Browse Questions

# 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