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Question Number 122076 by peter frank last updated on 14/Nov/20

Answered by MJS_new last updated on 14/Nov/20

let t=tan A ((1+t−(√(t^2 +1)))/( (√(t^2 +1))+t−1))=((1+(√(t^2 +1))−t)/( (√(t^2 +1))+t+1)) (t+1−(√(t^2 +1)))(t+1+(√(t^2 +1)))=−(t−1−(√(t^2 +1)))(t−1+(√(t^3 +1))) (t+1)^2 −(t^2 +1)=−(t−1)^2 +(t^2 +1) 2t=2t true

lett=tanA1+t−t2+1t2+1+t−1=1+t2+1−tt2+1+t+1(t+1−t2+1)(t+1+t2+1)=−(t−1−t2+1)(t−1+t3+1)(t+1)2−(t2+1)=−(t−1)2+(t2+1)2t=2ttrue

Answered by MJS_new last updated on 14/Nov/20

((1+(s/c)−(1/c))/((1/c)+(s/c)−1))=((1+(1/c)−(s/c))/((1/c)+(s/c)+1)) (((c+s−1)/c)/((1+s−c)/c))=(((c+1−s)/c)/((1+s+c)/c)) ((c+s−1)/(−c+s+1))=((c−s+1)/(c+s+1)) (c+s−1)(c+s+1)=(c−s+1)(−c+s+1) c^2 +s^2 +2cs−1=−c^2 −s^2 +2cs+1 2cs=2cs true

1+sc−1c1c+sc−1=1+1c−sc1c+sc+1c+s−1c1+s−cc=c+1−sc1+s+ccc+s−1−c+s+1=c−s+1c+s+1(c+s−1)(c+s+1)=(c−s+1)(−c+s+1)c2+s2+2cs−1=−c2−s2+2cs+12cs=2cstrue

Answered by ajfour last updated on 14/Nov/20

let sec A=s, tan A=t s^2 −t^2 =1 ⇒ ((s−1)/t)=(t/(s+1)) l.h.s. = ((1+t−s)/(s+t−1))=((1−(((s−1)/t)))/(1+(((s−1)/t)))) =((1−((t/(s+1))))/(1+((t/(s+1)))))=((1+s−t)/(s+t+1))=r.h.s. ★

letsecA=s,tanA=ts2−t2=1⇒s−1t=ts+1l.h.s.=1+t−ss+t−1=1−(s−1t)1+(s−1t)=1−(ts+1)1+(ts+1)=1+s−ts+t+1=r.h.s.★

Commented by peter frank last updated on 14/Nov/20

thank you both

thankyouboth

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