document.write( "Question 836084: Sorry, but I don't understand how to do these sort of questions...
\n" ); document.write( "Prove the following identities
\n" ); document.write( "2tan^2/1+tan^2 =2 sin^2
\n" ); document.write( "Sin^2(1+sec^2)=sec^2-cos^2
\n" ); document.write( "1-sinA+cosA/1-sinA= 1+sinA+cosA/cosA\r
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\n" ); document.write( "\n" ); document.write( "Please help, Thank you!
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Algebra.Com's Answer #503996 by stanbon(75887)\"\" \"About 
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.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan
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