Prove that: Root (1 + sin A) / (1 - sin A) = sec A + tan A - Teachoo - Ex 8.3

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

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Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) √((1 + sin⁑𝐴 )/(1 βˆ’γ€– sin〗⁑𝐴 )) = sec A + tan A Solving L.H.S √((𝟏 + π’”π’Šπ’β‘π‘¨ )/(𝟏 βˆ’γ€– π’”π’Šπ’γ€—β‘π‘¨ )) Rationalizing denominator Multiplying (1 + sin A) in numerator and denominator = √(((𝟏 + 𝐬𝐒𝐧⁑𝑨)(𝟏 + π’”π’Šπ’β‘γ€–π‘¨)γ€— )/((𝟏 βˆ’ 𝐬𝐒𝐧⁑𝑨)(𝟏 + π’”π’Šπ’β‘γ€–π‘¨)γ€— )) = √(((1 + sin⁑𝐴 )2 )/(12 βˆ’ 𝑠𝑖𝑛2𝐴)) = √(((1 + sin⁑𝐴 )2 )/(1 βˆ’ 𝑠𝑖𝑛2𝐴)) =√(((1 + sin⁑𝐴)2 )/(π’„π’π’”πŸ 𝑨)) =√(((1 + sin⁑𝐴 )/(π‘π‘œπ‘  𝐴))^2 ) = (𝟏 + π’”π’Šπ’β‘γ€– 𝑨〗)/𝒄𝒐𝒔⁑〖 𝑨〗 = 1/cos⁑〖 𝐴〗 + sin⁑〖 𝐴〗/cos⁑〖 𝐴〗 = sec A + tan A = R.H.S Hence proved

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