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

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Slide69.JPG

  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

Transcript

Question 30 (OR 2nd question) Prove that (tan⁡〖𝐴 〗+〖 sin〗⁡𝐴)/(tan⁡〖𝐴 〗−〖 𝑠𝑖𝑛〗⁡𝐴 ) = (𝑠𝑒𝑐 𝐴 + 1)/(𝑠𝑒𝑐 𝐴 − 1) Solving LHS (tan⁡〖𝐴 〗+〖 sin〗⁡𝐴)/(tan⁡〖𝐴 〗−〖 𝑠𝑖𝑛〗⁡𝐴 ) Writing everything in terms of sin A and cos A = (sin⁡〖 𝐴〗/cos⁡〖 𝐴〗 + sin⁡〖 𝐴〗)/(sin⁡〖 𝐴〗/cos⁡〖 𝐴〗 −〖 sin〗⁡〖 𝐴〗 ) = ((sin⁡〖 𝐴〗 + sin⁡𝐴 cos⁡𝐴)/cos⁡〖 𝐴〗 )/((sin⁡〖 𝐴〗 − sin⁡𝐴 cos⁡𝐴)/cos⁡〖 𝐴〗 ) = (sin⁡〖 𝐴〗 + sin⁡𝐴 cos⁡𝐴)/(sin⁡〖 𝐴〗 − sin⁡𝐴 cos⁡𝐴 ) Taking sin A cos A common = (sin⁡〖 𝐴〗 cos⁡〖 𝐴〗 (1/cos⁡𝐴 + 1))/(sin⁡〖 𝐴〗 cos⁡〖 𝐴〗 (1/cos⁡𝐴 − 1) ) = ( (1/cos⁡𝐴 + 1))/( (1/cos⁡𝐴 − 1) ) = (𝑠𝑒𝑐 𝐴 + 1)/(𝑠𝑒𝑐 𝐴 − 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.