Prove: (1 + tan^2 A) / (1 + cot^2 A) = (1 - tan A) / (1 - cot A)^2 = - Ex 8.3

part 2 - Ex 8.3, 4 (x) - Ex 8.3 - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry
part 3 - Ex 8.3, 4 (x) - Ex 8.3 - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry

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Prove the following identities, where the angles involved are acute angles for which the expressions are defined. ((1 +π‘‘π‘Žπ‘›2 𝐴)/(1 + π‘π‘œπ‘‘2 𝐴))=((1 βˆ’tan⁑〖 𝐴〗)/(1 βˆ’cot⁑ 𝐴))^2=π‘‘π‘Žπ‘›2 𝐴 Solving ((𝟏 + π’•π’‚π’πŸ 𝑨)/(𝟏 + π’„π’π’•πŸ 𝑨)) ((1 + π‘‘π‘Žπ‘›2 𝐴)/(1 + π’„π’π’•πŸ 𝐴)) = ((1 + π‘‘π‘Žπ‘›2 𝐴))/(((1+ 𝟏/(π’•π’‚π’πŸ 𝑨)) ) = ((1 + π‘‘π‘Žπ‘›2 𝐴))/(((tan^2⁑𝐴 + 1))/(tan^2⁑𝐴 ))= (π‘‘π‘Žπ‘›2 𝐴 (1 + π‘‘π‘Žπ‘›2 𝐴))/((π‘‘π‘Žπ‘›2 𝐴 + 1)) = tan2 A = R.H.S Solving ((πŸβˆ’ 𝒕𝒂𝒏⁑𝑨)/(πŸβˆ’ 𝒄𝒐𝒕⁑𝑨 ))^𝟐 ((1βˆ’ tan⁑𝐴)/(1βˆ’ 𝒄𝒐𝒕⁑𝑨 ))^2 = ((1 βˆ’ tan⁑〖 𝐴〗)/(1 βˆ’ 𝟏/𝒕𝒂𝒏⁑〖 𝑨〗 ) " " )^2 = (((1 βˆ’ tan⁑〖 𝐴)γ€—)/(((tan⁑〖 𝐴 βˆ’1γ€— ))/tan⁑〖 𝐴〗 ))^2 = (tan⁑〖 𝐴(1 βˆ’ tan⁑〖 𝐴)γ€— γ€—/( (tan⁑〖 𝐴 βˆ’1)γ€— ))^2 = (tan⁑〖 𝐴(1 βˆ’ tan⁑〖 𝐴)γ€— γ€—/(βˆ’(1 βˆ’ tan⁑〖 𝐴)γ€— ))^2 = (βˆ’tan⁑𝐴 )^2 = tan2 A = RHS Therefore, ((1 + π‘‘π‘Žπ‘›2 𝐴)/(1 + π‘π‘œπ‘‘2 𝐴))=((1 βˆ’ tan⁑〖 𝐴〗)/(1 βˆ’ cot⁑ 𝐴))^2=π‘‘π‘Žπ‘›2 𝐴 H\ence proved

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