If $cos48°=\frac{m}{n}$, then $sec 48° − cot 42°$ is equal to: |
$\frac{m-\sqrt{n^2-m^2}}{m}$ $\frac{m-\sqrt{n^2-m^2}}{n}$ $\frac{n-\sqrt{n^2-m^2}}{n}$ $\frac{n-\sqrt{n^2-m^2}}{m}$ |
$\frac{n-\sqrt{n^2-m^2}}{m}$ |
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 ) } = \(\frac{n}{ m }\) - \(\frac{√(n² - m²))}{m}\) = \(\frac{ n - √(n² - m²))}{m}\) |