Math, asked by anuragpani100, 1 year ago

Plz....Help in solving Q. 27

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Answered by vrsdwarakanathpaksrq
0
we know that sec^2x-tan^x=1
then (secx+tanx)(secx-tanx)=1
secx+tanx=1/(secx-tanx)
then
(seca+tana)(secb+tanb)(secc+tanc)
= 1/(seca-tana)(secb-tanb)(secc-tanc) _eq1

(seca+tana)(secb+tanb)(secc+tanc)=
(seca-tana)(secb-tanb)(secc-tanc) _eq2

from eq1 and eq2

(seca-tana)(secb-tanb)(secc-tanc) =
1/(seca-tana)(secb-tanb)(secc-tanc)

(seca-tana)(secb-tanb)(secc-tanc) * (seca-tana)(secb-tanb)(secc-tanc) =1

(seca-tana)^2(secb-tanb)^2(secc-tanc)^2=1

((seca-tana)(secb-tanb)(secc-tanc) )^2 =1

let (seca-tana)(secb-tanb)(secc-tanc)=k

=> k^2=1
then k=+-1
therefore
(seca-tana)(secb-tanb)(secc-tanc) = 1 or -1

and from eq 1

(seca+tana)(secb+tanb)(secc+tanc) = 1 or -1
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