If theta = 2 tan^{-1} frac{1}{8} + 2 tan^{-1} | vedclass

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(B) Given,$	heta = 2 	an^{-1} \frac{1}{8} + 2 	an^{-1} \frac{1}{5} + 	an^{-1} \frac{1}{7}$.Using the formula $2 	an^{-1} x = 	an^{-1} \frac{2x}{

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$27$

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(B) Given,$	heta = 2 	an^{-1} \frac{1}{8} + 2 	an^{-1} \frac{1}{5} + 	an^{-1} \frac{1}{7}$.Using the formula $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$,we have:$2 	an^{-1} \frac{1}{8} = 	an^{-1} \frac{2/8}{1-1/64} = 	an^{-1} \frac{1/4}{63/64} = 	an^{-1} \frac{16}{63}$.$2 	an^{-1} \frac{1}{5} = 	an^{-1} \frac{2/5}{1-1/25} = 	an^{-1} \frac{2/5}{24/25} = 	an^{-1} \frac{5}{12}$.Now,$	heta = 	an^{-1} \frac{16}{63} + 	an^{-1} \frac{5}{12} + 	an^{-1} \frac{1}{7}$.Using $	an^{-1} x + 	an^{-1} y = 	an^{-1} \frac{x+y}{1-xy}$:$	an^{-1} \frac{16}{63} + 	an^{-1} \frac{5}{12} = 	an^{-1} \frac{16/63 + 5/12}{1 - (16/63)(5/12)} = 	an^{-1} \frac{(192+315)/756}{(756-80)/756} = 	an^{-1} \frac{507}{676} = 	an^{-1} \frac{3}{4}$.So,$	heta = 	an^{-1} \frac{3}{4} + 	an^{-1} \frac{1}{7} = 	an^{-1} \frac{3/4 + 1/7}{1 - (3/4)(1/7)} = 	an^{-1} \frac{25/28}{25/28} = 	an^{-1} (1) = \frac{\pi}{4}$.Thus,$	an \frac{	heta}{2} = 	an \frac{\pi}{8} = \sqrt{2} - 1$.Given $	an \frac{	heta}{2} = \sqrt{m} + \sqrt{n}$ with $m This implies $m = 1$ and $n = 2$ (since $m Then $(m^n + n^m)^{m+n} = (1^2 + 2^1)^{1+2} = (1 + 2)^3 = 3^3 = 27$.

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 $(m^n + n^m)^{m+n}$ is equal to

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