\dfrac{3}{4}\right)^{2} }{4\left[1 - \left(\dfrac{3}{4}\right)^{2}\right]}$$
$$=\displaystyle \dfrac{16+ 9}{4(16 - 9)}$$
$$=\displaystyle \dfrac{25}{28}$$
If $$\displaystyle \tan \theta =\frac{x}{y},$$ then $$\displaystyle \frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$$ is equal to
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$$\displaystyle \frac{x^{2}+y^{2}}{x^{2}-y^{2}}$$
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$$\displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$$
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$$\displaystyle \frac{x}{\sqrt{x^{2}+y^{2}}}$$
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$$\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}$$
Explanation
$$\displaystyle \cfrac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$$
Multiply and divide by $$\cos \theta$$
$$=\cfrac{x\cfrac{\sin \theta }{\cos \theta }+y\cfrac{\cos \theta }{\cos \theta }}{x\cfrac{\sin \theta }{\cos \theta }-y\cfrac{\cos \theta}{\cos \theta }}$$
$$=\cfrac{x\times\cfrac{x}{y}+y}{x\times\cfrac{x}{y}-y}$$
$$=\cfrac{\cfrac{x^{2}+y^{2}}{y}}{\cfrac{x^{2}-y^{2}}{y}}$$
$$=\cfrac{x^{2}+y^{2}}{x^{2}-y^{2}}$$
If $$x = a \sec$$ $$\displaystyle \theta $$ + b tan $$\displaystyle \theta $$ and $$y = b \sec$$ $$\displaystyle \theta $$ + a tan
$$\displaystyle \theta $$, then $$\displaystyle x^{2}-y^{2}$$ is equal to
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$$\displaystyle 4ab \sec \theta \tan \theta $$
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$$\displaystyle a^{2}-b^{2}$$
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$$\displaystyle b^{2}-a^{2}$$
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$$\displaystyle a^{2}+b^{2}$$
Explanation
Given, $$x=a\sec \theta+b\tan \theta, y=b\sec \theta +a \tan \theta$$
$$\displaystyle x^{2}-y^{2}$$ $$=\left ( a\sec\theta+b\tan\theta \right )^{2}-\left ( b\sec\theta+a\tan\theta \right )^{2}$$
$$=a^{2}\sec^{2}\theta+2ab\sec\theta\tan\theta+b^{2}tan^2 \theta-\left ( b^{2}\sec^{2}\theta+2ab\sec\theta\tan\theta+a^{2}\tan^{2}\theta \right )$$
$$=\left ( a^{2}-b^{2} \right )\sec^{2}\theta+\left ( b^{2}-a^{2} \right )\tan^{2}\theta$$
$$=\left ( a^{2} - b^{2} \right )\left ( \sec^{2}\theta-\tan^{2}\theta \right )\left ......[ \because a^{2}-b^{2} =-\left ( b^{2}-a^{2} \right ) \right ]$$
$$=a^{2}-b^{2}$$ $$......\left ( \because\sec^{2}\theta-\tan^{2}\theta=1 \right )$$
Hence option $$'B'$$ is the answer.
Consider the following:
1.
$$\displaystyle \tan ^{2}\theta -\sin ^{2}\theta =\tan ^{2}\theta \sin ^{2}\theta $$
2.
$$\displaystyle (1+\cot ^{2}\theta )(1-\cos \theta )(1+\cos \theta )=1$$
Which of the statements given below is correct ?
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1 only is the identity
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2 only is the identity
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Both 1 and 2 are identities
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Neither 1 nor 2 is the identity
Explanation
$$\displaystyle 1.\tan^{2}\theta-\sin^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}-\sin^{2}\theta$$
$$=\dfrac{\sin^{2}\theta-\sin^{2}\theta\cos^{2}\theta}{\cos^{2}\theta}$$
$$=\dfrac{\sin^{2}\theta\left ( 1-\cos^{2}\theta \right )}{\cos^{2}\theta}$$
$$=\dfrac{\sin^{2}\theta}{\cos^{2}\theta}\times\sin^{2}\theta=\tan^{2}\theta\sin^{2}\theta\:$$
$$2.\left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+cos\theta \right )$$
$$=\left ( 1+\cot^{2}\theta \right )\left ( 1-\cos^{2}\theta \right )$$
$$=\left ( 1+\frac{\cos^{2}\theta}{\sin^{2}\theta} \right )\left ( 1-\cos^{2}\theta \right )$$
$$= \left ( \dfrac{\sin^{2}\theta+\cos^{2}\theta}{\sin^{2}\theta} \right )\times\sin^{2}\theta$$ .......[Since, $$(1-\cos^{2}\theta=\sin^{2}\theta )$$]
$$=1$$ .......[Since, $$\left ( \cos^{2}\theta+\sin^{2}\theta=1 \right )$$]
Therefore, both are Identities.
