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Question-209456




Question Number 209456 by peter frank last updated on 10/Jul/24
Answered by Ghisom last updated on 10/Jul/24
c=cos α     s=sin α     t=tan α =(s/c)  c=(√(1−s^2 ))=(1/( (√(t^2 +1))))  s=(√(1−c^2 ))=(t/( (√(t^2 +1))))  t=((√(1−c^2 ))/c)=(s/( (√(1−s^2 ))))  should be easy
c=cosαs=sinαt=tanα=scc=1s2=1t2+1s=1c2=tt2+1t=1c2c=s1s2shouldbeeasy
Commented by peter frank last updated on 11/Jul/24
thank you
thankyou
Answered by Spillover last updated on 10/Jul/24
  Since (A+B)=90°    Then    TanA=opposite/adjacent    Tan A=√s/r    Also    Tan B=opp/adj               =r/√s   By considering angle B    adjacent=√s    Opposite =r    from a²+b²=c²  r²+(√s)²=c²  c²=r²+s  c=√(r²+s)    Hyp=√(r²+s)    Now     Cos B=adj/hyp    Cos B=√s/√(r²+s)    We know that      √x/√y=√(x/y)    Hence    √s/√(r²+s)=√(s/r²+s)      Therefore :CosB=√(s/r²+s)
Since (A+B)=90°

Then

TanA=opposite/adjacent

Tan A=√s/r

Also

Tan B=opp/adj

=r/√s
By considering angle B

adjacent=√s

Opposite =r

from a²+b²=c²
r²+(√s)²=c²
c²=r²+s
c=√(r²+s)

Hyp=√(r²+s)

Now

Cos B=adj/hyp

Cos B=√s/√(r²+s)

We know that

√x/√y=√(x/y)

Hence

√s/√(r²+s)=√(s/r²+s)

Therefore :CosB=√(s/r²+s)

Commented by Spillover last updated on 10/Jul/24

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