need help with trig proof thing!!!!!!!!!!!!

dood · July 5th 2007, 01:08 PM

how do u prove : sec(A-B) = cos(A+B) / ( cosA^2 - sinB^2 )

**galactus** · July 5th 2007, 01:16 PM

Use the subtraction and addition formulas.

rewrite the left side as:

$\frac{1}{cosAcosB+sinAsinB}$

Multiply top and bottom by $cosAcosB-sinAsinB$

and you should be able to hammer it into the right side.

$\frac{1}{cosAcosB+sinAsinB}\cdot\frac{cosAcosB-sinAsinB}{cosAcosB-sinAsinB}$

$\frac{cosAcosB-sinAsinB}{cos^{2}Acos^{2}B-sin^{2}A\cdot{sin^{2}}B}$

$\frac{\overbrace{cosAcosB-sinAsinB}^{\text{cos(A+B)}}}{cos^{2}A-sin^{2}B}$

$\frac{cos(A+B)}{cos^{2}A-sin^{2}B}$

**janvdl** · July 5th 2007, 01:34 PM

Originally Posted by dood

how do u prove : sec(A-B) = cos(A+B) / ( cosA^2 - sinB^2 )

Haven't checked this, but logic tells me that:

$\frac{ cos(A+B) }{( cos A^2 - Sin B^2 )} = \frac{ cos (A+B) }{ (cos (A-B)) \ (cos (A + B)) } = \frac{1}{ cos (A-B)} = sec (A - B)$

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