Question

If cosec theta - sin theta=m and sec theta - cos theta=n, then prove that (m^n)^2/3 + (mn^2)^2/3 = 1

Answer 1

$\textbf{Given:}$

$m=cosec\theta-sin\theta$

$n=sec\theta-cos\theta$

$\textbf{To prove:}$

$(m^2n)^\frac{2}{3}+(mn^2)^\frac{2}{3}=1$

$\textbf{Solution:}$

$m=cosec\theta-sin\theta$

$m=\dfrac{1}{sin\theta}-sin\theta$

$m=\dfrac{1-sin^2\theta}{sin\theta}$

$m=\dfrac{cos^2\theta}{sin\theta}$

$n=sec\theta-cos\theta$

$n=\dfrac{1}{cos\theta}-cos\theta$

$n=\dfrac{1-cos^2\theta}{cos\theta}$

$n=\dfrac{sin^2\theta}{cos\theta}$

$m^2n=\dfrac{cos^4\theta}{sin^2\theta}{\times}\dfrac{sin^2\theta}{cos\theta}$

$\implies\bf\,m^2n=cos^3\theta$

$mn^2=\dfrac{cos^2\theta}{sin\theta}{\times}\dfrac{sin^4\theta}{cos^2\theta}$

$\implies\bf\,mn^2=sin^3\theta$

$\text{Now,}$

$(m^2n)^\frac{2}{3}+(mn^2)^\frac{2}{3}$

$=(cos^3\theta)^\frac{2}{3}+(sin^3\theta)^\frac{2}{3}$

$=cos^2\theta+sin^2\theta$

$=1$

$\therefore\boxed{\bf(m^2n)^\frac{2}{3}+(mn^2)^\frac{2}{3}=1}$

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