Question

Write each of the following in the simplest form:

(i) ${\cot }^{-1}\frac{a}{\sqrt{x^2-a^2}},\left|\mathit{}x\right|>a$
(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) ${\tan }^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\},x\ne 0$

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

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

Answer 1

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

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

(ii) Let $x=\cot \theta$
Now,
$$\begin{gathered} {\tan }^{-1}\left\{x+\sqrt{1+x^2}\right\}={\tan }^{-1}\left\{\cot \theta +\sqrt{1+{\cot }^2\theta }\right\} \\ ={\tan }^{-1}\left\{\cot \theta +\mathrm{cosec}\theta \right\} \\ ={\tan }^{-1}\left\{\frac{\cos \theta +1}{\sin \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{2{\cos }^2{\displaystyle \frac{\theta }{2}}}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}\right\} \\ ={\tan }^{-1}\left\{\cot \frac{\theta }{2}\right\} \\ ={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\theta }}{2}\right)\right\} \\ =\left(\frac{\mathrm{\pi }}{2}-\frac{\theta }{2}\right) \\ =\frac{\mathrm{\pi }}{2}-\frac{{\cot }^{-1}x}{2} \end{gathered}$$



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

$$\begin{gathered} {\tan }^{-1}\left\{\sqrt{1+x^2}-x\right\}={\tan }^{-1}\left\{\sqrt{1+{\cot }^2\theta }-\cot \theta \right\} \\ ={\tan }^{-1}\left\{\mathrm{cosec}\theta -\cot \theta \right\} \\ ={\tan }^{-1}\left\{\frac{1-\cos \theta }{\sin \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{2{\sin }^2{\displaystyle \frac{\theta }{2}}}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}\right\} \\ ={\tan }^{-1}\left\{\tan \left(\frac{\theta }{2}\right)\right\} \\ =\frac{\theta }{2} \\ =\frac{{\cot }^{-1}x}{2} \end{gathered}$$

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

$$\begin{gathered} {\tan }^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\}={\tan }^{-1}\left\{\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{\sqrt{{\sec }^2\theta }-1}{\tan \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{\sec \theta -1}{\tan \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{1-\cos \theta }{\sin \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{2{\sin }^2{\displaystyle \frac{\theta }{2}}}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}\right\} \\ ={\tan }^{-1}\left\{\tan \left(\frac{\theta }{2}\right)\right\} \\ =\frac{\theta }{2} \\ =\frac{{\tan }^{-1}x}{2} \end{gathered}$$

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

$$\begin{gathered} {\tan }^{-1}\left\{\frac{\sqrt{1+x^2}+1}{x}\right\}={\tan }^{-1}\left\{\frac{\sqrt{1+{\tan }^2\theta }+1}{\tan \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{\sqrt{{\sec }^2\theta }+1}{\tan \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{\sec \theta +1}{\tan \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{\cos \theta +1}{\sin \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{2{\cos }^2{\displaystyle \frac{\theta }{2}}}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}\right\} \\ ={\tan }^{-1}\left\{\cot \frac{\theta }{2}\right\} \\ ={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\theta }}{2}\right)\right\} \\ =\left(\frac{\mathrm{\pi }}{2}-\frac{\theta }{2}\right) \\ =\frac{\mathrm{\pi }}{2}-\frac{{\tan }^{-1}x}{2} \end{gathered}$$


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

$$\begin{gathered} {\tan }^{-1}\sqrt{\frac{a-x}{a+x}}={\tan }^{-1}\sqrt{\frac{a-a\cos \theta }{a+a\cos \theta }} \\ ={\tan }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }} \\ ={\tan }^{-1}\sqrt{\frac{2{\sin }^2{\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2{\cos }^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ ={\tan }^{-1}\left(\tan \frac{\theta }{2}\right) \\ =\frac{\theta }{2} \\ =\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right) \end{gathered}$$
$\therefore {\tan }^{-1}\sqrt{\frac{a-x}{a+x}}=\frac{{\cos }^{-1}\left(\frac{x}{a}\right)}{2}$

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

$$\begin{gathered} {\tan }^{-1}\left\{\frac{x}{a+\sqrt{a^2-x^2}}\right\}={\tan }^{-1}\left\{\frac{a\sin \theta }{a+\sqrt{a^2-a^2{\cos }^2\theta }}\right\} \\ ={\tan }^{-1}\left\{\frac{a\sin \theta }{a+a\sqrt{{\cos }^2\theta }}\right\} \\ ={\tan }^{-1}\left\{\frac{\sin \theta }{1+\cos \theta }\right\} \\ ={\tan }^{-1}\left\{\frac{2\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta }{2}}\right)}{2{\cos }^2{\displaystyle \frac{\theta }{2}}}\right\} \\ ={\tan }^{-1}\left\{\tan \frac{\theta }{2}\right\} \\ =\frac{\theta }{2} \\ =\frac{1}{2}{\sin }^{-1}\left(\frac{x}{a}\right) \end{gathered}$$


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

$$\begin{gathered} {\sin }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\}={\sin }^{-1}\left\{\frac{\sin \theta +\sqrt{1-{\sin }^2\theta }}{\sqrt{2}}\right\} \\ ={\sin }^{-1}\left\{\frac{\sin \theta +\cos \theta }{\sqrt{2}}\right\} \\ ={\sin }^{-1}\left\{\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta \right\} \\ ={\sin }^{-1}\left\{\cos \frac{\mathrm{\pi }}{4}\sin \theta +\sin \frac{\mathrm{\pi }}{4}\cos \theta \right\} \\ ={\sin }^{-1}\left\{\sin \left(\theta +\frac{\mathrm{\pi }}{4}\right)\right\} \\ =\theta +\frac{\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{4}+{\sin }^{-1}x \end{gathered}$$
$\therefore {\sin }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\}={\cos }^{-1}x+\frac{\pi }{4}$

(ix) Let $x=\cos \theta$
Now,
$$\begin{gathered} {\sin }^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}={\sin }^{-1}\left\{\frac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{2}\right\} \\ ={\sin }^{-1}\left\{\frac{\sqrt{2{\cos }^2{\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\sqrt{2{\sin }^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2}\right\} \\ ={\sin }^{-1}\left\{\frac{\cos {\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\sin {\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{2}}\right\} \\ ={\sin }^{-1}\left\{\frac{1}{\sqrt{2}}\sin \frac{\theta }{2}+\frac{1}{\sqrt{2}}\cos \frac{\theta }{2}\right\} \\ ={\sin }^{-1}\left\{\sin \left(\frac{\theta }{2}+\frac{\mathrm{\pi }}{4}\right)\right\} \\ =\frac{\theta }{2}+\frac{\mathrm{\pi }}{4} \\ =\frac{{\cos }^{-1}x}{2}+\frac{\mathrm{\pi }}{4} \end{gathered}$$
$\therefore {\sin }^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}=\frac{{\cos }^{-1}x}{2}+\frac{\pi }{4}$

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

$$\begin{gathered} \sin \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}=\sin \left\{2{\tan }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right\} \\ =\sin \left\{2{\tan }^{-1}\sqrt{\frac{2{\sin }^2{\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2{\cos }^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right\} \\ =\sin \left\{2{\tan }^{-1}\left(\tan \frac{\theta }{2}\right)\right\} \\ =\sin \theta \\ =\sin \left({\cos }^{-1}x\right) \\ =\sin \left({\sin }^{-1}\left(\sqrt{1-x^2}\right)\right) \\ =\sqrt{1-x^2} \end{gathered}$$