If theta = cot^{-1}(7) + cot^{-1}(8) + cot^{- | vedclass

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(B) Given that,$	heta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$.Using the property $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,we have:$

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$3$

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(B) Given that,$	heta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$.Using the property $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,we have:$	heta = 	an^{-1}(\frac{1}{7}) + 	an^{-1}(\frac{1}{8}) + 	an^{-1}(\frac{1}{18})$.First,apply the formula $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}(\frac{x+y}{1-xy})$ for the first two terms:$	an^{-1}(\frac{1}{7}) + 	an^{-1}(\frac{1}{8}) = 	an^{-1}(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7 	imes 8}}) = 	an^{-1}(\frac{\frac{15}{56}}{\frac{55}{56}}) = 	an^{-1}(\frac{15}{55}) = 	an^{-1}(\frac{3}{11})$.Now,add the third term:$	heta = 	an^{-1}(\frac{3}{11}) + 	an^{-1}(\frac{1}{18}) = 	an^{-1}(\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11 	imes 18}}) = 	an^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = 	an^{-1}(\frac{65}{195}) = 	an^{-1}(\frac{1}{3})$.Thus,$	an 	heta = \frac{1}{3}$,which implies $\cot 	heta = 3$.

If $\theta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$,then $\cot \theta$ is equal to

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