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Question Number 148773 by mim24 last updated on 31/Jul/21

Answered by som(math1967) last updated on 31/Jul/21

$$\frac{1}{\mathit{s}\mathit{i}\mathit{n}20}+\frac{\sqrt{3}}{\mathit{c}\mathit{o}\mathit{s}20}$$$$=\frac{\mathit{c}\mathit{o}\mathit{s}20+\sqrt{3}\mathit{s}\mathit{i}\mathit{n}20}{\mathit{s}\mathit{i}\mathit{n}20\mathit{c}\mathit{o}\mathit{s}20}$$$$=\frac{2(\frac{1}{2}\mathit{c}\mathit{o}\mathit{s}20+\frac{\sqrt{3}}{2}\mathit{s}\mathit{i}\mathit{n}20)}{\mathit{s}\mathit{i}\mathit{n}20\mathit{c}\mathit{o}\mathit{s}20}$$$$=\frac{2.2(\mathit{c}\mathit{o}\mathit{s}60\mathit{c}\mathit{o}\mathit{s}20+\mathit{s}\mathit{i}\mathit{n}60\mathit{s}\mathit{i}\mathit{n}20)}{2\mathit{s}\mathit{i}\mathit{n}20\mathit{c}\mathit{o}\mathit{s}20}$$$$=\frac{4\mathit{c}\mathit{o}\mathit{s}(60-20)}{\mathit{s}\mathit{i}\mathit{n}40}$$$$=\frac{4\mathit{c}\mathit{o}\mathit{s}40}{\mathit{s}\mathit{i}\mathit{n}40}=4\mathit{c}\mathit{o}\mathit{t}40=4\mathit{t}\mathit{a}\mathit{n}(90-40)=4\mathit{t}\mathit{a}\mathit{n}50$$

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