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Question

If $\mathrm{tan}\mathrm{\theta }=\frac{a}{b},$ show that $\frac{a\mathrm{sin}\mathrm{\theta }-b\mathrm{cos}\mathrm{\theta }}{\alpha \mathrm{sin}\mathrm{\theta }+b\mathrm{cos}\mathrm{\theta }}=\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$

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Solution

$\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}\frac{a\mathrm{sin}\theta -b\mathrm{cos}\theta }{a\mathrm{sin}\theta +b\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{by}\mathit{}b\mathrm{cos\theta }:\phantom{\rule{0ex}{0ex}}=\frac{\frac{a\mathrm{tan}\theta }{b}-1}{\frac{a\mathrm{tan}\theta }{b}+1}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$ Hence proved.

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