# 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 $$\;tan^{-1} \large \frac{\sqrt{sec^2\theta}-1}{tan\theta}$$ =$$tan^{-1}(\large\frac{sec\theta-1}{tan\theta})$$