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Simplify $\tan^{-1}\big(\large \frac{\cos x}{1+\sin x}\big)$

$\begin{array}{1 1} \large \frac{\pi}{4}-\frac{x}{2} \\ \large \frac{\pi}{4}+\frac{x}{2} \\ \large \frac{\pi}{4}-\frac{x}{4} \\ \large \frac{\pi}{4}+\frac{x}{4} \end{array} $

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Toolbox:
  • \( cosx=cos^2\large\frac{x}{2}-sin^2\large\frac{x}{2}\)
  • \(\large\frac{1-tanx}{1+tanx}=tan(\large\frac{\pi}{4}-x)\)
  • \((a+b)^2=a^2+b^2+2ab\)
  • \(sinx=2sin\large\frac{x}{2}cos\large\frac{x}{2}\)
  • \(a^2-b^2=(a+b)(a-b)\)
\(In\:(1+sinx),\:take\:1=cos^2\large\frac{x}{2}+sin^2\large\frac{x}{2}\: and\)
\(sinx=2sin\large\frac{x}{2}cos\large\frac{x}{2}\)
 
Take \( a=cos\large\frac{x}{2}\: and\:b=sin\large\frac{x}{2}\) in the above formula of (a+b)^2 then
\((cos\large\frac{x}{2}+sin\large\frac{x}{2})^2=cos^2\large\frac{x}{2}+sin^2\large\frac{x}{2}+2sin\large\frac{x}{2}cos\large\frac{x}{2}\)
\(1+sinx=(cos\large\frac{x}{2}+sin\large\frac{x}{2})^2\)
\( cosx=cos^2\large\frac{x}{2}-sin^2\large\frac{x}{2}\)
\(=(cos\large\frac{x}{2}-sin\large\frac{x}{2})(cos\large\frac{x}{2}+sin\large\frac{x}{2})\)
 
Substituting the values in the given expression we get
\(tan^{-1}\bigg(\large\frac{cosx}{1+sinx}\bigg)\)=
\( tan^{-1} \bigg(\large \frac{cos^2\large\frac{x}{2}-sin^2\large\frac{x}{2}}{(cos\large\frac{x}{2}+sin\large\frac{x}{2})^2} \bigg)\)
\(= tan^{-1} \bigg (\large\frac{(cos\large\frac{x}{2}+sin\large\frac{x}{2})(cos\large\frac{x}{2}-sin\large\frac{x}{2})}{(cos\large\frac{x}{2}+sin\large\frac{x}{2})^2} \bigg)\)
\(= tan^{-1} \bigg(\large \frac{cos\large\frac{x}{2}-sin\large\frac{x}{2}}{cos\large\frac{x}{2}+sin\large\frac{x}{2}} \bigg)\)
 
Divide num. and then. by \( cos\large\frac{x}{2} \)
\( \large\frac{cos\large\frac{x}{2}-sin\large\frac{x}{2}}{cos\large\frac{x}{2}+sin\large\frac{x}{2}}=\large\frac{1-\large\frac{sin\large\frac{x}{2}}{cos\large\frac{x}{2}}}{1-\frac{sin\large\frac{x}{2}}{cos\large\frac{x}{2}}}=\frac{1-tan\large\frac{x}{2}}{1+tan\large\frac{x}{2}}\)
\(\Rightarrow\: tan^{-1} \bigg( \large\frac{cos\large\frac{x}{2}-sin\large\frac{x}{2}}{cos\large\frac{x}{2}+sin\large\frac{x}{2}} \bigg)=tan^{-1} \bigg(\large \frac{1-tan\large\frac{x}{2}}{1+tan\large\frac{x}{2}} \bigg)\)
\(=tan^{-1}\bigg(\large\frac{tan\large\frac{\pi}{4}-tan\large\frac{x}{2}}{tan\large\frac{\pi}{4}+tan\large\frac{x}{2}} \bigg)\)
 
By taking \(\large\frac{x}{2}\) in the place of x in the above formula of \(\large\frac{1-tanx}{1+tanx}=tan(\large\frac{\pi}{4}-x)\) we get
\( = tan^{-1} \: tan \bigg( \large\frac{\pi}{4}-\large\frac{x}{2} \bigg) = \large\frac{\pi}{4}-\large\frac{x}{2} \)

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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