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

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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)$
$\Rightarrow \; tan^{-1} \bigg(\large \frac{cos\: x - sin\: x}{cos\: x + sin\: x} \bigg)$ $= \:tan^{-1}\: tan \bigg( \frac{\pi}{4}-x \bigg)$ \( = \frac{\pi}{4}-x\)
answered Feb 22, 2013 by thanvigandhi_1
edited Mar 14, 2013 by balaji.thirumalai
 

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