Answer
$\dfrac {\tan \theta -\cot \theta }{\tan \theta +\cot \theta }=2\sin ^{2}\theta -1$
Work Step by Step
$\dfrac {\tan \theta -\cot \theta }{\tan \theta +\cot \theta }=\dfrac {\dfrac {\sin \theta }{\cos \theta }-\dfrac {\cos \theta }{\sin \theta }}{\dfrac {\sin \theta }{\cos a}+\dfrac {\cos \theta }{\sin \theta }}=\dfrac {\dfrac {\sin ^{2}\theta -\cos ^{2}\theta }{\cos \theta \sin \theta }}{\dfrac {\sin ^{2}\theta +\cos ^{2}\theta }{\cos \theta \sin \theta }}=\sin ^{2}\theta -\cos ^{2}\theta =\sin ^{2}\theta -\left( 1-\sin ^{2}\theta \right) =2\sin ^{2}\theta -1$
