Prove that: (cos A - sin A + 1) / (cos A + sin A - 1) = cosec A + cot - Ex 8.3

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

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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. "cos A – sin A + 1" /"cos A + sin A – 1" = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A. Solving L.H.S (cos⁑𝐴 βˆ’ sin⁑𝐴 + 1)/(cos⁑𝐴 + sin⁑𝐴 βˆ’ 1) Since we need to use cosec and cot identity Dividing both numerator and denominator by sin A = (𝟏/π’”π’Šπ’β‘γ€– 𝑨〗 (cos⁑〖 𝐴 βˆ’ sin⁑〖𝐴 + 1γ€— γ€— ))/(𝟏/π’”π’Šπ’β‘γ€– 𝑨〗 (cos⁑〖 𝐴 + sin⁑〖 𝐴 βˆ’ 1γ€— γ€— ) ) = (cos⁑〖 𝐴〗/sin⁑〖 𝐴〗 βˆ’ sin⁑〖 𝐴〗/sin⁑〖 𝐴〗 + 1/sin⁑〖 𝐴〗 )/(cos⁑〖 𝐴〗/sin⁑〖 𝐴〗 + sin⁑〖 𝐴〗/sin⁑〖 𝐴〗 βˆ’ 1/sin⁑〖 𝐴〗 ) = cot⁑〖 𝐴 βˆ’ 1 + π‘π‘œπ‘ π‘’π‘ 𝐴〗/cot⁑〖 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴〗 = ((cot⁑〖 𝐴 + π‘π‘œπ‘ π‘’π‘ 𝐴) βˆ’ πŸγ€—)/((cot⁑〖 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴) γ€— ) = ((co𝑑⁑〖 𝐴 + π‘π‘œπ‘ π‘’π‘ 𝐴) βˆ’ (π’„π’π’”π’†π’„πŸ 𝑨 βˆ’ π’„π’π’•πŸ 𝑨)γ€—)/((cot⁑ 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴)) = ((co𝑑⁑〖 𝐴 + π‘π‘œπ‘ π‘’π‘ 𝐴) βˆ’ (πœπ¨π’”π’†π’„β‘π‘¨ βˆ’ 𝒄𝒐𝒕 𝑨)(πœπ¨π’”π’†π’„β‘π‘¨ + 𝒄𝒐𝒕 𝑨)γ€—)/((cot⁑ 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴)) = ((co𝑑⁑〖 𝐴 + π‘π‘œπ‘ π‘’π‘ 𝐴) [𝟏 βˆ’ (𝒄𝒐𝒔𝒆𝒄 𝑨 βˆ’ 𝒄𝒐𝒕 𝑨 )]γ€—)/([cot⁑ 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴]) = ((co𝑑⁑〖 𝐴 + π‘π‘œπ‘ π‘’π‘ 𝐴) [1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴 + π‘π‘œπ‘‘ 𝐴]γ€—)/([cot⁑ 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴]) = ((co𝑑⁑〖 𝐴 + π‘π‘œπ‘ π‘’π‘ 𝐴)[cot⁑ 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴]γ€—)/([cot⁑ 𝐴 + 1 βˆ’ π‘π‘œπ‘ π‘’π‘ 𝐴]) = cot A + cosec A = R.H.S Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo