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 ) }

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

= \(\frac{ n - √(n² - m²))}{m}\)