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 = \frac{s}{c}

c = \sqrt{1 - s^{2}} = \frac{1}{\sqrt{t^{2} + 1}}

s = \sqrt{1 - c^{2}} = \frac{t}{\sqrt{t^{2} + 1}}

t = \frac{\sqrt{1 - c^{2}}}{c} = \frac{s}{\sqrt{1 - s^{2}}}

should be easy

Commented by peter frank last updated on 11/Jul/24

thank 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