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In any Triangle ABC, if $ A=\tan^{-1}2\;and\;B=\tan^{-1}3,$prove that $C=\frac{\pi}{4}$

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  • In \( \Delta\: ABC \: A+B+C = \pi \)
  • \( tan(A+B)=\large\frac{tanA+tanB}{1-tanA.tanB}\)
  • \(tan(\pi-C)=-tanC\)
  • \(tan\large\frac{\pi}{4}=1\)
Given ABC is a triangle and \( tanA = 2 \: and\: tanB=3\)
\(\Rightarrow A+B+C=\pi\)
\( \Rightarrow A+B=\pi-C\)
Taking tan on both the sides and using the formula of \(tan(\pi-C)\)
\( \Rightarrow\:tan(A+B)=tan(\pi-C)=-tanC\)
From the above formula of tan(A+B)
\( \Rightarrow\:\large\frac{tanA+tanB}{1-tanAtanB}=-tanC\)
Substituting the values of tanA and tanB we get
\(\large \frac{2+3}{1-2.3}=\large\frac{5}{-5}=-1=-tanC\)
\( \Rightarrow tanC=1\)
\( C=\large\frac{\pi}{4}\)


answered Feb 19, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1

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