Question

sec B + tan B = p then what is the value of sec A - tan A

Answer 1

• 1 + tan²B = sec²B

⇒ sec²B - tan²B = 1

⇒ (sec B + tan B)(sec B - tan B) = 1

∴ sec B - tan B = 1/(sec B + tan B)

Ans = 1/p.

Answer 2

(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


Hope it helps