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Question Number 160746 by Rokon last updated on 05/Dec/21

Answered by som(math1967) last updated on 06/Dec/21

$$if\mathit{\theta }=45°\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{A}=\sqrt{2}\mathit{B}=0$$$$ii)A=\sqrt{2}(cos\theta +\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }-sin\mathit{\theta })$$$$\therefore \mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }=\sqrt{2}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$$$$\Rightarrow \mathit{s}\mathit{i}\mathit{n}\mathit{\theta }=(\sqrt{2}-1)\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$$$$\Rightarrow (\sqrt{2}+1)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }=(\sqrt{2}+1)(\sqrt{2}-1)\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$$$$\Rightarrow (\sqrt{2}+1)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }=\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$$$$\Rightarrow \sqrt{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }=\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }-\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$$$$\Rightarrow \sqrt{2}(c\mathit{o}\mathit{s}\mathit{\theta }+\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }-\mathit{c}\mathit{o}\mathit{s}\mathit{\theta })=\mathit{B}$$$$\therefore \mathit{B}=\sqrt{2}(\mathit{A}-\mathit{c}\mathit{o}\mathit{s}\mathit{\theta })$$

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