Question

✴️ AHOLA FRIENDS ✴️

⚪New question for u guys⚪

Prove the identity, where the angles involved are acute angles for which the expressions are defined-

(1+ Tan² A /1+ Cot² A) =(1- Tan A / 1-Cot A)² = Tan² A

➡️ Hoping for good answers ⬅️

Answer 1

[1st]

$= > \frac{1 + {tan}^{2} A}{1 + cot {}^{2}A } \\ \\ = > \frac{ {sec}^{2}A }{ {cosec}^{2}A } \\ \\ = > \frac{ \frac{1}{cos {}^{2}A } }{ \frac{1}{ {sin}^{2} A} } \\ \\ = > \frac{ {sin}^{2}A }{ { {cos}^{2}A }^{} } \\ \\ = > {tan}^{2} A$
_____________[PROVED]

=======================

[2nd]

$= > ( { \frac{1 - tanA}{1 - cotA} })^{2} \\ \\ = > \frac{ {(1 - tanA)}^{2} }{ {(1 - cotA)}^{2} } \\ \\ = > \frac{ {(1 - tanA)}^{2} }{ ( { \frac{tanA - 1}{tanA}) }^{2} } \\ \\ = > \frac{ {(1 - tanA)}^{2} }{ \frac{ {(1 - tanA)}^{2} }{ - {tan}^{2} A} } \\ \\ = > {tan}^{2} A$
______________[PROVED]

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Answer 2

Hi there !!

⬆️Look at the attachment ⬆️

Hope it will helpful .

Thanks :)