Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blank.
If $\text{k}=\sin\Big(\frac{\pi}{18}\Big)\sin\Big(\frac{5\pi}{18}\Big)\sin\Big(\frac{7\pi}{18}\Big),$ then the numerical value of k is _______.

Answer

If $\text{k}=\sin\Big(\frac{\pi}{18}\Big)\sin\Big(\frac{5\pi}{18}\Big)\sin\Big(\frac{7\pi}{18}\Big),$ then the numerical value of k is $\frac{1}{8}.$
Solution:
Given that ,$\text{k}=\sin\Big(\frac{\pi}{18}\Big)\sin\Big(\frac{5\pi}{18}\Big)\sin\Big(\frac{7\pi}{18}\Big)$
$\Rightarrow\text{k}=\sin10^\circ.\sin50^\circ.\sin70^\circ$
$\Rightarrow\text{k}=\sin10^\circ\sin(90^\circ-40^\circ)\sin(90^\circ-20^\circ)$
$\Rightarrow\text{k}=\sin10^\circ\cos40^\circ\cos20^\circ$
$\Rightarrow\text{k}=\sin10^\circ.\frac{1}{2}[2\cos40^\circ\cos20^\circ] $
$\Rightarrow\text{k}=\sin10^\circ.\frac{1}{2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20\circ)]$
$[2\cos\text{x}\cos\text{y}=\cos(\text{x}+\text{y})+\cos(\text{x}-\text{y})]$
$\Rightarrow\text{k}=\frac{1}{2}\sin10^\circ[\cos60^\circ+\cos20^\circ]$]
$\Rightarrow\text{k}=\frac{1}{2}\sin10^\circ\Big(\frac{1}{2}+\cos20^\circ\Big)$
$\Rightarrow\text{k}=\frac{1}{4}\sin10^\circ+\frac{1}{2}\sin10^\circ.\cos20^\circ$
$\Rightarrow\text{k}\frac{1}{4}\sin10^\circ+\frac{1}{4}(2\sin10^\circ\cos20^\circ)$
$\Rightarrow\text{k}=\frac{1}{4}\sin10^\circ+\frac{1}{4}[\sin(10^\circ+20^\circ)+\sin(10^\circ-20^\circ)]$
$[2\sin\text{x}\cos\text{y}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})]$
$\Rightarrow\text{k}=\frac{1}{4}\sin10^\circ+\frac{1}{4}[\sin30^\circ+\sin(-10^\circ)]$
$\Rightarrow\text{k}=\frac{1}{4}\sin10^\circ+\frac{1}{4}\sin30^\circ-\frac{1}{4}\sin10^\circ$
$\Rightarrow\text{k}=\frac{1}{4}\sin30^\circ=\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$
Hence ,the value of the filler is $\frac{1}{8}$.

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Question 21 Mark

Fill in the blank.
The maximum distance of a point on the graph of the function $\text{y}=\sqrt{3}\sin\text{x}+\cos\text{x}$ from x-axis is _______.

Answer

The maximum distance of a point on the graph of the function $\text{y}=\sqrt{3}\sin\text{x}+\cos\text{x}$ from x-axis is 2units.
Solution:
Given that $\text{y}=\sqrt3\sin\text{x}+\cos\text{x}\dots(1)$
$\therefore$ The maximum distance from a point on the graph of eqn. (1) from x-axis
$=\sqrt{(\sqrt3)^2+(1)^2}=\sqrt{3+1}=2\cdot$
Hence , the value of the filler is 2 units.

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Question 31 Mark

Fill in the blank.
$3(\sin\text{x}-\cos\text{x})^4+6(\sin\text{x}+\cos\text{x})^2+4(\sin^6\text{x}+\cos^6\text{x})=$ _______.

