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If $\theta=2 \tan ^{-1} \frac{1}{8}+2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}$ and $\tan \frac{\theta}{2}=\sqrt{m}+\sqrt{n}$, where $m$ and $n$ are positive integers such that $m < n$, then $\left(m^n+n^m\right)^{m+n}$ is equal to
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Verified Answer
The correct answer is:
27
Given,
$\begin{aligned}
\theta & =2 \tan ^{-1}\left(\frac{1}{8}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\
\Rightarrow \theta & =2\left[\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right]+\tan ^{-1} \frac{1}{7} \\
& =2 \tan ^{-1}\left[\frac{1 / 8+1 / 5}{1-1 / 40}\right]^{-1}+\tan ^{-1} 1 / 7 \\
& =2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7} \\
& =\tan ^{-1}\left(\frac{2 / 3}{1-1 / 9}\right)+\tan ^{-1} 1 / 7 \\
& =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1} \frac{1}{7} \\
& =\tan ^{-1}\left(\frac{3 / 4+1 / 7}{1-3 / 28}\right) \\
& =\tan ^{-1}(1)=\frac{\pi}{4} \\
\Rightarrow \tan \frac{\theta}{2} & =\tan _{\frac{\pi}{8}}=\sqrt{m}+\sqrt{n}
\end{aligned}$
$\begin{aligned}
& \tan \left(\frac{\pi}{8}\right)=\sqrt{2}-1=\sqrt{m}+\sqrt{n}, \text { when } m < n \\
& \Rightarrow \sqrt{m}=\sqrt{2} \text { and } \sqrt{n}=-1 \\
& \Rightarrow m=2, n=1 \\
& \left(m^n+n^m\right)^{m+n}=\left(1^2+2^1\right)^3=3^3=27
\end{aligned}$
$\begin{aligned}
\theta & =2 \tan ^{-1}\left(\frac{1}{8}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\
\Rightarrow \theta & =2\left[\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right]+\tan ^{-1} \frac{1}{7} \\
& =2 \tan ^{-1}\left[\frac{1 / 8+1 / 5}{1-1 / 40}\right]^{-1}+\tan ^{-1} 1 / 7 \\
& =2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7} \\
& =\tan ^{-1}\left(\frac{2 / 3}{1-1 / 9}\right)+\tan ^{-1} 1 / 7 \\
& =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1} \frac{1}{7} \\
& =\tan ^{-1}\left(\frac{3 / 4+1 / 7}{1-3 / 28}\right) \\
& =\tan ^{-1}(1)=\frac{\pi}{4} \\
\Rightarrow \tan \frac{\theta}{2} & =\tan _{\frac{\pi}{8}}=\sqrt{m}+\sqrt{n}
\end{aligned}$
$\begin{aligned}
& \tan \left(\frac{\pi}{8}\right)=\sqrt{2}-1=\sqrt{m}+\sqrt{n}, \text { when } m < n \\
& \Rightarrow \sqrt{m}=\sqrt{2} \text { and } \sqrt{n}=-1 \\
& \Rightarrow m=2, n=1 \\
& \left(m^n+n^m\right)^{m+n}=\left(1^2+2^1\right)^3=3^3=27
\end{aligned}$
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