Fill In The Blanks[1 Marks ]

Question 11 Mark

Simplest form of $(1-\cos^2\text{A})(1+\cos^2\text{A})$ is __________.

Answer

Simplest form of $(1-\cos^2\text{A})(1+\cos^2\text{A})$ is 1.Solution:
We know that
$1-\cos^2\text{A}=\sin^2\text{A}$
$1-\cot^2\text{A}=\text{cosec}^2\text{ A}$
$=(1-\cos^2\text{A})(1+\cos^2\text{A})$
$=\sin^2\text{A}.\text{cosec}^2\text{ A}$
$=\sin^2\text{A}.\frac{1}{\sin^2\text{A}}[\text{cosec}^2\text{ A}=\frac{1}{\sin^2\text{A}}]$
$=1$

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Question 21 Mark

Simplest form of $\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$ is _______.

Answer

Simplest form of $\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$ is $\tan^2\text{A}$.
Solution:
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}=\Big(1+\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)$
$\Big(1+\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}=\Big(\frac{\sin^2\text{A}+\cos^2\text{A}}{\cos^2\text{A}}\Big)\Big(\frac{\sin^2\text{A}}{\sin^2\text{A}+\cos^2\text{A}}\Big)$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}=\tan^2\text{A}$

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Question 31 Mark

If $\tan\text{A}=1,$ then $2\sin\text{A}\cos\text{A}=$ ________.

Answer

If $\tan\text{A}=1,$ then $2\sin\text{A}\cos\text{A}=1.$
Solution:
Value of $2\sin\text{A}\cos\text{A}=1$
Given,
$\tan\text{A}=1$
$\Rightarrow\tan\text{A}=\tan45$
$\Rightarrow\text{A}=45$
Value of $2\sin\text{A}\cos\text{A}=1$
$=\sin2\text{A}$
$=\sin(2\times45)$
$=\sin90$
$=1$
$\therefore$ Value of $2\sin\text{A}\cos\text{A}=1$

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Question 41 Mark

The distance of the point (–3, 4) from Y – axis is _________.

Answer

By graph, the distance of the point (-3, 4) from y-axis is -3 units.

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Question 51 Mark

Value of $\frac{2\tan^260^\circ}{1+\tan^230^\circ}$ is __________.

Answer

$\frac{2\tan^260^\circ}{1+\tan^230^\circ}\ \dots(1)$
We know that
$\tan60^\circ=\sqrt{3}$
$\tan30^\circ=\frac{1}{\sqrt{3}}$
From eq. (1)
$=\frac{2(\sqrt3)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{2\times3}{1+\Big(\frac{1}{3}\Big)}$
$=\frac{6}{\frac{3+1}{3}}$
$=\frac{6}{\frac{4}{3}}$
$=\frac{18}{4}=\frac{2}{2}$

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Question 61 Mark

If $\tan\text{A}=\cot\text{B},$ then A + B = _________.

Answer

$\tan\text{A}=\cot\text{B},$ $\frac{\sin\text{A}}{\cos\text{A}}=\frac{\cos\text{B}}{\sin\text{B}}$ $\sin\text{A}\sin\text{B}=\cos\text{A}\cos\text{B}$ $\cos\text{A}\cos\text{B}=\sin\text{A}\sin\text{B}=0$ $\cos(\text{A}+\text{B})=0$ $\cos(\text{A}+\text{B})=\cos90^\circ$ $(\text{A}+\text{B})=90^\circ$

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Question 71 Mark

$\frac{3\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)=$ ________.

Answer

$\frac{3\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)=$ $\frac{5}{2}$ Solution: $\Rightarrow\frac{3\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)$ $\Rightarrow\frac{3\cot(90^\circ-40^\circ)}{\tan50^\circ}-\frac{\cot35^\circ}{2\sin(90^\circ-55^\circ)}$ $\Rightarrow\frac{3\tan50^\circ}{\tan50^\circ}-\frac{\cot35^\circ}{2\cot35^\circ}$ $\Rightarrow3-\frac{1}{2}$ $\Rightarrow\frac{6-1}{2}$$\Rightarrow\frac{5}{2}$

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Question 81 Mark

The value of $\sin 23^\circ\cos 67^\circ + \cos 23^\circ \sin 67^\circ$ is _______.

