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Question

# If $\mathrm{cosec}x-\mathrm{sin}x={a}^{3},\mathrm{sec}x-\mathrm{cos}x={b}^{3}$, then prove that ${a}^{2}{b}^{2}\left({a}^{2}+{b}^{2}\right)=1$.

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Solution

## $\mathrm{cosec}x-\mathrm{sin}x={a}^{3}\phantom{\rule{0ex}{0ex}}\therefore \frac{1}{\mathrm{sin}x}-\mathrm{sin}={a}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{1-{\mathrm{sin}}^{2}x}{\mathrm{sin}x}={a}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{cos}}^{2}x}{\mathrm{sin}x}={a}^{3}\phantom{\rule{0ex}{0ex}}⇒a={\left(\frac{{\mathrm{cos}}^{2}x}{\mathrm{sin}x}\right)}^{\frac{1}{3}}....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{sec}x-\mathrm{cos}x={b}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cos}x}-\mathrm{cos}={b}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{1-{\mathrm{cos}}^{2}x}{\mathrm{cos}x}={b}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{sin}}^{2}x}{\mathrm{cos}x}={b}^{3}\phantom{\rule{0ex}{0ex}}⇒b={\left(\frac{{\mathrm{sin}}^{2}x}{\mathrm{cos}x}\right)}^{\frac{1}{3}}.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{LHS}={a}^{2}{b}^{2}\left({a}^{2}+{b}^{2}\right)={\left(ab\right)}^{2}\left({a}^{2}+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}={\left[{\left(\frac{{\mathrm{cos}}^{2}x}{\mathrm{sin}x}\right)}^{\frac{1}{3}}{\left(\frac{{\mathrm{sin}}^{2}x}{\mathrm{cos}x}\right)}^{\frac{1}{3}}\right]}^{2}\left[{\left({\left(\frac{{\mathrm{cos}}^{2}x}{\mathrm{sin}x}\right)}^{\frac{1}{3}}\right)}^{2}+{\left({\left(\frac{{\mathrm{sin}}^{2}x}{\mathrm{cos}x}\right)}^{\frac{1}{3}}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}={\left(\mathrm{sin}x\mathrm{cos}x\right)}^{\frac{2}{3}}\left[\frac{{\left({\mathrm{cos}}^{2}x\right)}^{\frac{2}{3}}}{{\left(\mathrm{sin}x\right)}^{\frac{2}{3}}}+\frac{{\left({\mathrm{sin}}^{2}x\right)}^{\frac{2}{3}}}{{\left(\mathrm{cos}x\right)}^{\frac{2}{3}}}\right]\phantom{\rule{0ex}{0ex}}={\left(\mathrm{sin}x\mathrm{cos}x\right)}^{\frac{2}{3}}\left[\frac{{\left({\mathrm{cos}}^{3}x\right)}^{\frac{2}{3}}+{\left({\mathrm{sin}}^{3}x\right)}^{\frac{2}{3}}}{{\left(\mathrm{sin}x\right)}^{\frac{2}{3}}{\left(\mathrm{cos}x\right)}^{\frac{2}{3}}}\right]\phantom{\rule{0ex}{0ex}}={\left(\mathrm{sin}x\mathrm{cos}x\right)}^{\frac{2}{3}}\left[\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\left(\mathrm{sin}x\mathrm{cos}x\right)}^{\frac{2}{3}}}\right]\phantom{\rule{0ex}{0ex}}=1=\mathrm{RHS}$

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