যদি \(cos43°= \cfrac{x}{\sqrt{x^2+y^2}}\) হয় তবে \(tan47°\) এর মান কত?
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\(cos43°= \cfrac{x}{\sqrt{x^2+y^2}}\)
বা, \(cos^243°= \cfrac{x^2}{x^2+y^2}\)
বা, \(1-cos^243°= 1-\cfrac{x^2}{x^2+y^2}\)
বা, \(sin^243°= \cfrac{\cancel{x^2}+y^2-\cancel{x^2}}{x^2+y^2}\)
বা, \(sin^2(90°-47°)= \cfrac{y^2}{x^2+y^2}\)
বা, \(cos^2 47°= \cfrac{y^2}{x^2+y^2}\)
বা, \(\cfrac{1}{cos^2 47°}= \cfrac{1}{\cfrac{y^2}{x^2+y^2}}\)
বা, \(sec^2 47°= \cfrac{x^2+y^2}{y^2}\)
বা, \(sec^2 47°-1= \cfrac{x^2+y^2}{y^2}-1\)
বা, \(tan^2 47°= \cfrac{x^2+\cancel{y^2}-\cancel{y^2}}{y^2}\)
বা, \(tan 47°= \sqrt{\cfrac{x^2}{y^2}}\)
বা, \(tan 47°=\cfrac{x}{y}\) [Answer]

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