document.write( "Question 749317: 1/(sec(θ)-tan(θ))=sec(θ)+tan(θ) how do I prove this? \n" ); document.write( "

Algebra.Com's Answer #456113 by FrankM(1040) $\"\"$ $\"About$

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1/(sec(θ)-tan(θ))=sec(θ)+tan(θ)\r
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\n" ); document.write( "\n" ); document.write( "(sec(θ)-tan(θ)) (sec(θ)+tan(θ)) = 1 (we cross multiplied)\r
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\n" ); document.write( "\n" ); document.write( "(sec^2(θ)-tan^2(θ)) = 1 ( (a+b) * (a-b) = a^2 - b^2 )\r
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\n" ); document.write( "\n" ); document.write( "tan^2(θ) + 1 = sec^2(θ) (rearranged)\r
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\n" ); document.write( "\n" ); document.write( "sin^2(θ)/cos^2(θ) + cos^2(θ)/cos^2(θ)= 1/cos^2(θ) \r
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\n" ); document.write( "\n" ); document.write( "[Here, Tan = Sin/Cos, Substitute Cos/Cos for 1 and 1/Cos for Sec]\r
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\n" ); document.write( "\n" ); document.write( "sin^2(θ)+cos^2(θ)^2=1 (multiply both sides by cos^2)\r
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\n" ); document.write( "\n" ); document.write( "Above equation is the pythagorean theorum. Solved.\r
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