\(\cfrac{sinθ}{x}=\cfrac{cosθ}{y}\) হলে, প্রমাণ করো যে , \(sinθ−cosθ=\cfrac{x−y}{\sqrt{x^2+y^2}}\)। Madhyamik 2017


\(\cfrac{sinθ}{x}=\cfrac{cosθ}{y}\)
বা, \(\cfrac{sinθ}{cosθ}=\cfrac{x}{y}\)
বা, \(tanθ=\cfrac{x}{y}\)
বা, \(tan^2⁡θ=\cfrac{x^2}{y^2}\)
বা, \(1+ tan^2⁡θ=1+\cfrac{x^2}{y^2}\)
বা, \(sec^2⁡θ=\cfrac{y^2+x^2}{y^2}\)
বা, \(secθ=\cfrac{\sqrt{y^2+x^2}}{y}\)
বা, \(cosθ=\cfrac{y}{\sqrt{x^2+y^2}} \)

এখন \(\cfrac{sinθ}{x}=\cfrac{cosθ}{y}\) সমীকরনে
\(cosθ=\cfrac{y}{\sqrt{x^2+y^2}} \) বসিয়ে পাই
\(\cfrac{sinθ}{x}=\cfrac{\cfrac{y}{\sqrt{x^2+y^2}}}{y}\)
বা, \(\cfrac{sinθ}{x}=\cfrac{1}{\sqrt{x^2+y^2}}\)
বা, \(sinθ=\cfrac{x}{\sqrt{x^2+y^2}}\)

\(\therefore sinθ−cosθ=\cfrac{x}{\sqrt{x^2+y^2}}-\cfrac{y}{\sqrt{x^2+y^2}}\)
\(=\cfrac{x−y}{\sqrt{x^2+y^2}}\)(প্রমাণিত)

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