Question

If x= cosecA - sinA and y= secA - cosA, then show that x^2/3 + y^2/3 = ( xy)^-2/3
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Answer 1

$\textbf{Given:}$

$x=cosecA-sinA\;\;\text{and}\;\;y=secA-cosA$

$\textbf{To prove:}$

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=(xy)^{\frac{-2}{{3}}$

$x=cosecA-sinA$

$x=\dfrac{1}{sinA}-sinA$

$x=\dfrac{1-sin^2A}{sinA}$

$x=\dfrac{cos^2A}{sinA}$

$y=secA-cosA$

$y=\dfrac{1}{cosA}-cosA$

$y=\dfrac{1-cos^2A}{cosA}$

$y=\dfrac{sin^2A}{cosA}$

$\text{Now,}$

$\bf\,x^{\frac{2}{3}}+y^{\frac{2}{3}}$

$=(\dfrac{cos^2A}{sinA})^{\frac{2}{3}}+(\dfrac{sin^2A}{cosA})^{\frac{2}{3}}$

$=\dfrac{cos^\frac{4}{3}A}{sin^{\frac{2}{3}}A}+\dfrac{sin^\frac{4}{3}A}{cos^{\frac{2}{3}}A}$

$=\dfrac{cos^\frac{6}{3}A+sin^\frac{6}{3}A}{sin^{\frac{2}{3}}A\;cos^{\frac{2}{3}}A}$

$=\dfrac{cos^2A+sin^2A}{sin^{\frac{2}{3}}A\;cos^{\frac{2}{3}}A}$

$=\dfrac{1}{sinA\;cosA)^{\frac{2}{3}}}$........(1)

$\bf\,xy$

$=(\dfrac{cos^2A}{sinA})(\dfrac{sin^2A}{cosA})$

$=cosA\;sinA$

$\bf\,(xy)^{\frac{2}{3}}$

$=(cosA\;sinA)^{\frac{2}{3}}$

$\text{Now, (1) becomes}$

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=\frac{1}{\,(xy)^{\frac{2}{3}}}$

$\implies\boxed{\bf\,x^{\frac{2}{3}}+y^{\frac{2}{3}}=(xy)^{\frac{-2}{3}}}$

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