question_answer

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        \[\frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{4} \right)=0\]

\[\Rightarrow \]   \[{{\sin }^{-1}}\left( \frac{3}{4} \right)=2\theta \] \[\Rightarrow \] \[\frac{3}{4}=\sin 2\theta \]

\[\Rightarrow \]   \[\frac{3}{4}=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }\]

\[\Rightarrow \]   \[3+3{{\tan }^{2}}\theta =8\tan \theta \]

\[\Rightarrow \]   \[3{{\tan }^{2}}\theta -8\tan \theta +3=0\]

\[\Rightarrow \]   \[\tan \theta =\frac{8\pm \sqrt{64-36}}{6}=\frac{8\pm \sqrt{28}}{6}\]

\[\Rightarrow \]   \[\tan \theta =\frac{8\pm 2\sqrt{7}}{6}=\frac{4\pm \sqrt{7}}{3}\]

\[\Rightarrow \]   \[\tan \theta =\frac{4+\sqrt{7}}{3}\,\,\text{or}\,\,\tan \theta =\frac{4-\sqrt{7}}{3}\]

\[\Rightarrow \]   \[\tan \theta =\frac{4-\sqrt{7}}{3}\]

\[\left[ \because \tan \theta =\frac{4+\sqrt{7}}{3}\,\,\text{rejected},\,\,\theta >45{}^\circ  \right]\]

\[\Rightarrow \]   \[\tan \left( \frac{1}{2}{{\sin }^{-1}}\frac{3}{4} \right)=\frac{4-\sqrt{7}}{3}\]