Answer

$3(\sin\text{x}-\cos\text{x})^4+6(\sin\text{x}+\cos\text{x})^2+4(\sin^6\text{x}+\cos^6\text{x})=$ 13.
Solution:
Given expression is
$3(\sin\text{x}-\cos\text{x})^4+6(\sin\text{x}+\cos\text{x})^2+4(\sin^6\text{x}+\cos^6\text{x})$
$$$=3[\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}]^2+6(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x})\\+4[(\sin^2\text{x})^3+(\cos^2\text{x})^3]$
$=3[1-2\sin\text{x}\cos\text{x}]^2+6(1+2\sin\text{x}\cos\text{x})+4[(\sin^2\text{x}+\cos^2\text{x})^3\\-3\sin^2\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^2\text{x})]$
$=3(1+4\sin^2\text{x}\cos^2\text{x}-4\sin\text{x}\cos\text{x})+6+12\sin\text{x}\cos\text{x}+4-12\sin^2\text{x}\cos^2\text{x}$
$=3+12\sin^2\text{x}\cos^2\text{x}-12\sin\text{x}\cos\text{x}+6+12\sin\text{x}\cos\text{x}+4-12\sin^2\text{x}\cos^2\text{x}$
$=3+6+4=13$
Hence, the value of the filler is $13$.$$

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Question 41 Mark

Fill in the blank.
In a triangle ABC with $\angle\text{C}=90^\circ$ the equation whose roots are tan A and tan B is _______.
[Hint: $\text{A + B}=90^\circ\Rightarrow\tan\text{A}\tan\text{B}=1$ and $\tan\text{A}+\tan\text{B}=\frac{2}{\sin2\text{A}}$ ]

Answer

In a triangle ABC with $\angle\text{C}=90^\circ$ the equation whose roots are tan A and tan B is $\text{x}^2-\Big(\frac{2}{\sin2\text{A}}\Big)\text{x}+1=0.$
Solution:
Given $\text{a}\triangle\text{ABC}$ with$\angle\text{C}=90^\circ$
$\text{x}^2-(\tan\text{A}+\tan\text{B})\text{x}+\tan\text{A}.\tan\text{B}=0$
$\text{A+B}=90^\circ[\because\angle\text{C}]$
$\Rightarrow\tan(\text{A+B})=\tan90^\circ$
$\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=\frac{1}{0}$
$\Rightarrow 1-\tan\text{A}\tan\text{B}=0$
$\Rightarrow\tan\text{A}\tan\text{B}=1\dots(1)$
$$Now $\tan\text{A}+\tan\text{B}=\frac{\sin\text{A}}{\cos\text{B}}+\frac{\sin\text{B}}{\cos\text{B}}$
$=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\frac{\sin\text{(A+B)}}{\cos\text{A}\cos\text{B}}=\frac{\sin90^\circ}{\cos\text{A}.\cos(90^\circ-\text{A})}$
$=\frac{1}{\cos\text{A}\sin\text{A}}$
$\therefore\tan\text{A}+\tan\text{B}=\frac{2}{2\sin\text{A}\cos\text{A}}=\frac{2}{\sin2\text{A}}\dots(2)$
$$From (1) and (2) we get
$\text{x}^2-\Big(\frac{2}{\sin2\text{A}}\Big)\text{x}+1=0$
Hence, the value of the filler is $\text{x}^2-\Big(\frac{2}{\sin2\text{A}}\Big)\text{x}+1=0.$

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Question 51 Mark

Fill in the blank.
If $\sin\text{x}+\cos\text{x}=\text{a},$ then,

$\sin^6\text{x}+\cos^6\text{x}=...........$
$|\sin\text{x}-\cos\text{x}|=...........$

Answer

If $\sin\text{x}+\cos\text{x}=\text{a},$ then,

$\sin^6\text{x}+\cos^6\text{x}=$ $=\frac{1}{4}[4-3(\text{a}^2-1)^2]$
$|\sin\text{x}-\cos\text{x}|=$ $\sqrt{2-\text{a}^2}.$

Solution:
Given that, $\sin\text{x}+\cos\text{x}=\text{a}$
squaring both sides, we get,
$(\sin\text{x}+\cos\text{x})^2=\text{a}^2$
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\text{a}^2$
$\Rightarrow1+2\sin\text{x}\cos\text{x}=\text{a}^2$
$\Rightarrow\sin\text{x}\cos\text{x}=\frac{\text{a}^2-1}{2}\dots(1)$

