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Show that $\tan({\frac{1}{2}}\sin^{-1}{\frac{3}{4}})=\frac{4-\sqrt 7}{3}$

1 Answer

Toolbox:
  • Take $( sin^{-1}\frac{3}{4}=x)$ and proceed
  • $( cosx=2cos^2\frac{x}{2}-1)$
  • $(cosx=\sqrt{1-sin^2x})$
L.H.S.
Let $( sin^{-1}\frac{3}{4}=x \Rightarrow sinx=\frac{3}{4})$
L.H.S$=(tan\frac{x}{2})$
$( cos x=\sqrt{1-sin^2x}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt 7}{4})$
$( \Rightarrow\:cosx= 2cos^2\frac{x}{2}-1=\frac{\sqrt 7}{4})$
$( \Rightarrow 2cos^2\frac{x}{2}=\frac{\sqrt{7}}{4}+1=\frac{4+\sqrt 7}{4})$
$( \Rightarrow cos\frac{x}{2}=\sqrt{\frac{4+\sqrt 7}{8}})$
$(sin\frac{x}{2}=\sqrt{1-cos^2\frac{x}{2}}=\sqrt{1-\big(\frac{4+\sqrt7}{8}\big)}=\sqrt{\frac{4-\sqrt 7}{8}})$
$(\Rightarrow\: tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{4-\sqrt 7}{4+\sqrt 7})$
$(=\frac{4-\sqrt 7}{3})$ (by rationalising the denominator)
$( tan\frac{x}{2}=\frac{4-\sqrt 7}{3})$
$(\Rightarrow\:L.H.S=R.H.S.)$
answered Jun 12, 2014 by balaji.thirumalai
edited Nov 30 by meena.p
 
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