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Question-215696




Question Number 215696 by BaliramKumar last updated on 15/Jan/25
Answered by MATHEMATICSAM last updated on 15/Jan/25
sec^4 x − cosec^4 x − 2sec^2 x + 2cosec^2 x = ((63)/8)  ⇒ sec^2 x(sec^2 x − 2) + cosec^2 x(2 − cosec^2 x) = ((63)/8)  ⇒ (1 + tan^2 x)(tan^2 x − 1) + (1 + cot^2 x)(1 − cot^2 x) = ((63)/8)  ⇒ (tan^4 x − 1) + (1 − cot^4 x) = ((63)/8)  ⇒ tan^4 x − 1 + 1 − cot^4 x = ((63)/8)  ⇒ tan^4 x − cot^4 x = ((63)/8)
sec4xcosec4x2sec2x+2cosec2x=638sec2x(sec2x2)+cosec2x(2cosec2x)=638(1+tan2x)(tan2x1)+(1+cot2x)(1cot2x)=638(tan4x1)+(1cot4x)=638tan4x1+1cot4x=638tan4xcot4x=638
Commented by MATHEMATICSAM last updated on 15/Jan/25
I got something like this
Igotsomethinglikethis
Answered by A5T last updated on 15/Jan/25
sec^4 x−cosec^4 x−2sec^2 x+2cosec^2 x=((63)/8)  ⇒(sec^2 x−1)^2 −(cosec^2 x−1)^2 =((63)/8)  sec^2 x=tan^2 x+1; cosec^2 x=1+cot^2 x  ⇒tan^4 x−(1/(tan^4 x))=((63)/8)⇒8tan^8 x−63tan^4 x−8=0  tan^4 x=((63+_− (√(63^2 −4(−8)8)))/(16))>0⇒tan^4 x=((63+65)/(16))=8  ⇒tan^2 x=2(√2)⇒cot^2 x=(1/(2(√2)))=((√2)/4)  ⇒tan^2 x+cot^2 x=((9(√2))/4)=(9/(2(√2)))
sec4xcosec4x2sec2x+2cosec2x=638(sec2x1)2(cosec2x1)2=638sec2x=tan2x+1;cosec2x=1+cot2xtan4x1tan4x=6388tan8x63tan4x8=0tan4x=63+6324(8)816>0tan4x=63+6516=8tan2x=22cot2x=122=24tan2x+cot2x=924=922
Commented by BaliramKumar last updated on 15/Jan/25
also   tan^4 x−cot^4 x = ((63)/8)  tan^4 x − (1/(tan^4 x)) = 8 − (1/8)  tan^4 x=8  tan^2 x=(√8)             (   (√8) + (1/( (√8))) = (9/(2(√2))))
alsotan4xcot4x=638tan4x1tan4x=818tan4x=8tan2x=8(8+18=922)

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