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Questions  

If (secAtanA)(secBtanB)(secCtanC) =(secA+tanA)(secB+tanB)(secC+tanC) then each side is equal to

a
0
b
1
c
-1
d
±1

detailed solution

Correct option is D

We have,(sec⁡A−tan⁡A)(sec⁡B−tan⁡B)(sec⁡C−tan⁡C)=(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C)⇒sec2⁡A−tan2⁡Asec2⁡B−tan2⁡Bsec2⁡C−tan2⁡C={(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(scc⁡C+tan⁡C)}2⇒1={(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C)}2⇒(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C)=±1Hence, LHS=RHS=±1.

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