Question

If ( secA + tanA ) ( secB + tanB) ( secC + tanC ) = ( secA - tanA ) ( secB - tanB ) ( secC - tanC ). Prove that each of the side is equal to + or - 1

Answer 1

=(1/cosA+sinA/cosA)(1/cosB+sinB/cosB)(1/cosC+sinC/cosC)=(1/cosA-sinA/cosA)(1/cosB-sinB/cosB)(1/cosC-sinC/cosC)

=(1+sinA/cosA)(1+sinB/cosB)(1+sinC/cosC)=(1-sinA/cosA)(1-sinB/cosB)(1-sinC/cosC)

Now, [squaring both the sides]

=

${(1+sinA/cosA)}^{2} {(1+sinb/cosb)}^{2} {(1+sinc/cosc)}^{2} = {(1 - sinA/cosA)}^{2} {(1 - sinb/cosb)}^{2} {(1 - sinc/cosc)}^{2}$

Therefore after opening bracket it will be

$({ - cosA/Cosa})^{2} ({ - cosb/cosb})^{2}( { - cosc/cosc})^{2}= ( {cosA/cosA})^{2}( cosb/cosb)^{2} ( {cosc/cosc})^{2}$

Therefore cos/cos will be canel so, now

=

$- {1}^{2} \times - {1}^{2} \times - {1}^{2} = {1}^{2} \times {1}^{2} \times {1}^{2}$

= - 1=1

Hence, proved

Answer 2

$\boxed{answer}$

see in attachment...............