If $cos48°=\frac{m}{n}$, then $sec 48° − cot 42°$ is equal to:

$\frac{n-\sqrt{n^2-m^2}}{m}$

We are given that :-

cos 48° = $\frac{m}{n}$

{ we know, cosA = $\frac{B}{H}$ }

By using pythagoras theorem,

P² + B² = H²

P² + m² = n²

P = $\sqrt {n² - m² }$

Now,

sec 48° - cot 42°

= sec 48° - cot ( 90° - 48°)

= sec 48° - tan 48°      { tan A = cot ( 90° - A ) }

= $\frac{n}{ m }$ - $\frac{√(n² - m²))}{m}$

= $\frac{ n - √(n² - m²))}{m}$