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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

• $1-cos\theta=2sin^2\large\frac{\theta}{2}$
• $sin\theta=2sin\large\frac{\theta}{2}\: cos\large\frac{\theta}{2}$
• $1+tan^2\theta=sec^2\theta$
• $tan \theta = \large\frac{\sin \theta}{\cos \theta}$
• $sec \theta = \large\frac{1}{cos \theta}$
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 $\;tan^{-1} \large \frac{\sqrt{sec^2\theta}-1}{tan\theta}$ =$tan^{-1}(\large\frac{sec\theta-1}{tan\theta})$