document.write( "Question 62235: Prove that:
\n" ); document.write( "1 + cos A/ 1 - cos A = tan²A/(secA - 1)² \n" ); document.write( "

Algebra.Com's Answer #42949 by kapilsinghi(68) $\"\"$ $\"About$

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1 + cos A/ 1 - cos A = tan²A/(secA - 1)²\r
\n" ); document.write( "\n" ); document.write( "LHS = (1 + cos A)/ (1 - cos A)\r
\n" ); document.write( "\n" ); document.write( "RHS = tan²A/(secA - 1)²
\n" ); document.write( " = [tanA/(sec-1)]^2
\n" ); document.write( " = [tanA/((1/cosA)-1)]^2
\n" ); document.write( " = [tanA/(1-cosA)/cosA]^2
\n" ); document.write( " = [(tanA*cosA)/(1-cosA)]^2
\n" ); document.write( " = [sinA/(1-cosA)]^2
\n" ); document.write( " = sinA^2/(1-cosA)^2
\n" ); document.write( " = (1-cosA^2)/(1-cosA)^2
\n" ); document.write( " = [(1-cosA)*(1+cosA)]/[(1-cosA)*(1-cosA)]
\n" ); document.write( " = (1+cosA)/(1-cosA)\r
\n" ); document.write( "\n" ); document.write( "LHS = RHS\r
\n" ); document.write( "\n" ); document.write( "Regards
\n" ); document.write( "Kapil
\n" ); document.write( "kapilsinghi123@gmail.com \n" ); document.write( "

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