प्रश्नों के उत्तर लिखिए। (प्रत्येक प्रश्न 2 अंक का हे)

Question 12 Marks

फलन को सरलतम रूप में लिखिए: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}},|x| < a$

Answer

मान लीजिए x = a sin $ \theta$ $\Rightarrow $ $\frac{x}{a}$ = sin $\theta$ $\Rightarrow \sin^{-1} \left(\frac{x}{a}\right)=\theta$ ...(i)
$\therefore $ $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}$ = $\tan ^{-1}\left(\frac{a}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right)$ = $\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right) $
= $\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)$ [$\because \sin^2 x + \cos^2 x = 1 \Rightarrow$ cos x = $\sqrt{1-\sin ^{2} x}$]
= $\tan ^{-1}(\tan \theta)$ = $\theta$ = $\sin ^{-1}\left(\frac{x}{a}\right) $ [समी (i) से]

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Question 22 Marks

फलन को सरलतम रूप में लिखिए: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$, $ \frac{-\pi}{4}<x<\frac{3 \pi}{4}$

Answer

$tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$ = $ \tan ^{-1}\left(\frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}\right)$
कोष्ठक के भीतर अंश तथा हर को cos x द्वारा भाग देने पर,
= $\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$ = $\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-x\right)\right] $ $\left[\because \tan \left(\frac{\pi}{4}-x\right)=\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}=\frac{1-\tan x}{1+\tan x}\right]$
$=\frac{\pi}{4}-x$

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Question 32 Marks

फलन को सरलतम रूप में लिखिए: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0

Answer

$\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) = \tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2}(x / 2)}{2 \cos ^{2}(x / 2)}}\right) [\therefore 1 - \cos x = 2 \sin^2(\frac{x}{2})$ तथा $1 + \cos x = 2 \cos ^2(\frac{x}{2})]$
$= \tan ^{-1}\left(\sqrt{\tan ^{2} \frac{x}{2}}\right)= \tan ^{-1}\left(\tan \frac{x}{2}\right) = \frac{x}{2}$
नोट $\sqrt{\tan ^{2} \frac{x}{2}} = \tan \frac{x}{2}$ यदि $x$ धनात्मक है अथवा $\sqrt{\tan ^{2} \frac{x}{2}} = -\tan \frac{x}{2}$ यदि $x$ ऋणात्मक है।

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Question 42 Marks

फलन को सरलतम रूप में लिखिए: $\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}},|x|>1$

Answer

मान लीजिए x = sec $ \theta $, तब $\theta = sec^{-1} x$ ...(i)
$\therefore$ $\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}}$ = $\tan ^{-1}\left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)$ = $\tan ^{-1}\left(\frac{1}{\sqrt{\tan ^{2} \theta}}\right) $ $\left(\because \sec ^{2} \theta-\tan ^{2} \theta=1\right)$
= $\tan ^{-1}\left(\frac{1}{\tan \theta}\right)$
= $\tan ^{-1}(\cot \theta)$ = $\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\theta\right)\right]$ $\left[\because \tan \left(\frac{\pi}{2}-\theta\right)=\cot \theta\right]$
= $\frac{\pi}{2}-\theta$ = $\frac{\pi}{2}-\sec ^{-1} x $ [समी (i) से]

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Question 52 Marks

फलन को सरलतम रूप में लिखिए: $\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0$

Answer

मान लीजिए x = tan $\theta$, तब $\theta = \tan^{-1} x$ ...(i)
$\therefore$ $\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}$ = $ \tan ^{-1} \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}$ = $\tan ^{-1} \frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}$ = $\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) $ $\left(\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right)$
= $\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)$= $\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$($\because$ 1 - cos $\theta = 2 \sin^2 \frac{\theta}{2}$ तथा sin $\theta$ = 2 sin $\frac{\theta}{2}$ cos $\frac{\theta}{2}$)
= $\tan ^{-1}\left(\frac{2 \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}}\right)$ = $\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$ = $\frac{\theta}{2}$ $\frac{\tan ^{-1} x}{2}$ [समी (i) से]
$\therefore$ $ \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}$= $\frac{1}{2} \tan ^{-1} x$

