If $ heta=2 an ^{-1} \frac{1}{8}+2 an ^{-1} \frac{1}{5}+ an ^{-1} \frac{1}{7}$ and $ an \frac{ heta}{2}=\sqrt{m}+\sqrt{n}$, where $m$ and $n$ are positive integers such that $m

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If $	heta=2 	an ^{-1} \frac{1}{8}+2 	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{7}$ and $	an \frac{	heta}{2}=\sqrt{m}+\sqrt{n}$, where $m$ and $n$ are positive integers such that $m < n$, then $\left(m^n+n^might)^{m+n}$ is equal to

Aicharya · Accepted Answer

Given, $\begin{aligned} 	heta & =2 	an ^{-1}\left(\frac{1}{8}ight)+2 	an ^{-1}\left(\frac{1}{5}ight)+	an ^{-1}\left(\frac{1}{7}ight) \ \Rightarrow 	heta & =2\left[	an ^{-1} \frac{1}{8}+	an ^{-1} \frac{1}{5}ight]+	an ^{-1} \frac{1}{7} \ & =2 	an ^{-1}\left[\frac{1 / 8+1 / 5}{1-1 / 40}ight]^{-1}+	an ^{-1} 1 / 7 \ & =2 	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{7} \ & =	an ^{-1}\left(\frac{2 / 3}{1-1 / 9}ight)+	an ^{-1} 1 / 7 \ & =	an ^{-1}\left(\frac{3}{4}ight)+	an ^{-1} \frac{1}{7} \ & =	an ^{-1}\left(\frac{3 / 4+1 / 7}{1-3 / 28}ight) \ & =	an ^{-1}(1)=\frac{\pi}{4} \ \Rightarrow 	an \frac{	heta}{2} & =	an _{\frac{\pi}{8}}=\sqrt{m}+\sqrt{n} \end{aligned}$ $\begin{aligned} & 	an \left(\frac{\pi}{8}ight)=\sqrt{2}-1=\sqrt{m}+\sqrt{n}, 	ext { when } m < n \ & \Rightarrow \sqrt{m}=\sqrt{2} 	ext { and } \sqrt{n}=-1 \ & \Rightarrow m=2, n=1 \ & \left(m^n+n^might)^{m+n}=\left(1^2+2^1ight)^3=3^3=27 \end{aligned}$

If $\theta=2 \tan ^{-1} \frac{1}{8}+2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}$ and $\tan…

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