If $$\sin{x}+\sin^{2}{x}=1$$, then $$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$$ is equals to
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$$0$$
0%
$$1$$
0%
$$-1$$
0%
$$2$$
Explanation
Given $$\sin{x}+\sin^{2}{x}=1$$
$$\Rightarrow \sin x=\cos^2 x$$
Now,$$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$$
$$= \sin^{6}{x}+3\sin^{5}{x}+3\sin^{4}{x}+\sin^{3}{x}-2$$
$$={ (\sin ^{ 2 }{ x } ) }^{ 3 }+3{ (\sin ^{ 2 }{ x } ) }^{ 2 }\sin { x } +3{ (\sin ^{ 2 }{ x } ) }^{ 2 }+\sin ^{ 2 }{ x } \sin { x } -2$$
$$={ (\sin ^{ 2 }{ x } ) }^{ 3 }+3{ (\sin ^{ 2 }{ x } ) }^{ 2 }\sin { x } +3\sin ^{ 2 }{ x } { (\sin { x } ) }^{ 2 }+\sin ^{ 3 }{ x } -2$$
$$={(\sin^{2}{x}+\sin{x})}^{3}-2$$
$$=1-2=-1$$
If $$\displaystyle \tan 2A= \cot (A-60^{\circ}),$$ where 2A is an acute angle, then the value of A is
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$$\displaystyle 30^{\circ}$$
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$$\displaystyle 60^{\circ}$$
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$$\displaystyle 50^{\circ}$$
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$$\displaystyle 24^{\circ}$$
Explanation
$$\displaystyle \tan 2A= \cot (A-60^{\circ})$$
$$\cot (90 - 2A) = \cot (A - 60)$$
$$90 - 2 A = A - 60$$
$$3A = 150$$
$$A = 50^{\circ}$$
If $$\displaystyle x = a\cos ^{3}\theta $$ and y = b $$\displaystyle \sin ^{3}\theta ,$$ then the value of $$\displaystyle \left ( \frac{x}{a} \right )^{2/3}+\left ( \frac{y}{b} \right )^{2/3}$$ is
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1
0%
-2
0%
2
0%
-1
Explanation
Given: $$x = a \cos^3 \theta$$ and $$y = b \sin^3 \theta$$
$$\therefore \ \dfrac xa = \cos^3 \theta$$ and $$\dfrac yb = \sin^3 \theta$$
Hence $$\displaystyle \left ( \dfrac{x}{a} \right )^{\dfrac{2}{3}}+\left ( \dfrac{y}{b} \right )^{\dfrac{2}{3}}$$
$$=\left ( {{\cos^{3}\theta}} \right )^{\dfrac{2}{3}}+\left ( {{\sin^{3}\theta}} \right )^{\dfrac{2}{3}}$$
$$=\left ( \cos\theta \right )^{2}+\left ( \sin\theta \right )^{2}$$
$$=\sin^{2}\theta+\cos^{2}\theta$$
$$=1$$
The value of $$\displaystyle \cot 15^{\circ}\cot 16^{\circ}\cot 17^{\circ}.....\cot 73^{\circ}\cot 74^{\circ}\cot 75^{\circ}$$ is
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$$\displaystyle \frac{1}{2}$$
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$$0$$
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$$1$$
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$$-1$$
Explanation
$$\cot { 15° } \cot { 16° } \cot { 17° } .........\cot { 73° } \cot { 74° } \cot { 75° } $$
$$=\cot { \left( 90°-75° \right) } \cot { \left( 90°-74° \right) } \cot { \left( 90°-73° \right) } ......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° } $$
$$=\tan { 75° } \tan { 74° } \tan { 73° } .......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° } $$
$$=(\tan { 75° }.\cot75^o) (\tan { 74° }.\cot74^o)( \tan { 73° }.\cot73^o) .......\cot45^o $$
$$=1\times1\times1\times.........\times1$$
$$=1$$
Hence, the answer is $$1.$$
$$\displaystyle \dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}$$ equals
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$$\displaystyle \sec ^{2}A$$
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-1
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$$\displaystyle \cot ^{2}A$$
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$$\displaystyle \tan ^{2}A$$
Explanation
$$\displaystyle \frac{1+\tan ^{2}A}{1+\cot ^{2}A}$$
= $$\displaystyle \dfrac{1+\dfrac{sin ^{2}A}{cos ^{2}A}}{1+\dfrac{cos ^{2}A}{sin ^{2}A}}$$
= $$\displaystyle \dfrac{\dfrac{cos^2 A + sin ^{2}A}{cos ^{2}A}}{\dfrac{ sin^2 A + cos ^{2}A}{sin ^{2}A}}$$
= $$\displaystyle \dfrac{\sin^2 A}{\cos^2 A}$$
= $$\displaystyle \tan^2 A$$
Using trigonometric identities $$\displaystyle 5 \text{ cosec} ^{2}\theta -5\text{ cot} ^{2}\theta -3$$ expressed as an integer is
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5
0%
3
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2
0%
0
Explanation
The given equation is
$$\displaystyle 5 \text{ cosec} ^{2}\theta -5\cot ^{2}\theta -3$$
