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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Prove that \( tan \bigg( \frac{\pi}{4}+\frac{1}{2} cos^{-1}\frac{a}{b} \bigg)+tan \bigg(\frac{\pi}{4}-\frac{1}{2} cos^{-1}\frac{a}{b} \bigg) = \frac{2b}{a} \)

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Toolbox:
  • Take \( cos^{-1}\large\frac{a}{b}=\theta\)\( \Rightarrow\:\) \( cos\theta=\large\frac{a}{b}\)
  • \( tan(A \pm B) = \large\frac{tanA \pm tanB}{1 \mp tanAtanB}\)
  • \( cos\theta = \large\frac{1-tan^2\large\frac{\theta}{2}}{1+tan^2\large\frac{\theta}{2}}\)
Put \(cos^{-1}\large\frac{a}{b}=\theta\)\( \Rightarrow\:\) \( cos\theta=\large\frac{a}{b}\)
Then,L.H.S = \( tan \bigg[ \large\frac{\pi}{4}+\large\frac{\theta}{2} \bigg] + tan\bigg[ \large\frac{\pi}{4}-\large\frac{\theta}{2} \bigg] \)
Using the formula of tan(A+B) and tan(A-B) by taking A=\(\frac{\pi}{4}\:and\:B=\frac{\theta}{2}\) we have
\(tan \bigg[ \large\frac{\pi}{4}+\large\frac{\theta}{2}\bigg]=\frac{tan\frac{\pi}{4}+tan\frac{\theta}{2}}{1-tan\frac{\pi}{4}.tan\frac{\theta}{2}}=\large \frac{1+tan\large\frac{\theta}{2}}{1-tan\large\frac{\theta}{2}}\)
\(tan \bigg[ \large\frac{\pi}{4}-\large\frac{\theta}{2}\bigg]=\frac{tan\frac{\pi}{4}-tan\frac{\theta}{2}}{1+tan\frac{\pi}{4}.tan\frac{\theta}{2}}=\large \frac{1-tan\large\frac{\theta}{2}}{1+tan\large\frac{\theta}{2}}\)
Substituting the values in L.H.S. we get
\(L.H.S. =\large \frac{1+tan\large\frac{\theta}{2}}{1-tan\large\frac{\theta}{2}}+\large\frac{1-tan\large\frac{\theta}{2}}{1+tan\large\frac{\theta}{2}}\)
\(=\large\frac{(1+tan\frac{\theta}{2})^2+\large(1-tan\frac{\theta}{2})^2}{1-\large\:tan^2\frac{\theta}{2}}\)
\( = \bigg( \large\frac{2+2tan^2\large\frac{\theta}{2}}{1-tan^2\large\frac{\theta}{2}} \bigg)\)
\( = 2\bigg( \large\frac{1+tan^2\large\frac{\theta}{2}}{1-tan^2\large\frac{\theta}{2}} \bigg)\)
But we know that \( cos\theta = \large\frac{1-tan^2\large\frac{\theta}{2}}{1+tan^2\large\frac{\theta}{2}}\)
\(\Rightarrow\:\frac{1}{ cos\theta} = \large\frac{1-tan^2\large\frac{\theta}{2}}{1+tan^2\large\frac{\theta}{2}}\)
Therefore, L.H.S.becomes \( = \large\frac{2}{cos\theta}=2\large\frac{b}{a}=R.H.S\)
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 20, 2013 by rvidyagovindarajan_1
 

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