Write each of the following in the simplest form:
Question:

Write each of the following in the simplest form:

(i) $\cot ^{-1} \frac{a}{\sqrt{x^{2}-a^{2}}},|x|>a$

(ii) $\tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}, x \in R$

(iii) $\tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}, x \in R$

(iv) $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\}, x \neq 0$

(v) $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}+1}{x}\right\}, x \neq 0$

(vi) $\tan ^{-1} \sqrt{\frac{a-x}{a+x}},-a<x<a$

(vii) $\tan ^{-1}\left\{\frac{x}{a+\sqrt{a^{2}-x^{2}}}\right\},-a<x<a$

(viii) $\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\},-1<x<1$

(ix) $\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}, 0<x<1$

(x) $\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}$

Solution:

(i) Let $x=a \sec \theta$

Now,

$\cot ^{-1} \frac{a}{\sqrt{x^{2}-a^{2}}}=\cot ^{-1}\left(\frac{a}{\sqrt{a^{2} \sec ^{2} \theta-a^{2}}}\right)$

$=\cot ^{-1} \frac{a}{a \sqrt{\tan ^{2} \theta}}$

$=\cot ^{-1}(\cot \theta)$

$=\theta$

$=\sec ^{-1} \frac{x}{a}$

(ii) Let $x=\cot \theta$

Now,

$\tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}=\tan ^{-1}\left\{\cot \theta+\sqrt{1+\cot ^{2} \theta}\right\}$

$=\tan ^{-1}\{\cot \theta+\operatorname{cosec} \theta\}$

$=\tan ^{-1}\left\{\frac{\cos \theta+1}{\sin \theta}\right\}$

$=\tan ^{-1}\left\{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right\}$

$=\tan ^{-1}\left\{\cot \frac{\theta}{2}\right\}$

$=\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}$

$=\left(\frac{\pi}{2}-\frac{\theta}{2}\right)$

$=\frac{\pi}{2}-\frac{\cot ^{-1} x}{2}$

(iii) Let $x=\cot \theta$

Now,

$\tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}=\tan ^{-1}\left\{\sqrt{1+\cot ^{2} \theta}-\cot \theta\right\}$

$=\tan ^{-1}\{\operatorname{cosec} \theta-\cot \theta\}$

$=\tan ^{-1}\left\{\frac{1-\cos \theta}{\sin \theta}\right\}$

$=\tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{2}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right\}$

$=\tan ^{-1}\left\{\tan \left(\frac{\theta}{2}\right)\right\}$

$=\frac{\theta}{2}$

$=\frac{\cot ^{-1} x}{2}$

(iv) Let $x=\tan \theta$

Now,

$\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\}=\tan ^{-1}\left\{\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right\}$

$=\tan ^{-1}\left\{\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right\}$

$=\tan ^{-1}\left\{\frac{\sec \theta-1}{\tan \theta}\right\}$

$=\tan ^{-1}\left\{\frac{1-\cos \theta}{\sin \theta}\right\}$

$=\tan ^{-1}\left\{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right\}$

$=\tan ^{-1}\left\{\tan \left(\frac{\theta}{2}\right)\right\}$

$=\frac{\theta}{2}$

$=\frac{\tan ^{-1} x}{2}$

(v) Let $x=\tan \theta$

Now,

$\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}+1}{x}\right\}=\tan ^{-1}\left\{\frac{\sqrt{1+\tan ^{2} \theta}+1}{\tan \theta}\right\}$

$=\tan ^{-1}\left\{\frac{\sqrt{\sec ^{2} \theta}+1}{\tan \theta}\right\}$

$=\tan ^{-1}\left\{\frac{\sec \theta+1}{\tan \theta}\right\}$

$=\tan ^{-1}\left\{\frac{\cos \theta+1}{\sin \theta}\right\}$

$=\tan ^{-1}\left\{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{2}{2} \cos \frac{\theta}{2}}\right\}$

$=\tan ^{-1}\left\{\cot \frac{\theta}{2}\right\}$

$=\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}$

$=\left(\frac{\pi}{2}-\frac{\theta}{2}\right)$

$=\frac{\pi}{2}-\frac{\tan ^{-1} x}{2}$

(vi) Let $x=a \cos \theta$

Now,

$\tan ^{-1} \sqrt{\frac{a-x}{a+x}}=\tan ^{-1} \sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}$

$=\tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$

$=\tan ^{-1} \sqrt{\frac{2 \sin ^{2} \%}{2 \cos ^{2} \theta / 2}}$

$=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$

$=\frac{\theta}{2}$

$=\frac{1}{2} \cos ^{-1}\left(\frac{x}{a}\right)$

$\therefore \tan ^{-1} \sqrt{\frac{a-x}{a+x}}=\frac{\cos ^{-1}\left(\frac{x}{a}\right)}{2}$

(vii) Let $x=a \sin \theta$

Now,

$\tan ^{-1}\left\{\frac{x}{a+\sqrt{a^{2}-x^{2}}}\right\}=\tan ^{-1}\left\{\frac{a \sin \theta}{a+\sqrt{a^{2}-a^{2} \cos ^{2} \theta}}\right\}$

$=\tan ^{-1}\left\{\frac{a \sin \theta}{a+a \sqrt{\cos ^{2} \theta}}\right\}$

$=\tan ^{-1}\left\{\frac{\sin \theta}{1+\cos \theta}\right\}$

$=\tan ^{-1}\left\{\frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos ^{2} \frac{\theta}{2}}\right\}$

$=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\}$

$=\frac{\theta}{2}$

$=\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)$

(viii) Let $x=\sin \theta$

Now,

$\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}=\sin ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\}$

$=\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}$

$=\sin ^{-1}\left\{\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta\right\}$

$=\sin ^{-1}\left\{\cos \frac{\pi}{4} \sin \theta+\sin \frac{\pi}{4} \cos \theta\right\}$

$=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$

$=\theta+\frac{\pi}{4}$

$=\frac{\pi}{4}+\sin ^{-1} x$

$\therefore \sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}=\cos ^{-1} x+\frac{\pi}{4}$

(ix) Let $x=\cos \theta$

Now,

$\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}=\sin ^{-1}\left\{\frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{2}\right\}$

$=\sin ^{-1}\left\{\frac{\sqrt{2 \cos ^{2} \theta} / 2+\sqrt{2 \sin ^{2} \theta / 2}}{2}\right\}$

$=\sin ^{-1}\left\{\frac{\cos ^{\theta} / 2+\sin ^{\theta} / 2}{\sqrt{2}}\right\}$

$=\sin ^{-1}\left\{\frac{1}{\sqrt{2}} \sin \frac{\theta}{2}+\frac{1}{\sqrt{2}} \cos \frac{\theta}{2}\right\}$

$=\sin ^{-1}\left\{\sin \left(\frac{\theta}{2}+\frac{\pi}{4}\right)\right\}$

$=\frac{\theta}{2}+\frac{\pi}{4}$

$=\frac{\cos ^{-1} x}{2}+\frac{\pi}{4}$

$\therefore \sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}=\frac{\cos ^{-1} x}{2}+\frac{\pi}{4}$

$(x)$ Let $x=\cos \theta$

Now,

$\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}=\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right\}$

$=\sin \left\{2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \%}{2 \cos ^{2} \theta / 2}}\right\}$

$=\sin \left\{2 \tan ^{-1}\left(\tan \frac{\theta}{2}\right)\right\}$

$=\sin \theta$

$=\sin \left(\cos ^{-1} x\right)$

$=\sin \left(\sin ^{-1}\left(\sqrt{1-x^{2}}\right)\right)$

$=\sqrt{1-x^{2}}$

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