= $$ 5 ( \text{cosec} ^{2}\theta - \cot ^{2}\theta) -3$$
= $$ 5 - 3$$ (since, $$\text{cosec}^2 \theta - \text{cot}^2 \theta = 1$$)
= $$2$$
$$\displaystyle \frac{4}{3}\cot ^{2}30^{\circ}+3\sin ^{2}60^{\circ}-2\text{cosec} ^{2}60^{\circ}-\frac{3}{4}\tan ^{2}30^{\circ}$$ is
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$$1$$
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$$\displaystyle -\frac{20}{3}$$
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$$\displaystyle \frac{10}{3}$$
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$$5$$
Explanation
$$\displaystyle \frac{4}{3}\cot ^{2}30^{\circ}+3\sin
^{2}60^{\circ}-2\text{cosec}^{2}60^{\circ}-\frac{3}{4}\tan
^{2}30^{\circ}$$
$$= \displaystyle \frac{4}{3}(\sqrt 3)^2+3(\frac {\sqrt 3}{2})^2-2(\frac {2}{\sqrt 3})^2-\frac{3}{4}(\frac {1}{\sqrt 3})^2$$
$$=\displaystyle \frac{4}{3}\times 3+3\times \frac {3}{4}-2\times \frac {4}{3}-\frac{3}{4}\times \frac {1}{3}$$
$$=\displaystyle 4+ \frac {9}{4}- \frac {8}{3}-\frac{1}{4}$$
$$=\displaystyle \frac {10}{3}$$
Option C is correct.
In the given figure, $$tan x^o = \dfrac{4}{3}$$ and "T" is the mid-point of PR, calculate the length of PQ.
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0%
$$\sqrt 8$$ m
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$$9$$ m
0%
$$\sqrt{59}$$ m
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$$10$$ m
Explanation
In $$\triangle RST$$
$$\Rightarrow \tan { x=\dfrac { RT }{ SR } } $$
$$\Rightarrow \dfrac { 4 }{ 3 } =\dfrac { RT }{ 3 } $$
$$\Rightarrow RT=4m$$
$$\therefore PR=2RT=8m$$
Now, $$\triangle PSQ$$ is a right angled triangle, such that
$$\Rightarrow PQ{ PR }^{ 2 }+{ PQ }^{ 2 }={ PQ }^{ 2 }$$
$${ \Rightarrow PQ }^{ 2 }+{ PR }^{ 2 }={ RQ }^{ 2 }={ \left( 8 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 }=64+364$$
$$\Rightarrow PQ=\sqrt { 100 } =10m$$
Hence, the answer is $$10m.$$
Evaluate:
$$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$$
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$$\displaystyle \sqrt{3}$$
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$$2$$
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$$1$$
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$$\displaystyle \sqrt{2}$$
Explanation
$$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$$
$$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\tan (90^{\circ}-23^{\circ}).\tan (90^{\circ}-7^{\circ})$$ $$\displaystyle \left [ \because \tan (90^{\circ}-\theta )=\cot \theta \right ]$$
=$$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\cot 23^{\circ}.\cot 7^{\circ}$$
$$\displaystyle \tan 7^{\circ}.\cot 7^{\circ}\tan 23^{\circ}.\cot 23^{\circ}.\sqrt{3}$$ $$\displaystyle (\because \tan \theta .\cot \theta =1)$$
$$\displaystyle =1\times 1\times \sqrt{3}=\sqrt{3}$$
Evaluate: $$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$$
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0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
$$3$$
Explanation
$$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$$
$$=\sin { \left[ { 90 }^{ o }-{ 40 }^{ o }+\theta \right] } -\cos { \left( { 40 }^{ o }-\theta \right) } +(\tan { { 1 }^{ o } } \tan {{89}^{o}})(\tan{{10}^{o}}\tan{{80}^{o}})(\tan {{20}^{o}}\tan {{70}^{o}})$$
$$=\sin {\left[{90}^{o}-({40}^{o}-\theta) \right]} -\cos {\left({40}^{o}-\theta \right)} +\tan {({90}^{o}} -{ 89 }^{ o })\tan { { 89 }^{ o } } \tan { { (90 }^{ o } } -{ 80 }^{ o })\tan { { 80}^{ o } } \tan { { (90 }^{ o } } -{ 70 }^{ o })\tan { { 70 }^{ o } } $$
$$=\cos { \left( { 40 }^{ o }-\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\cot { { 89 }^{ o } } \tan { { 89 }^{ o } } \cot { { 80 }^{ o } } \tan { { 80 }^{ o } } \cot { { 70 }^{ o } } \tan { { 70 }^{ o } } $$
$$=0+1$$
$$\because \cot {\theta} \tan {\theta}=1$$]
$$=1$$
Hence, $$\sin { \left( { 50 }^{ o }+\theta \right) } -\cos { \left( { 40 }^{ o }-\theta \right) } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}=1$$
0:0:1
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Practice Class 10 Maths Quiz Questions and Answers
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