Question

if $\cot \alpha +\tan \alpha =m$ and $\frac{1}{\cos \alpha }-\cos \alpha =n$ then eliminate $\alpha$.

Answer 1

$co{t}^2\alpha +ta{n}^2\alpha$

$se{c}^2\alpha -1+se{c}^2\alpha -1+2=m^2$

$se{c}^2\alpha +se{c}^2\alpha =m^2$

$\frac{1}{si{n}^2\alpha co{s}^2\alpha }=m^2$

$\frac{1}{na{s}^2\alpha }=m^2$

$\frac{si{n}^2\alpha }{as\alpha }=n$

$\frac{1}{m^{2}n}=a{s}^2\alpha$

$as\alpha ={\left[\frac{1}{m^{2}n}\right]}^{\frac{1}{3}}$

$\alpha ={\left(n{m}^2\right)}^{\frac{1}{3}}$

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

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