Check sibling questions

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


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

  1. Class 10
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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 and Computer Science at Teachoo