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

Question 30 (OR 2nd question).jpg

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  1. Class 10
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Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.