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(B) Given $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$. Since $a>0$, $f(x)$ is an increasing function. Thus, $f(-1)=0$ and $f(1)=2$. $-a+b

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$f(1)$

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(B) Given $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$. Since $a>0$, $f(x)$ is an increasing function. Thus, $f(-1)=0$ and $f(1)=2$. $-a+b=0 \implies a=b$. $a+b=2 \implies 2a=2 \implies a=1, b=1$. So, $f(x)=x+1$. Now, evaluate the expression: $	an ^{-1} \frac{1}{8}+	an ^{-1} \frac{1}{5} = 	an ^{-1} \left( \frac{\frac{1}{8}+\frac{1}{5}}{1-\frac{1}{8} 	imes \frac{1}{5}} ight) = 	an ^{-1} \left( \frac{13/40}{39/40} ight) = 	an ^{-1} \frac{1}{3}$. Then, $	an ^{-1} \frac{1}{7}+	an ^{-1} \frac{1}{3} = 	an ^{-1} \left( \frac{\frac{1}{7}+\frac{1}{3}}{1-\frac{1}{7} 	imes \frac{1}{3}} ight) = 	an ^{-1} \left( \frac{10/21}{20/21} ight) = 	an ^{-1} \frac{1}{2}$. Finally, $\cot \left( 	an ^{-1} \frac{1}{2} ight) = \cot \left( \cot ^{-1} 2 ight) = 2$. Since $f(1)=1+1=2$, the result is $f(1)$.

For $a>0$, if $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$, then $\cot \left[\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right]=$

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