Solve it please Need correct answer (1 + tan^{2} A)÷(1 + cot^{2} A)�

1.	Solve it please Need correct answer (1 + tan^{2} A)÷(1 + cot^{2} A)
Answer» _______________ ♧Question♧ $\dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ \sf(1 + cot^{2} A)}\bigg\}}$ ♧Answer♧ $\dashrightarrow{\Bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ \sf(1 + cot^{2} A)}\bigg\}} \\ \\ \sf{As \: we \: know \: that \: \: \: \: \: cot^{2} = \frac{1}{tan^{2} } } \\ \\ \dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ \sf(1 + \dfrac{1}{tan^{2}A } )}\bigg\}} \\ \\ \sf{Now,By \: taking \: \red{LCM}} \\ \\ \dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ (\sf{\frac{tan ^{2}A + 1 }{tan^{2}A })}}\bigg\}} \\ \\\dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ (\sf{tan ^{2}A + 1 } )} \times \sf tan ^{2}A \bigg\}} \\ \\ \dashrightarrow{ \bigg\{ \cancel\dfrac{ \sf(1 + tan^{2}A )}{ (\sf{tan ^{2}A + 1 } )} \times \sf tan ^{2}A \bigg\}} \\ \\ {\dashrightarrow{ \boxed{ \bigg\{{ \sf \red{tan ^{2}A }} \bigg\}}}} \sf{ \longleftarrow{Ans}}$ ___________________

Solve it please Need correct answer (1 + tan^{2} A)÷(1 + cot^{2} A)

Answer»

_______________

♧Question♧

$\dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ \sf(1 + cot^{2} A)}\bigg\}}$

♧Answer♧

$\dashrightarrow{\Bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ \sf(1 + cot^{2} A)}\bigg\}} \\ \\ \sf{As \: we \: know \: that \: \: \: \: \: cot^{2} = \frac{1}{tan^{2} } } \\ \\ \dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ \sf(1 + \dfrac{1}{tan^{2}A } )}\bigg\}} \\ \\ \sf{Now,By \: taking \: \red{LCM}} \\ \\ \dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ (\sf{\frac{tan ^{2}A + 1 }{tan^{2}A })}}\bigg\}} \\ \\\dashrightarrow{ \bigg\{\dfrac{ \sf(1 + tan^{2}A )}{ (\sf{tan ^{2}A + 1 } )} \times \sf tan ^{2}A \bigg\}} \\ \\ \dashrightarrow{ \bigg\{ \cancel\dfrac{ \sf(1 + tan^{2}A )}{ (\sf{tan ^{2}A + 1 } )} \times \sf tan ^{2}A \bigg\}} \\ \\ {\dashrightarrow{ \boxed{ \bigg\{{ \sf \red{tan ^{2}A }} \bigg\}}}} \sf{ \longleftarrow{Ans}}$