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Question 62 Marks

सिद्ध कीजिए: $2 \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{7} = \tan^{-1} \frac{31}{17}$

Answer

ज्ञात है, $2 \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{7} = \tan^{-1} \frac{31}{17}$
बायाँ पक्ष$ = 2 \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left[\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right] + \tan^{-1}\left(\frac{1}{7}\right) [\because 2 \tan^{-1}x = \tan^{-1} \left(\frac{2 x}{1-x^{2}}\right)]$
$= \tan^{-1}\frac{1}{1-\frac{1}{4}} + \tan^{-1} \frac{1}{7} = \tan^{-1} \left(\frac{4}{3}\right) + \tan^{-1} \left(\frac{1}{7}\right)$
$= \tan^{-1} \left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right) [\because \tan^{-1}x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-x y}\right)]$
$= \tan^{-1} \left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right) = \tan^{-1} \left(\frac{\frac{31}{21}}{\frac{17}{21}}\right) = \tan^{-1} (\frac{31}{21} \times \frac{21}{17}) = \tan^{-1} \left(\frac{31}{17}\right) =$ दायाँ पक्ष इति सिद्धम्

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Question 72 Marks

सिद्ध कीजिए: $\tan ^{-1} \frac{2}{11}$ + $ \tan ^{-1} \frac{7}{24}$ = $\tan ^{-1} \frac{1}{2}$

Answer

ज्ञात है, $2 \tan ^{-1}\left(\frac{1}{2}\right)$ + $\tan ^{-1}\left(\frac{1}{7}\right)$= $\tan ^{-1}\left(\frac{31}{17}\right) $
बायाँ पक्ष = 2$ \tan ^{-1}\left(\frac{1}{2}\right)$ + $\tan ^{-1}\left(\frac{1}{7}\right)$ = $\tan ^{-1}\left[\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right]$ + $\tan ^{-1}\left(\frac{1}{7}\right)$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
= $\tan ^{-1} \frac{1}{1-\frac{1}{4}}$ + $ \tan ^{-1} \frac{1}{7}$ = $ \tan ^{-1}\left(\frac{4}{3}\right)$ + $ \tan ^{-1}\left(\frac{1}{7}\right)$
= $\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right)$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
= $\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right)$ = $ \tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)$= $\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right)$ = $\tan ^{-1}\left(\frac{31}{17}\right)$ = दायाँ पक्ष इति सिद्धम्

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Question 82 Marks

सिद्ध कीजिए: $3 \cos^{-1} x = \cos ^{-1} \left(4 x^{3}-3 x\right), x \in\left[\frac{1}{2}, 1\right]$

Answer

मान लीजिए $x = \cos \theta$
बायाँ पक्ष $= 3 \cos^{-1}x = 3 \cos^{-1} (\cos \theta) = 3 \theta$
दायाँ पक्ष $= \cos^{-1} \left(4 x^{3}-3 x\right) $
$= \cos^{-1} \left(4 \cos ^{3} \theta-3 \cos \theta\right)$
$= \cos^{-1} (\cos 3 \theta)=3 \theta \left(\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right)$
बायाँ पक्ष = दायाँ पक्ष इति सिद्धम्

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Question 92 Marks

$\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$ व्यंजक की गणना कीजिए।

Answer

$\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right) $ $\Rightarrow $ $\tan \left[\tan ^{-1} \frac{\frac{3}{5}}{\sqrt{1-\left(\frac{3}{5}\right)^{2}}}+\tan ^{-1} \frac{2}{3}\right]$
($\because$ $\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}$ तथा $cot ^{-1} \frac{x}{y}=\tan ^{-1} \frac{y}{x}$)
= $\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$ = tan $\left[\tan ^{-1} \frac{\left(\frac{3}{4}+\frac{2}{3}\right)}{1-\frac{3}{4} \times \frac{2}{3}}\right]$ = $\tan \left(\tan ^{-1} \frac{\frac{17}{12}}{\frac{1}{2}}\right)$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
= $\tan \left(\tan ^{-1} \frac{17}{6}\right)$ = $\frac{17}{6}$