Answer

The value of $\sin 23^\circ\cos 67^\circ + \cos 23^\circ \sin 67^\circ$ is 1
Solution:
$\sin23^\circ \cos67^\circ+\cos23° \sin67^\circ$
$=\sin23^\circ \cos(90^\circ−23^\circ)+\cos23^\circ \sin(90^\circ−23^\circ)$
$=\sin23^\circ \sin23^\circ+\cos23^\circ \cos23^\circ$
$ [∵\sin(90^\circ−\text{x})=\cos\text{x}, \cos(90^\circ−\text{x})=\sin\text{x}]$
$=\sin2^23^\circ+\cos2^23^\circ=1$

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Question 91 Mark

The value of $(\tan1^\circ\tan^\circ...\tan89^\circ)$ is equal to _________.

Answer

The value of $(\tan1^\circ\tan^\circ...\tan89^\circ)$ is equal to 1
$=\tan 1^\circ\tan2^\circ\tan3^\circ...\tan89^\circ$
$=\tan (90^\circ-89^\circ)\tan (90^\circ-88^\circ) \\ \tan (90^\circ-87^\circ)...\tan 87^\circ \tan 88^\circ\tan 89^\circ$
$=\cot 89^\circ\cot 88^\circ \cot 87^\circ...\tan 87^\circ \tan 88^\circ\tan 89^\circ$
$=(\cot 89^\circ\tan 89^\circ) (\cot 88^\circ\tan 88^\circ)\\ (\cot 87^\circ\tan 87^\circ)...(\cot 44^\circ\tan44^\circ)(\tan 45^\circ)$
$=1\times1\times1...\times1 \ \ (\because\cot\theta\tan\theta=1\text{and}\tan45^\circ=1)$
$\therefore\tan1^\circ\tan2^\circ\tan3^\circ...\tan89^\circ=1$

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Question 101 Mark

$\frac{2\cos67^\circ}{\sin23^\circ}-\frac{\tan40^\circ}{\cot50^\circ}-\cos0^\circ=$ __________.

Answer

$\frac{2\cos67^\circ}{\sin23^\circ}-\frac{\tan40^\circ}{\cot50^\circ}-\cos0^\circ=$ 0.
Solution:
$=\frac{2\cos67^\circ}{\sin23^\circ}-\frac{\tan40^\circ}{\cot50^\circ}-\cos0^\circ$
$=\frac{2\sin(90-67)}{\sin23}-\frac{\cot(90-40)}{\cot50}-1$
$=2-1-1$
$=2-2=0.$

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Question 111 Mark

$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ=$ ____________.

Answer

$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ=$ 1.
Solution:
$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ$
$\Rightarrow\Big[\frac{\cos(90-35^\circ)}{\cos55^\circ}\Big]^2+\Big[\frac{\sin(90-43)}{\sin47^\circ}\Big]^2-2\cos60^\circ$
$\Rightarrow\Big(\frac{\cos55^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\sin47^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ$
$\Rightarrow(1)^2+(1)^2-2\cos60^\circ\ \Big[\therefore\cos60^\circ=\frac12\Big]$
$\Rightarrow1+1-2\times\frac12$
$\Rightarrow1+1-1=1.$

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Question 121 Mark

$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec}31^\circ=$ _______.

Answer

$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec}31^\circ=$ 2.Solution:
$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec}31^\circ$ $=\frac{\cos(90^\circ-10^\circ)}{\sin10^\circ}+\cos59^\circ\text{cosec}(90^\circ-59^\circ)$ $=\frac{\sin10^\circ}{\sin10^\circ}+\cos59^\circ\sec59^\circ$ $=1+1 = 2$

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