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# Write the following functions in the simplest form : $(i) tan^{-1} \bigg( \frac{\sqrt{1+x^2} - 1}{x} \bigg), x \neq 0$ $(ii) tan^{-1} \frac{1}{\sqrt{x^2-1}} ,| x | > 1$

Q: (i) is Q.No. 5 of sec.2 in chapter 2 Q.(ii) is Q.No. 6 of sec 2 in chapter 2

Toolbox:
• $1-cos\theta=2sin^2\large\frac{\theta}{2}$
• $sin\theta=2sin\large\frac{\theta}{2}\: cos\large\frac{\theta}{2}$
• $1+tan^2\theta=sec^2\theta$
• $tan \theta = \large\frac{\sin \theta}{\cos \theta}$
• $sec \theta = \large\frac{1}{cos \theta}$
Ans:  for  Q   (i)
Given $tan^{-1} \large \frac{\sqrt {1 + x^2} - 1}{x}$:
Let $x=tan\theta \Rightarrow \theta = tan^{-1}x$
Substituting for $x = tan \theta$ and using the properties of $sec \theta, cos \theta, cos \theta$, we can reduce this to:
Substituting for $x$, we get, $tan^{-1}\large \frac{\sqrt{1+x^2}-1}{x}\:$$= \:tan^{-1} \bigg[ \large\frac{\sqrt{1+tan^2\theta}-1}{tan\theta}\bigg]$
Substituting, $1+tan^2\theta=sec^2\theta$, this reduces to $\;tan^{-1} \large \frac{\sqrt{sec^2\theta}-1}{tan\theta}$ =$tan^{-1}(\large\frac{sec\theta-1}{tan\theta})$
Substituting, $tan \theta = \large \frac{\sin \theta}{\cos \theta}$ and $sec \theta = \large \frac{1}{cos \theta}$, this reduces to: $\tan^{1} \bigg( \Large \frac{ \Large\frac{1}{cos \theta} -1}{\Large \frac{sin \theta}{cos \theta}} \bigg)$$= tan^{-1} \bigg( \large \frac{1-cos\theta}{sin\theta} \bigg)$
Substituting for $cos \theta$ and $sin\theta$, this reduces to $tan^{-1} \bigg( \large \frac{2sin^2\frac{\theta}{2}}{2sin\frac{\theta}{2}cos\frac{\theta}{2}} \bigg)$=$\;tan^{-1}\large\frac{sin\frac{\theta}{2}}{cos\large\frac{\theta}{2}}$ = $tan^{-1}\: tan\large\frac{\theta}{2}$
$tan^{-1} \large \frac{\sqrt {1 + x^2} - 1}{x}$ = $tan^{-1}\: tan\large\frac{\theta}{2}$
$\Rightarrow$ $tan^{-1}\: tan\large\frac{\theta}{2} = \frac{\theta}{2}=\large\frac{1}{2}tan^{-1}x$

Ans:  for   Q.  (ii)

Toolbox:
• $sec^2\theta-1=tan^2\theta$
• $cot\theta = tan ( \large\frac{\pi}{2}-\theta)$
• $\large \large\frac{1}{tan \theta}$ = $cot \theta$
Given, $tan^{-1} \large \frac{1}{\sqrt{x^2 - 1}}$, $| x | > 1$:
We need to remove the square root sign to help simplify the identity:
Let $\;x = sec\theta \Rightarrow \theta = sec^{-1}x$. Substiting for $x$ and using known trignometric identities, we can further reduce this:
Substituting, $tan^{-1}\big(\large \frac{1}{\sqrt{x^2-1}}\big)\:$ becomes$\: tan^{-1} \bigg( \large \frac{1}{\sqrt{sec^2\theta-1}} \bigg)$
Substituting for $sec^2\theta-1=tan^2\theta$, this reduces to $tan^{-1}\large \frac{1}{tan\theta}$
Substituting for $\large \frac{1}{tan \theta}$ = $cot \theta$, this reduces to $tan^{-1}(cot\theta)$
Substituting for $cot\theta = tan (\large \frac{\pi}{2}-\theta)$, we get $tan^{-1}\: tan \bigg( \large\frac{\pi}{2}-\theta \bigg)$
$tan^{-1} \large \frac{1}{\sqrt{x^2 - 1}}$ $= \large\frac{\pi}{2}-\theta=\large\frac{\pi}{2}-sec^{-1}x$
edited Mar 17, 2013