Question

If $cot\alpha +tan\alpha =m$ and $\frac{1}{cos\alpha }-cos\alpha =n$, then

• A. $m(m{n}^2{)}^{1/3}-n(n{m}^2{)}^{1/3}=1$
• B. $m(m^{2}n{)}^{1/3}-n(n{m}^2{)}^{1/3}=1$
• C. $n(m{n}^2{)}^{1/3}-m(n{m}^2{)}^{1/3}=1$
• D. $n(m^{2}n{)}^{1/3}-m(n{m}^2{)}^{1/3}=1$

Answer 1

The correct option is A $m(m{n}^2{)}^{1/3}-n(n{m}^2{)}^{1/3}=1$

$cot\alpha +tan\alpha =m\Rightarrow \frac{1}{tan\alpha }+tan\alpha =m$

$\Rightarrow 1+{tan}^2\alpha =mtan\alpha \Rightarrow {sec}^2\alpha =mtan\alpha ...\left(1\right)$

$\frac{1}{cos\alpha }-cos\alpha =n\Rightarrow sec\alpha -\frac{1}{sec\alpha }=n$

$\Rightarrow {sec}^2\alpha -1=nsec\alpha \Rightarrow {tan}^2\alpha =nsec\alpha$

$\Rightarrow {tan}^4\alpha =n^2{sec}^2\alpha$

Substituting value from (1)

${tan}^4\alpha =n^{2}mtan\alpha \Rightarrow {tan}^3\alpha =n^{2}m\Rightarrow tan\alpha ={\left(n^{2}m\right)}^{\frac{1}{3}}...\left(2\right)$

Substituting (2) in (1), we get

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

From (2) and (3)

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

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