CUET Preparation Today
CUET
General Test
Quantitative Reasoning
Trigonometry
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} |