Question

If (sec A +tanA)(sec B+tanB)(secC+tanC)=k=(secA-tanA)(sdcB-tanB)(sec C-tanC),then k=

Answer 1

( secA + tanA)( secB + tanB)( secC + tanC) = K( secA - tanA) (secB -tanB)(secC -tanC)

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

( 1+sinA)(1+sinB)(1+sinC) = K( 1-sinA)(1 -sinB)(1 -sinC)

this is possible only when

sinA = sinB = sinC = 0

so, (1 +0)(1+0)(1+0) = K(1-0)(1-0)(1-0)
K = 1

Answer 2

( secA + tanA)( secB + tanB)( secC + tanC) = K( secA - tanA) (secB -tanB)(secC -tanC)

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

( 1+sinA)(1+sinB)(1+sinC) = K( 1-sinA)(1 -sinB)(1 -sinC)

this is possible only when 

sinA = sinB = sinC = 0

so, (1 +0)(1+0)(1+0) = K(1-0)(1-0)(1-0)
K = 1