Question Number 209456 by peter frank last updated on 10/Jul/24
Answered by Ghisom last updated on 10/Jul/24
$${c}=\mathrm{cos}\:\alpha\:\:\:\:\:{s}=\mathrm{sin}\:\alpha\:\:\:\:\:{t}=\mathrm{tan}\:\alpha\:=\frac{{s}}{{c}} \\ $$$${c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${s}=\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${t}=\frac{\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }}{{c}}=\frac{{s}}{\:\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\ $$
Commented by peter frank last updated on 11/Jul/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
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)
Commented by Spillover last updated on 10/Jul/24
