Question

if (secA+tanA)(secB+tanB)(secC+tanC)=(secA-tanA)(secB-tanB)(secC-tanC) then prove that each of the side is equal to ±1

Answer 1

HELLO DEAR,



(secA + tanA)(secB + tanB)(secC + tan C)


=> (secA - tanA)(secB - tanB)(secC - tanC)



{ Mulitply both sides with }
(secA + tanA)(secB + tanB)(secC + tan C)",


we get,


(secA + tanA)2(secB + tanB)2(secC + tan C)2 


=> (sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C)
        

        = (1)(1)(1) = 1


=> [(secA + tanA)(secB + tanB)(secC + tanC)]2=1






(secA + tanA)(secB + tanB)(secC + tan C) = ± 1



Similarly, we get



(secA – tanA)(secB – tanB)(secC – tan C) = ± 1



I hope its help you dear,
THANKS

Answer 2

Answer:

Step-by-step explanation:

L.H.S. (secA+tanA)(secB+tanB)(secC+ tanC) = (secA-tanA)(secB-tanC)( secC-tanC )  

Multiply both sides by (secA-tanA)(secB-tanC)( secC-tanC), that is RHS  

(secA+tanA) (secA-tanA) (secB+tanB) (secB-tanC) (secC+ tanC) ( secC-tanC) = [(secA-tanA)(secB-tanC)( secC-tanC)] 2  

(sec 2 A-tan 2 A) (sec 2 B-tan 2 C) ( sec 2 C-tan 2 C) = [(secA-tanA)(secB-tanC)( secC-tanC)] 2  

1 = [(secA-tanA)(secB-tanC)( secC-tanC)] 2   [since sec 2 A-tan 2 A = 1 ]  

+- 1 = [(secA-tanA)(secB-tanC)( secC-tanC)  

therefore RHS = +-1  

Similarly LHS = +-1

Hence proved.