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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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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
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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 \(tan^{-1}\: tan \bigg( \large\frac{\pi}{2}-\theta \bigg) \)
$tan^{-1} \large \frac{1}{\sqrt{x^2 - 1}}$ \( = \large\frac{\pi}{2}-\theta=\large\frac{\pi}{2}-sec^{-1}x\)
answered Mar 1, 2013 by thanvigandhi_1
edited Mar 17, 2013 by rvidyagovindarajan_1
 

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