\( cos41°= \cfrac{x}{\sqrt{x^2+y^2}}\) হলে tan49° এর মান নির্ণয় করো।
Loading content...

\(cos41°=\cfrac{x}{\sqrt{x^2+y^2}}\)

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)

🚫 Don't Click. Ad Inside 😈

Similar Questions