Find the minimum value of n for which |
\[{{\tan }^{-1}}\frac{n}{\pi }>\frac{\pi }{4},\] \[n\in N.\] |
OR |
Show that \[\tan \left( \frac{1}{2}{{\sin }^{-1}}\frac{3}{4} \right)=\frac{4-\sqrt{7}}{3}.\] |
Answer:
We have, \[{{\tan }^{-1}}\frac{n}{\pi }>\frac{\pi }{4}\] \[\Rightarrow \] \[\tan \left( {{\tan }^{-1}}\frac{n}{\pi } \right)>\tan \frac{\pi }{4}\] \[\Rightarrow \] \[\frac{n}{\pi }>1\] \[\Rightarrow \] \[n>\pi \] \[\Rightarrow \] \[n>3.14\] \[\Rightarrow \] n = 4, 5, 6,?. \[[\because \,\,\,n\in N]\] Hence, the minimum value of n is 4. OR Let \[\frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{4} \right)=0\] \[\Rightarrow \] \[{{\sin }^{-1}}\left( \frac{3}{4} \right)=2\theta \] \[\Rightarrow \] \[\frac{3}{4}=\sin 2\theta \] \[\Rightarrow \] \[\frac{3}{4}=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }\] \[\Rightarrow \] \[3+3{{\tan }^{2}}\theta =8\tan \theta \] \[\Rightarrow \] \[3{{\tan }^{2}}\theta -8\tan \theta +3=0\] \[\Rightarrow \] \[\tan \theta =\frac{8\pm \sqrt{64-36}}{6}=\frac{8\pm \sqrt{28}}{6}\] \[\Rightarrow \] \[\tan \theta =\frac{8\pm 2\sqrt{7}}{6}=\frac{4\pm \sqrt{7}}{3}\] \[\Rightarrow \] \[\tan \theta =\frac{4+\sqrt{7}}{3}\,\,\text{or}\,\,\tan \theta =\frac{4-\sqrt{7}}{3}\] \[\Rightarrow \] \[\tan \theta =\frac{4-\sqrt{7}}{3}\] \[\left[ \because \tan \theta =\frac{4+\sqrt{7}}{3}\,\,\text{rejected},\,\,\theta >45{}^\circ \right]\] \[\Rightarrow \] \[\tan \left( \frac{1}{2}{{\sin }^{-1}}\frac{3}{4} \right)=\frac{4-\sqrt{7}}{3}\]
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