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Question

# Prove the following identities (1-16) $\mathrm{cosec}x\left(\mathrm{sec}x-1\right)-\mathrm{cot}x\left(1-\mathrm{cos}x\right)=\mathrm{tan}x-\mathrm{sin}x$

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Solution

## $\mathrm{LHS}=\mathrm{cosec}x\left(\mathrm{sec}x-1\right)-\mathrm{cot}x\left(1-\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}x}\left(\frac{1}{\mathrm{cos}x}-1\right)-\frac{\mathrm{cos}x}{\mathrm{sin}x}\left(1-\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}x}\left(\frac{1-\mathrm{cos}x}{\mathrm{cos}x}\right)-\frac{\mathrm{cos}x}{\mathrm{sin}x}\left(1-\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-\mathrm{cos}x}{\mathrm{sin}x}\right)\left(\frac{1}{\mathrm{cos}x}-\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-\mathrm{cos}x}{\mathrm{sin}x}\right)\left(\frac{1-{\mathrm{cos}}^{2}x}{\mathrm{cos}x}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-\mathrm{cos}x}{\mathrm{sin}x}\right)\left(\frac{{\mathrm{sin}}^{2}x}{\mathrm{cos}x}\right)\phantom{\rule{0ex}{0ex}}=\left(1-\mathrm{cos}x\right)\left(\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}x}{\mathrm{cos}x}-\mathrm{sin}x\phantom{\rule{0ex}{0ex}}=\mathrm{tan}x-\mathrm{sin}x\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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