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If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have, - Mathematics

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Sum

If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,

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Solution

(secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC)

Multiplying both sides by

(secA – tanA)(secB – tanB)(secC – tanC) we get

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

`(secC – tanC) = (secA – tanA)^2 (secB – tanB)^2 (secC – tanC)^2`

`(sec^2 A – tan^2 A)(sec^2 B – tan^2 B) (sec^2 C – tan^2 C) = (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)(secB + tanB)(secC + tanC),

we get

(secA + tanA)(secB + tanB)(secC + tanC) = ±1

Concept: Trigonometric Identities
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