# Write the following function in the simplest form: $tan^{-1} \bigg( \frac{cos\: x - sin\: x}{cos\: x + sin\: x} \bigg) , 0 < x < \pi$

Toolbox:
• $$\large \frac{cosx-sinx}{cosx+sinx}$$ $$=\large \frac{1-tanx}{1+tanx}$$
• $$tan(A-B)=\large \frac{tanA-tanB}{1+tanA.tanB}$$
• $$tan\frac{\pi}{4}=1$$
• $tan x = \large \frac{sinx}{cosx}$
Given $\; tan^{-1} \bigg(\large \frac{cos\: x - sin\: x}{cos\: x + sin\: x} \bigg)$$, 0 < x < \pi: Dividing the numerator by cosx, we get \large\frac{cosx-sinx}{cosx} =$$1-$$\large \frac{sinx}{cosx} = 1 - tanx. Dividing the denominator by cosx, we get \large\frac{cosx+sinx}{cosx}$$ = 1+$$\large \frac{sinx}{cosx} = 1 +tanx. Substituting for numerator and denominator, \; tan^{-1} \bigg(\large \frac{cos\: x - sin\: x}{cos\: x + sin\: x} \bigg) reduces to tan^{-1} \bigg(\large \frac{1-tanx}{1+tanx} \bigg) Substituting $$tan \large \frac{\pi}{4}$$ $$=1$$, this reduces to $$tan^{-1} \bigg( \large \frac{tan\frac{\pi}{4}-tanx}{1+tan\frac{\pi}{4}.tanx} \bigg)$$ We know that $$tan(A-B)=\large \frac{tanA-tanB}{1+tanA.tanB}$$ Let A=$$\large \frac{\pi}{4}$$ and B= x \Rightarrow \large \frac{tan\frac{\pi}{4}-tanx}{1+tan\frac{\pi}{4}.tanx}$$=tan(\frac{\pi}{4}-x)$
$\Rightarrow \; tan^{-1} \bigg(\large \frac{cos\: x - sin\: x}{cos\: x + sin\: x} \bigg)$ $= \:tan^{-1}\: tan \bigg( \frac{\pi}{4}-x \bigg)$ $$= \frac{\pi}{4}-x$$
edited Mar 14, 2013