$\sin^6\text{x}+\cos^6\text{x}=(\sin^2\text{x})^3+(\cos^2\text{x})3$

$=(\sin^2\text{x}+\cos^2\text{x})^3-3\sin^2\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^2\text{x})$
$=(1)^3-3\Big(\frac{\text{a}^2-1}{2}\Big)^2\cdot1$
$=1-\frac{3(\text{a}^2-1)^2}{4}$
$=\frac{1}{4}[4-3(\text{a}^2-1)^2]$
Hence ,the value of the filler is $=\frac{1}{4}[4-3(\text{a}^2-1)^2]$

$|\sin\text{x}-\cos\text{x}|^2=\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}$

$=1-2\Big(\frac{\text{a}^2-1}{2}\Big)$
$=1-(\text{a}^2-1)=1-\text{a}^2+1$
$=2-\text{a}^2$
$\therefore|\sin\text{x}-\cos\text{x}|=\sqrt{2-\text{a}^2}$
$[\therefore|\sin\text{x}-\cos\text{x}|>0]$
Hence , the value of the filler is $\sqrt{2-\text{a}^2}\cdot$

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Question 61 Mark

Fill in the blank.
Given x > 0, the values of $\text{f(x)}=-3\cos\sqrt{3+\text{x + x}^2}$ lie in the interval _______.

Answer

Given x > 0, the values of $\text{f(x)}=-3\cos\sqrt{3+\text{x + x}^2}$ lie in the interval [-3, 3].
Solution:
Given that ,$\text{f(x)}=-3\cos\sqrt{3+\text{x}+\text{x}^2}$
put $\sqrt{3+\text{x}+\text{x}^2}=\text{y}$$$
$\therefore\text{f(x)}=-3\cos\text{y}$
$\because-1\leq\cos\text{y}\leq1$
$3\geq-3\cos\text{y}\geq-3$
$\Rightarrow-3\leq-3\cos\text{y}\leq3$
$\therefore-3\leq-3\cos\sqrt{3+\text{x}+\text{x}^2}\leq3,\text{x}>0$
Hence , the value of the filler is [-3, 3].

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Question 71 Mark

Fill in the blank.
If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then $\tan2\text{A}=$ _______.

Answer

If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then $\tan2\text{A}=$ $\tan\text{B}.$
Solution:
Given that, $\tan\text{A}=\frac{1-\cos}{\sin\text{B}}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\Big(\frac{1-\cos\text{B}}{\sin\text{B}}\Big)}{1-\Big(\frac{1-\cos\text{B}}{\sin\text{B}}\Big)^2}$
$=\frac{2\Bigg(\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2}}\Bigg)}{1-\Bigg(\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2}}\Bigg)^2}$$\begin{bmatrix}\because1-\cos\text{B}=2\sin^2\frac{\text{B}}{2}\\\sin\text{B}=2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2} \end{bmatrix}$
$=\frac{2\Bigg(\frac{\sin\frac{\text{B}}{2}}{\cos\frac{\text{B}}{2}}\Bigg)}{1-\Bigg(\frac{\sin\frac{\text{B}}{2}}{\cos\frac{\text{B}}{2}}\Bigg)^2}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}=\tan\text{B}$
So, $\tan2\text{A}=\tan\text{B}$
Hence ,the value of the filler is $\tan\text{B}.$

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Question 81 Mark

Fill in the blank.
The value of $\frac{\sin50^\circ}{\sin130^\circ}$ is _______.

Answer

The value of $\frac{\sin50^\circ}{\sin130^\circ}$ is 1.
Solution:
$\frac{\sin50^\circ}{\sin130^\circ}=\frac{\sin50^\circ}{\sin(180^\circ-50^\circ)}=\frac{\sin50^\circ}{\sin50^\circ}=1$
Hence , the value of filler is 1.

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MATHS Fill In The Blanks[1 Marks ]

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