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Question Number 212477 by Durganand last updated on 14/Oct/24

Answered by mehdee7396 last updated on 14/Oct/24

$$\frac{sin2acosa-sinacos2a}{sin2asina}$$$$=\frac{sina}{sin2asina}=\frac{1}{sin2a}✓$$

Answered by A5T last updated on 14/Oct/24

$$\frac{cosA}{sinA}-\frac{cos2A}{sin2A}=\frac{2{cos}^{2}A-({cos}^{2}A-{sin}^{2}A)}{sin2A}$$$$=\frac{{cos}^{2}A+{sin}^{2}A}{sin2A}=\frac{1}{sin2A}$$

Answered by Nadirhashim last updated on 15/Oct/24

$$\mathit{l}\mathit{e}\mathit{f}\mathit{t}\mathit{s}\mathit{i}\mathit{d}\mathit{e}$$$$\frac{\mathit{c}\mathit{o}\mathit{x}\mathit{A}}{\mathit{s}\mathit{i}\mathit{n}\mathit{A}}-\frac{\mathit{c}\mathit{o}\mathit{s}2\mathit{A}}{\mathit{s}\mathit{i}\mathit{n}2\mathit{A}}$$$$\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{A}-\mathit{c}\mathit{o}\mathit{s}2\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathit{A}}{\mathit{s}\mathit{i}\mathit{n}2A\mathit{s}\mathit{i}\mathit{n}\mathit{A}}$$$$$$$$\frac{\mathit{s}\mathit{i}\mathit{n}(2\mathit{A}-A)}{\mathit{s}\mathit{i}\mathit{n}2\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathit{A}}=\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{A}}{\mathit{s}\mathit{i}\mathit{n}2\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathit{A}}$$$$=\frac{1}{\mathit{s}\mathit{i}\mathit{n}2\mathit{A}}$$$$$$

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