Question
Evaluate:
$\sec\Big\{\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$

Answer

$\sec\Big\{\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
$\sec\Big\{-\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$\big[\because\ \cot^{-1}(-\text{x})=-\cot^{-1}(\text{x})\text{ for all }\text{x}\in(-1,1)\big]$
$=\sec\Big\{-\sec^{-1}\Big(\frac{13}{5}\Big)\Big\}$ $\Big[\because\ \cot^{-1}\Big(\frac{\text{b}}{\text{p}}\Big)=\sec^{-1}\Big(\frac{\text{h}}{\text{b}}\Big)\Big]$
$=\sec\Big\{\sec^{-1}\Big(\frac{13}{5}\Big)\Big\}$ $[\because\ \sec(\text{-x})=\sec\text{x}]$
$=\frac{13}{5}$

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