Answer
$\dfrac {\tan ^{2}t-1}{sec^{2}t}=\dfrac {\tan ^{2}t-1}{\tan ^{2}t+1}=\dfrac {\cot t\left( \tan ^{2}t-1\right) }{\cot t\left( \tan ^{2}t+1\right) }=\dfrac {\tan t-\cot t}{\tan t+\cot t}$
Work Step by Step
$\dfrac {\tan ^{2}t-1}{sec^{2}t}=\dfrac {\tan ^{2}t-1}{\tan ^{2}t+1}=\dfrac {\cot t\left( \tan ^{2}t-1\right) }{\cot t\left( \tan ^{2}t+1\right) }=\dfrac {\tan t-\cot t}{\tan t+\cot t}$