যদি \(sin\alpha=\cfrac{5}{13}\) হয়, তবে দেখাও যে, \(tan\alpha+sec\alpha=\cfrac{3}{2}\)।
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\(sin\alpha=\cfrac{5}{13}\)
বা, \(sin^2\alpha=\cfrac{25}{169}\)
বা, \(1-sin^2\alpha =1-\cfrac{25}{169}\)
বা, \(cos^2\alpha=\cfrac{169-25}{169}\)
বা, \(\cos\alpha=\sqrt{\cfrac{144}{169}}=\cfrac{12}{13}\)

\(tan\alpha+sec\alpha\)
\(=\cfrac{sin \alpha}{cos\alpha}+\cfrac{1}{cos\alpha}\)
\(=\cfrac{sin\alpha+1}{cos\alpha}\)
\(=\cfrac{\cfrac{5}{13}+1}{\cfrac{12}{13}}\)
\(=\cfrac{\cfrac{5+13}{13}}{\cfrac{12}{13}}\)
\(=\cfrac{\cancel{18}3}{\cancel{13}}\times \cfrac{\cancel{13}}{\cancel{12}2}\)
\(=\cfrac{3}{2}\) [প্রমানিত]

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