#### Answer

$\tan ^{2}\theta $

#### Work Step by Step

$\dfrac {1+\tan ^{2}\theta }{1+\cot ^{2}\theta }=\dfrac {1+\dfrac {\sin ^{2}\theta }{\cos ^{2}\theta }}{1+\dfrac {\cos ^{2}\theta }{\sin ^{2}\theta }}=\dfrac {\dfrac {\cos ^{2}\theta +\sin ^{2}\theta }{\cos ^{2}\theta }}{\dfrac {\sin ^{2}\theta +\cos ^{2}\theta }{\sin ^{2}\theta }}=\dfrac {\sin ^{2}\theta }{\cos ^{2}\theta }=\tan ^{2}\theta $