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Question Number 65724 by bshahid010@gmail.com last updated on 02/Aug/19

Answered by mr W last updated on 03/Aug/19

S=sin x+cos x+(1/(sin x))+(1/(cos x))+((sin x)/(cos x))+((cos x)/(sin x))  =sin x+cos x+((sin x+cos x)/(sin x cos x))+(1/(sin x cos x))  =sin x+cos x+((sin x+cos x+1)/(sin x cos x))  let sin x+cos x=t  ⇒sin x cos x=((t^2 −1)/2)  ⇒S=t+((2(t+1))/(t^2 −1))=t+(2/(t−1))=1+(t−1)+(2/(t−1))  since −(√2)≤t≤(√2)  −(√2)−1≤t−1≤(√2)−1    for t−1>0:  S=1+(t−1)+(2/(t−1))≥1+2(√2)=S_(min)   S_(min)  at t=(√2)+1>(√2) ⇒impossible    for t−1<0:  S=1−[(1−t)+(2/(1−t))]≤1−2(√2)=S_(max)   S_(max)  at t=1−(√2)>−(√2) ⇒possible    ⇒mininmum of ∣S∣ is ∣1−2(√2)∣=2(√2)−1≈1.828  at t=1−(√2) or x=nπ−(−1)^n (sin^(−1) ((2−(√2))/2))−(π/4)

S=sinx+cosx+1sinx+1cosx+sinxcosx+cosxsinx=sinx+cosx+sinx+cosxsinxcosx+1sinxcosx=sinx+cosx+sinx+cosx+1sinxcosxletsinx+cosx=tsinxcosx=t212S=t+2(t+1)t21=t+2t1=1+(t1)+2t1since2t221t121fort1>0:S=1+(t1)+2t11+22=SminSminatt=2+1>2impossiblefort1<0:S=1[(1t)+21t]122=SmaxSmaxatt=12>2possiblemininmumofSis122∣=2211.828att=12orx=nπ(1)n(sin1222)π4

Answered by Tanmay chaudhury last updated on 03/Aug/19

sinx+cosx=a  (√2) ((1/(√2))sinx+(1/(√2))cosx)=a  (√2) sin((π/4)+x)=a  +1≥sin((π/4)+x)≥−1  so (√2)≥ a≥−(√2)   tanx+cotx  ((sin^2 x+cos^2 x)/(cosxsinx))  (2/(2sinxcosx))  (2/(sin2x))  (2/(1+sin2x−1))  (2/((sinx+cosx)^2 −1))  (2/(a^2 −1))  secx+cosecx  =((sinx+cosx)/(sinxcosx))  =(sinx+cosx)×(1/(sinxcosx))  =(sinx+cosx)×(2/(2sinxcosx))  =a×(2/(a^2 −1))  so sinx+cosx+tanx+cotx+secx+cosecx  =a+(2/(a^2 −1))+((2a)/(a^2 −1))  =a+((2(a+1))/((a+1)(a−1)))  =a+(2/(a−1))  when a=(√2)   =((√2)/1)+(2/((√2) −1))  =(√2) +2((√2) +1)  =3(√2) +2  =3×1.41+2  =6.23  so mod of ∣6.23∣=6.23  when a=−(√2)   =((−(√2))/1)+(2/(−(√2) −1))  =((−(√2))/1)−((2((√2) −1))/(((√2) +1)((√2) −1)))  =−(√2) −2(√2) +2  =2−3(√2)   =−2.23  so mod of −2.23 ∣−2.23∣=2.23  hence minimum value=2.23

sinx+cosx=a2(12sinx+12cosx)=a2sin(π4+x)=a+1sin(π4+x)1so2a2tanx+cotxsin2x+cos2xcosxsinx22sinxcosx2sin2x21+sin2x12(sinx+cosx)212a21secx+cosecx=sinx+cosxsinxcosx=(sinx+cosx)×1sinxcosx=(sinx+cosx)×22sinxcosx=a×2a21sosinx+cosx+tanx+cotx+secx+cosecx=a+2a21+2aa21=a+2(a+1)(a+1)(a1)=a+2a1whena=2=21+221=2+2(2+1)=32+2=3×1.41+2=6.23somodof6.23∣=6.23whena=2=21+221=212(21)(2+1)(21)=222+2=232=2.23somodof2.232.23∣=2.23henceminimumvalue=2.23

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