Prove that (tan⁡ A  + sin ⁡A)/(tan⁡ A - sin ⁡A ) = (sec A + 1)/(sec A - 1)

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Question 30 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic - Part 2

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Question 30 (OR 2nd question) Prove that (tan⁑〖𝐴 γ€—+γ€– sin〗⁑𝐴)/(tan⁑〖𝐴 γ€—βˆ’γ€– 𝑠𝑖𝑛〗⁑𝐴 ) = (𝑠𝑒𝑐 𝐴 + 1)/(𝑠𝑒𝑐 𝐴 βˆ’ 1) Solving LHS (tan⁑〖𝐴 γ€—+γ€– sin〗⁑𝐴)/(tan⁑〖𝐴 γ€—βˆ’γ€– 𝑠𝑖𝑛〗⁑𝐴 ) Writing everything in terms of sin A and cos A = (sin⁑〖 𝐴〗/cos⁑〖 𝐴〗 + sin⁑〖 𝐴〗)/(sin⁑〖 𝐴〗/cos⁑〖 𝐴〗 βˆ’γ€– sin〗⁑〖 𝐴〗 ) = ((sin⁑〖 𝐴〗 + sin⁑𝐴 cos⁑𝐴)/cos⁑〖 𝐴〗 )/((sin⁑〖 𝐴〗 βˆ’ sin⁑𝐴 cos⁑𝐴)/cos⁑〖 𝐴〗 ) = (sin⁑〖 𝐴〗 + sin⁑𝐴 cos⁑𝐴)/(sin⁑〖 𝐴〗 βˆ’ sin⁑𝐴 cos⁑𝐴 ) Taking sin A common = (sin⁑𝐴 (1 + cos⁑𝐴 ))/(sin⁑𝐴 (1 βˆ’ cos⁑𝐴 ) ) = ((1 + cos⁑𝐴 ))/( (1 βˆ’ cos⁑𝐴 ) ) Taking cos A common = (cos⁑〖 𝐴〗 (1/cos⁑𝐴 + 1))/(cos⁑〖 𝐴〗 (1/cos⁑𝐴 βˆ’ 1) ) = ( (1/cos⁑𝐴 + 1))/( (1/cos⁑𝐴 βˆ’ 1) ) = (𝑠𝑒𝑐 𝐴 + 1)/(𝑠𝑒𝑐 𝐴 βˆ’ 1) = RHS Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.