\( cos41°= \cfrac{x}{\sqrt{x^2+y^2}}\) হলে tan49° এর মান নির্ণয় করো।
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\(cos41°=\cfrac{x}{\sqrt{x^2+y^2}}\)
এখন, \( sin^2 41°=1-cos^2 41°\)
\(=1-\cfrac{x^2}{x^2+y^2}\)
\(=\cfrac{x^2+y^2-x^2}{x^2+y^2}\)
\(=\cfrac{y^2}{x^2+y^2}\)
\(∴sin41°= \cfrac{y}{\sqrt{x^2+y^2}}\)
\(∴tan49°= tan(90°-41°)=cot41°\)
\(=\cfrac{cos41°}{sin41°}\)
\(=\cfrac{\cfrac{x}{\sqrt{x^2+y^2}}}{\cfrac{y}{\sqrt{x^2+y^2}}}\)
\(=\cfrac{x}{\sqrt{x^2+y^2}} × \cfrac{\sqrt{x^2+y^2}}{y}=\cfrac{x}{y}\) (Answer)
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