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

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

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
$= tan^{-1} \: tan \bigg( \large\frac{\pi}{4}-\large\frac{x}{2} \bigg) = \large\frac{\pi}{4}-\large\frac{x}{2}$

edited Mar 19, 2013