# 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