Question

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(1 + tan^{2} A)÷(1 + cot^{2} A)​

Answer 1

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♧Question♧

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

♧Answer♧

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

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