Question

$\cot \theta +\tan \theta =x$ and $\sec \theta -\cos \theta =y$ then ${\left(x^{2}y\right)}^{2/3}-{\left(x{y}^2\right)}^{2/3}.$

Answer 1

$m=cot\theta +tan\theta =\frac{cos\theta }{sin\theta }+\frac{sin\theta }{cos\theta }=\frac{co{s}^2\theta +si{n}^2\theta }{sin\theta cos\theta }=cosec\theta sec\theta =m$

$n=sec\theta -cos\theta =\frac{1}{cos\theta }-\frac{cos\theta }{1}=\frac{1-co{s}^2\theta }{cos\theta }=\frac{si{n}^2\theta }{cos\theta }=tan\theta sin\Theta =n$

$LHS=(m^{2}n{)}^{2/3}-(m{n}^2{)}^{2/3}=(m^4{n}^2{)}^{1/3}-(m^2{n}^4{)}^{1/3}$

$=(cose{c}^4\theta se{c}^4\theta ta{n}^2\theta si{n}^2\theta {)}^{1/3}-(cose{c}^2\theta se{c}^2\theta ta{n}^4\theta si{n}^4\theta {)}^{1/3}$

$=(\frac{1}{si{n}^4\theta }\frac{1}{co{s}^4\theta }\frac{si{n}^2\theta }{co{s}^2\theta }si{n}^2\theta {)}^{1/3}$ - $(\frac{1}{si{n}^2\theta }\frac{1}{co{s}^2\theta }\frac{si{n}^4\theta }{co{s}^4\theta }\frac{si{n}^4\theta }{1}{)}^{1/3}$

$=(\frac{1}{co{s}^6\theta }{)}^{1/3}-(\frac{si{n}^6\theta }{cos{5}^6\theta }{)}^{1/3}$ = $se{c}^2\theta -ta{n}^2\theta =1$