If cot ^{-1}(7)+cot ^{-1}(8)+cot ^{-1}(18)=co | vedclass

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(C) Given: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ Using the property $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ for $y > 0$: $	an ^

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$3$

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(C) Given: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ Using the property $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ for $y > 0$: $	an ^{-1}(\frac{1}{7})+	an ^{-1}(\frac{1}{8})+	an ^{-1}(\frac{1}{18})=	an ^{-1}(\frac{1}{x})$ First,apply $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}(\frac{A+B}{1-AB})$ for $	an ^{-1}(\frac{1}{7}) + 	an ^{-1}(\frac{1}{8})$: $	an ^{-1}(\frac{\frac{1}{7}+\frac{1}{8}}{1-(\frac{1}{7})(\frac{1}{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: $	an ^{-1}(\frac{3}{11}) + 	an ^{-1}(\frac{1}{18}) = 	an ^{-1}(\frac{\frac{3}{11}+\frac{1}{18}}{1-(\frac{3}{11})(\frac{1}{18})})$ $= 	an ^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = 	an ^{-1}(\frac{65}{195}) = 	an ^{-1}(\frac{1}{3})$ Thus,$	an ^{-1}(\frac{1}{3}) = 	an ^{-1}(\frac{1}{x})$,which implies $x = 3$.

If $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$,then the value of $x$ is

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