document.write( "Question 836084: Sorry, but I don't understand how to do these sort of questions...
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document.write( "Prove the following identities
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document.write( "2tan^2/1+tan^2 =2 sin^2
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document.write( "Sin^2(1+sec^2)=sec^2-cos^2
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document.write( "1-sinA+cosA/1-sinA= 1+sinA+cosA/cosA\r
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document.write( "Please help, Thank you! \n" );
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Algebra.Com's Answer #503996 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Prove the following identities \n" ); document.write( "2tan^2/1+tan^2 =2 sin^2 \n" ); document.write( "Note: 1+tan^2 = sec^2 \n" ); document.write( "---- \n" ); document.write( "So, 2tan^2/sec^2 = 2sin^2 \n" ); document.write( "Then 2[sin^2/cos^2]/[1/cos^2] = 2sin^2 \n" ); document.write( "Therefore: 2[sin^2/cos^2] = 2sin^2 \n" ); document.write( "Finally: 2sin^2 = 2sin^2\r \n" ); document.write( "\n" ); document.write( "--------------------\r \n" ); document.write( "\n" ); document.write( "Sin^2(1+sec^2)=sec^2-cos^2 \n" ); document.write( "--------- \n" ); document.write( "Multiply on the left side: \n" ); document.write( "Convert sec^2 to tan^2+1 on the right side \n" ); document.write( "sin^2 + tan^2 = tan^2+1 - cos^2 \n" ); document.write( "------ \n" ); document.write( "Rearrange the right side: \n" ); document.write( "sin^2 + tan^2 = tan^2 + [1-cos^2] \n" ); document.write( "------ \n" ); document.write( "Apply: sin^2 = 1-cos^2 to get \n" ); document.write( "sin^2 + tan^2 = tan^2 + sin^2 \n" ); document.write( "========================================\r \n" ); document.write( "\n" ); document.write( "1-sinA+cosA/1-sinA= 1+sinA+cosA/cosA \n" ); document.write( "Comment: This is confusing. \n" ); document.write( "Using parentheses, show where numerators and denominators begin and end. \n" ); document.write( "================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan \n" ); document.write( " \n" ); document.write( " |