Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blank.If $\text{y}=2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ then ____ < y < ____.

Answer

If $\text{y}=2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ then $-2\pi<\text{y}<2\pi.$Solution:
We are given that, $\text{y}=2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Let us suppose, $\text{x}=\tan\theta$
$\therefore\ \text{y}=2\tan^{-1}\tan\theta+\sin^{-1}\frac{2\tan\theta}{1+\tan^2\theta}$
$\Rightarrow\ \text{y}=2\theta+\sin^{-1}\sin2\theta$
$\Big[\because\ \tan^{-1}(\tan\text{x})=\text{x and }\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$\Rightarrow\ \text{y}=2\theta+2\theta$
$[\because\ \sin^{-1}(\sin\text{x})=\text{x}]$
$\Rightarrow\ \text{y}=4\theta\ [\because\ \theta=\tan^{-1}\text{x}]$
$\Rightarrow\ \text{y}=4\tan^{-1}\text{x}$
$\Rightarrow\ \tan^{-1}\text{x}=\frac{\text{y}}{4}$
$\Rightarrow\ -\frac{\pi}{2}<\frac{\text{y}}{4}<\frac{\pi}{2}$
$\Big[\because\ -\frac{\pi}{2}<\tan^{-1}\text{x}<\frac{\pi}{2}\Big]$
$=-2\pi<\text{y}<2\pi.$

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Question 21 Mark

Fill in the blank.If $\cos\Big(\tan^{-1}\text{x}+\cot^{-1}\sqrt{3}\Big)=0,$ then value of x is __________.

Answer

If $\cos\Big(\tan^{-1}\text{x}+\cot^{-1}\sqrt{3}\Big)=0,$ then value of x is $\sqrt{3}.$Solution:
We have, $\cos\Big(\tan^{-1}\text{x}+\cot^{-1}\sqrt{3}\Big)=0$
$\Rightarrow\ \tan^{-1}\text{x}+\cot^{-1}\sqrt{3}=\cos^{-1}0$
$\Rightarrow\ \tan^{-1}\text{x}+\cot^{-1}\sqrt{3}=\frac{\pi}{2}$
$\Rightarrow\ \tan^{-1}\text{x}=\frac{\pi}{2}-\cot^{-1}\sqrt{3}$
$\Rightarrow\ \tan^{-1}\text{x}=\tan^{-1}\sqrt{3}$
$\Big(\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big)$
$\therefore\ \text{x}=\sqrt{3}.$

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Question 31 Mark

Fill in the blank.The principal value of $\cos^{-1}\Big(-\frac{1}{2}\Big)$ is __________.

Answer

The principal value of $\cos^{-1}\Big(-\frac{1}{2}\Big)$ is $\frac{2\pi}{3}.$Solution:
We have, $\cos^{-1}\Big(-\frac{1}{2}\Big)$
$=\cos^{-1}\Big(\cos\frac{2\pi}{3}\Big)$
$\Big[\because\ \cos\frac{2\pi}{3}=-\frac{1}{2}\Big]$
$=\frac{2\pi}{3}\ \Big [\because\ \cos^{-1}(\cos\text{x})=\text{x},\ \text{x}\in[0,\pi]\Big]$

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Question 41 Mark

Fill in the blank.
The set of values of $\sec^{-1}\Big(\frac{1}{2}\Big)$ is __________.

Answer

The set of values of $\sec^{-1}\Big(\frac{1}{2}\Big)$ is $\phi.$
Solution:
Since domain of $\sec^{-1}\Big(\frac{1}{2}\Big)$ is R - (-1, 1)
$\Rightarrow\ (-\infty, -1)\cup[1,\infty)$
So, there is no set of values exist for $\sec^{-1}\frac{1}{2}$
So, $\phi$ is the answer.

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Question 51 Mark

Fill in the blank.
The principal value of $\tan^{-1}\sqrt{3}$ is __________.

Answer

The principal value of $\tan^{-1}\sqrt{3}$ is $\frac{\pi}{3}.$
Solution:
$\because\ \tan^{-1}\sqrt{3}=\tan^{-1}\tan\Big(\frac{\pi}{3}\Big)$
$\Big[\because\ \tan^{-1}(\tan\text{x})=\text{x},\ \text{x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$=\Big(\frac{\pi}{3}\Big)$

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Question 61 Mark

Fill in the blank.
The value of $\cos(\sin^{-1}\text{x}+\cos^{-1}\text{x}),$ where $|\text{x}|\leq1,$ is ______.

Answer

The value of $\cos(\sin^{-1}\text{x}+\cos^{-1}\text{x}),$ where $|\text{x}|\leq1,$ is $0.$
Solution:
We have, $\cos(\sin^{-1}\text{x}+\cos^{-1}\text{x})$
$=\cos\frac{\pi}{2}$
$\Big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=0$

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Question 71 Mark

Fill in the blank.
The value of $\cot^{-1}(-x)$ for all $\text{x}\in\text{R}$ in terms of $\cot^{-1}x$ is $....$

Answer

We know that, $\cot^{-1}(-\text{x}) = \pi-\cot^{-1}\text{x},\ \text{x}\in\text{R}$

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Question 81 Mark

Fill in the blank.
The result $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$ is true when value of xy is _____.

Answer

The result $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$ is true when value of xy is -1.
Solution:
We know that $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$
Where, $\text{xy}>-1.$

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Question 91 Mark

Fill in the blank.
The value of expression $\tan\Big(\frac{\sin^{-1}\text{x}+\cos^{-1}\text{x}}{2}\Big),$ where $\text{x}=\frac{\sqrt{3}}{2},$ is ________.

Answer

The value of expression $\tan\Big(\frac{\sin^{-1}\text{x}+\cos^{-1}\text{x}}{2}\Big),$ where $\text{x}=\frac{\sqrt{3}}{2},$ is ________.
Solution:
$\tan\Big(\frac{\sin^{-1}\text{x}+\cos^{-1}\text{x}}{2}\Big)=\tan\bigg(\frac{\frac{\pi}{2}}{2}\bigg)$
$\Big(\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big)$
$=\tan\frac{\pi}{4}=1$

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Question 101 Mark

Fill in the blank.
The value of $\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big)$ is __________.

Answer

The value of $\sin^{-1}\Big(\sin\frac{3\pi}{5}\Big)$ is $\frac{2\pi}{5}.$
Solution:
$\because\ -\frac{\pi}{2}\leq\sin\text{x}\leq\frac{\pi}{2}$
$\therefore\ \sin^{-1}\Big(\sin\frac{3\pi}{5}\Big)=\sin^{-1}\sin\Big(\pi-\frac{2\pi}{5}\Big)$
$=\sin^{-1}\Big(\sin\frac{2\pi}{5}\Big)=\frac{2\pi}{5}$

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Question 111 Mark

Fill in the blank.
The value of $\cos^{-1}\Big(\cos\frac{14\pi}{3}\Big)$ is __________.

Answer

The value of $\cos^{-1}\Big(\cos\frac{14\pi}{3}\Big)$ is $\frac{2\pi}{3}.$
Solution:
We have, $\cos^{-1}\Big(\cos\frac{14\pi}{3}\Big)$
$=\cos^{-1}\cos\Big(4\pi+\frac{2\pi}{3}\Big)$
$=\cos^{-1}\cos\frac{2\pi}{3}$
$[\because\ \cos(2\text{n}\pi+\theta)=\cos\theta]$
$=\frac{2\pi}{3}\ \Big[\because\ \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]\Big]$

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MATHS Fill In The Blanks[1 Marks ]

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