Question

If (sec A + tan A)(sec B+ tan B)(sec C+ tan C) = (sec A - tan A)(sec B - tan B)(sec C - tan C) prove that each of the sides is equal to +-1

Answer 1

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) i.e RHS  

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

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

    (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  

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

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

∴ rh.s = +-1  

lly l.h.s= +-1

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