Question

$Solve \: It \: Fast..$
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Answer 1

Answer:

(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