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Question

# Prove that: (i) $\frac{1}{\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(x-b\right)}=\frac{\mathrm{cot}\left(x-a\right)-\mathrm{cot}\left(x-b\right)}{\mathrm{sin}\left(a-b\right)}$ (ii) $\frac{1}{\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}=\frac{\mathrm{cot}\left(x-a\right)+\mathrm{tan}\left(x-b\right)}{\mathrm{cos}\left(a-b\right)}$ (iii) $\frac{1}{\mathrm{cos}\left(x-a\right)\mathrm{cos}\left(a-b\right)}=\frac{\mathrm{tan}\left(x-b\right)-\mathrm{tan}\left(x-a\right)}{\mathrm{sin}\left(a-b\right)}$

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Solution

## $\left(\mathrm{i}\right)\mathrm{RHS} =\frac{\mathrm{cot}\left(x-a\right)-\mathrm{cot}\left(x-b\right)}{\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{\mathrm{cos}\left(x-a\right)}{\mathrm{sin}\left(x-a\right)}-\frac{\mathrm{cos}\left(x-b\right)}{\mathrm{sin}\left(x-b\right)}}{\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(x-b\right)\mathrm{cos}\left(x-a\right)-\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}{\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(x-b\right)\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(x-b-x+a\right)}{\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(x-b\right)\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(a-b\right)}{\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(x-b\right)\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{LHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\text{p}\mathrm{roved}.$ $\left(\mathrm{ii}\right)\mathrm{RHS}=\frac{\mathrm{cot}\left(x-a\right)+\mathrm{tan}\left(x-b\right)}{\mathrm{cos}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{\mathrm{cos}\left(x-a\right)}{\mathrm{sin}\left(x-a\right)}+\frac{\mathrm{sin}\left(x-b\right)}{\mathrm{cos}\left(x-b\right)}}{\mathrm{cos}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(x-b\right)\mathrm{cos}\left(x-a\right)+\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(x-b\right)}{\mathrm{cos}\left(a-b\right)\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(x-b-x+a\right)}{\mathrm{cos}\left(a-b\right)\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\left(\mathrm{Using}\mathrm{cos}\left(A-B\right)=\mathrm{cos}A\mathrm{cos}bB+\mathrm{sin}A\mathrm{sin}B\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(a-b\right)}{\mathrm{cos}\left(a-b\right)\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$ $\left(\mathrm{iii}\right)\mathrm{RHS}=\frac{\mathrm{tan}\left(x-b\right)-\mathrm{tan}\left(x-a\right)}{\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{\mathrm{sin}\left(x-b\right)}{\mathrm{cos}\left(x-b\right)}-\frac{\mathrm{sin}\left(x-a\right)}{\mathrm{cos}\left(x-a\right)}}{\mathrm{sin}\left(a-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(x-b\right)\mathrm{cos}\left(x-a\right)-\mathrm{sin}\left(x-a\right)\mathrm{cos}\left(x-b\right)}{\mathrm{sin}\left(a-b\right)\mathrm{cos}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(x-b-x+a\right)}{\mathrm{sin}\left(a-b\right)\mathrm{cos}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\left(\mathrm{Using}\mathrm{sin}\left(A-B\right)=\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(a-b\right)}{\mathrm{sin}\left(a-b\right)\mathrm{cos}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{cos}\left(x-a\right)\mathrm{cos}\left(x-b\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{LHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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