Question-211019

Question Number 211019 by peter frank last updated on 26/Aug/24

Answered by som(math1967) last updated on 26/Aug/24

l e t x = s i n α y = s i n β

c o s α + c o s β = a (s i n α - s i n β)

\Rightarrow \frac{2 c o s \frac{α + β}{2} c o s \frac{α - β}{2}}{2 s i n \frac{α - β}{2} c o s \frac{α + β}{2}} = a

\Rightarrow c o t \frac{α - β}{2} = a

\Rightarrow \frac{1}{2} (α - β) = \cot^{- 1} a

\Rightarrow \sin^{- 1} x - \sin^{- 1} y = {2 \cot}^{- 1} a

\Rightarrow \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} \times \frac{d y}{d x} = 0

∴ \frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}

Commented by peter frank last updated on 27/Aug/24

thank you som