Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Clearly $\alpha \ne$ 0
$\cot \alpha +\tan \alpha =m$
$\Rightarrow 1+{\tan }^2\alpha =m\tan \alpha \Rightarrow {\sec }^2\alpha =m\tan \alpha \left(1\right)\text{and}\frac{1}{\cos \alpha }-\cos \alpha =n\Rightarrow {\sec }^2\alpha -1=n\sec \alpha \Rightarrow {\tan }^2\alpha =n\sec \alpha \Rightarrow {\tan }^4\alpha =n^2{\sec }^2\alpha \Rightarrow {\tan }^4\alpha =n^{2}m\tan \alpha \left[\text{using (1)}\right]\Rightarrow {\tan }^3\alpha =n^{2}m\Rightarrow \tan \alpha =(n^{2}m{)}^{\frac{1}{3}}\text{and}{\sec }^2\alpha =m(n^{2}m{)}^{\frac{1}{3}}\left[\text{using (1)}\right]\text{Now}{\sec }^2\alpha -{\tan }^2\alpha =1\Rightarrow m(n^{2}m{)}^{\frac{1}{3}}-(n^{2}m{)}^{\frac{2}{3}}=1\Rightarrow m(m{n}^2{)}^{\frac{1}{3}}-n(n{m}^2{)}^{\frac{1}{3}}=1$