♧ Solve and also Justify your Answer :-
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\ \large \mathsf { \frac{1 + \tan {}^{2}A }{1 + \cot {}^{2}A } } \\ \end{gathered}[/tex]
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♧ Solve and also Justify your Answer :-
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\ \large \mathsf { \frac{1 + \tan {}^{2}A }{1 + \cot {}^{2}A } } \\ \end{gathered}[/tex]
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[tex]\colorbox{red}{★QUESTION★}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\ \large \mathsf { \frac{1 + \tan {}^{2}A }{1 + \cot {}^{2}A } } \\ \end{gathered}[/tex]
[tex]\huge\mathfrak\green{☟︎︎︎Answer✍︎}[/tex]
[tex]\frac{1 + tan \: ^{2} A }{1 + cot {}^{2}A} = \frac{sec {}^{2}A}{cosec {}^{2}A } = \frac{ \frac{1}{cos {}^{2}A } }{ \frac{1}{ {sin}^{2}A} } = \frac{sin {}^{2}A }{ {cos}^{2}A} = tan {}^{2}A [/tex][/tex]
[tex]\sf \colorbox{pink} {ANSWER BY ACHALMUCHHAL2}[/tex]
Verified answer
[tex]it \: can \: be \: written \: as \: follows \\ \frac{1 + {tan}^{2}a}{1 + {cos}^{2}a} = \frac{{sec}^{2}a}{{cosec}^{2}a} \\ \frac{1}{ {cos}^{2}a } \frac{1}{ {sin}^{2}a} = \frac{{sin}^{2} a}{{cos}^{2} a} = {tan}^{2}a \\ \\ final \: answer \: = {tan}^{2} a[/tex]