Question

If $(cot\theta +tan\theta )=m$ and $(sec\theta -cos\theta )=n$, prove that

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

Answer 1

$m=(cot\theta +tan\theta )=\frac{cos\theta }{sin\theta }+\frac{sin\theta }{cos\theta }=\frac{si{n}^2\theta +co{s}^2\theta }{sin\theta cos\theta }=\frac{1}{sin\theta cos\theta }=cosec\theta sec\theta n=(sec\theta -cos\theta )=\frac{1}{cos\theta }-\frac{cos\theta }{1}=\frac{1-co{s}^2\theta }{cos\theta }=\frac{si{n}^2\theta }{cos\theta }=tan\theta sin\theta$

$LHS=(m^{2}n{)}^{\frac{2}{3}}-(m{n}^2{)}^{\frac{2}{3}}=(m^4{n}^2{)}^{\frac{1}{3}}-(m^2{n}^4{)}^{\frac{1}{3}}=(cose{c}^4\theta se{c}^4\theta ta{n}^2\theta si{n}^2\theta {)}^{\frac{1}{3}}-(cose{c}^2\theta se{c}^2\theta ta{n}^4\theta si{n}^4\theta {)}^{\frac{1}{3}}=(\frac{1}{si{n}^4\theta }\times \frac{1}{co{s}^4\theta }\times \frac{si{n}^2\theta }{co{s}^2\theta }\times si{n}^2\theta {)}^{\frac{1}{3}}-(\frac{1}{si{n}^2\theta }\times \frac{1}{co{s}^2\theta }\times \frac{si{n}^4\theta }{co{s}^4\theta }\times si{n}^4\theta {)}^{\frac{1}{3}}=(\frac{1}{co{s}^6\theta }{)}^{\frac{1}{3}}-(\frac{si{n}^6\theta }{co{s}^6\theta }{)}^{\frac{1}{3}}=se{c}^2\theta -ta{n}^2\theta =1=RHS$