Question

If (sec A+tan A)(secB+tanB)(secC+tanC)=(secA-tanA)(secB-tanB)(secC-tanC), prove that each of the side is equal to ±1.​

Answer 1

Answer:

Step-by-step explanation:

(secA+tanA)(secB+tanB)(sec C+tanC)=(secA-tanA)(secB-TAnB)(secC-tanC)

multiplying both sides by (SecA-TanA)(secB-TanB)(SecC-tanC)

we get

(secA+tannA)(secB+tanB)(secC+tanC)(secA-tanA)(secB-tanB)(secC-tanC)=(secA-tanA)²(SecB-tanB)²(SecC-tanC)²

⇒(Sec²A-tan²A)(sec²b-tan²B)(sec²c-tan²C)=(secA-tanA)²(SecB-tanB)²(secC-tanC)²

⇒1=[(secA-tanA)(secB-tanB)(secC-tanC)]²

⇒(secA-tanA)(SecB-tanB)(secC-tanC)=+/-1

Similarly multiplying both sides by(secA+tanA)(seB+tanB)(secC+tanC) we get (SecA+tanA)(SecB+tanB)(SecC+tanC)=+/-1

Answer 2

Answer:

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