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Question 102 Marks

यदि $\sin (\sin ^{-1} \frac{1}{5} + \cos^{-1} x) = 1,$ तब $x$ का मान ज्ञात कीजिए।

Answer

ज्ञात है, $(\sin ^{-1} \frac{1}{5} + \cos^{-1} x) = 1 \Rightarrow \sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1} 1 \left(\because \sin \theta=x \Rightarrow \theta=\sin ^{-1} x\right)$
$\Rightarrow \sin ^{-1} \frac{1}{5} + \cos ^{-1} x = \sin ^{-1}\left(\sin \frac{\pi}{2}\right) \left[\because \sin \left(\frac{\pi}{2}\right)=1\right]$
$\Rightarrow \sin ^{-1} \frac{1}{5} + \cos ^{-1} x = \frac{\pi}{2} \Rightarrow \sin ^{-1} \frac{1}{5} = \frac{\pi}{2}-\cos ^{-1} x$
$\Rightarrow \sin ^{-1} \frac{1}{5} = \sin ^{-1} x \left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$
$\Rightarrow \frac{1}{5} = x$

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Question 112 Marks

$\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$ में से प्रत्येक का मान ज्ञात कीजिए।

Answer

$\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$ = $\tan ^{-1}\left[2 \cos \left\{2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)\right\}\right]$ $\left(\because \sin \frac{\pi}{6}=\frac{1}{2}\right)$
= $\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$ = $\tan ^{-1}\left(2 \cos \frac{\pi}{3}\right)$
= $\tan ^{-1}\left(2 \times \frac{1}{2}\right)$ = $\tan ^{-1}(1) $ $\left(\because \cos \frac{\pi}{3}=\frac{1}{2}\right)$
= $\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$ = $\frac{\pi}{4}$ $\left(\because \tan \frac{\pi}{4}=1\right)$

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Question 122 Marks

फलन को सरलतम रूप में लिखिए: $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a > 0; \frac{-a}{\sqrt{3}}$

Answer

मान लीजिए $x = a \tan \theta \Rightarrow \frac{x}{a} = \tan \theta \Rightarrow \theta = \tan^{-1}\left(\frac{x}{a}\right) ...(i)$
$\therefore \tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right) = \tan ^{-1}\left(\frac{3 a^{2}(a \tan \theta)-(a \tan \theta)^{3}}{a^{3}-3 a(a \tan \theta)^{2}}\right)= \tan ^{-1}\left[\frac{a^{3}\left(3 \tan \theta-\tan ^{3} \theta\right)}{a^{3}\left(1-3 \tan ^{2} \theta\right)}\right]$
$\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)= \tan ^{-1}(\tan 3 \theta) \left(\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$
$= 3 \theta = 3tan ^{-1}\left(\frac{x}{a}\right) [$समी $(i)$ से$]$

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Question 132 Marks

सिद्ध कीजिए: $3 \sin ^{-1} x=\sin ^{-1}$ $\left(3 x-4 x^{3}\right)$, x $ \in\left[-\frac{1}{2}, \frac{1}{2}\right]$

Answer

मान लीजिए $ \sin^{-1} x = \theta$
$\Rightarrow x = \sin \theta,$ तब
बायाँ पक्ष $= 3 \sin^{-1} x = 3 \sin^{-1}(\sin \theta) = 3 \theta$
दायाँ पक्ष $= \sin^{-1} \left(3 x-4 x^{3}\right) = \sin^{-1} \left(3 \sin \theta-4 \sin ^{3} \theta\right)$
$\left(3 \sin \theta-4 \sin ^{3} \theta\right) \left[\because \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta\right]$
बायाँ पक्ष $=$ दायाँ पक्ष इति सिद्धम्

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गणित प्रश्नों के उत्तर लिखिए। (प्रत्येक प्रश्न 2 अंक का हे)

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