\(\sec A + \tan A = \frac {5} {2} \)হলে, \(\sec A – \tan A = ?\)
May 13, 2023 | , 42 BCS Preliminary
| - \(\frac{1} {2} \)
- \(\frac{1} {5} \)
- \(\frac{2} {5} \)
- \(\frac{5} {2}\)
দেওয়া আছে, \(\sec A + \tan A = \frac {5} {2}\)
আমরা জানি,
\(\sec^2 A - \tan^2 A = 1 \\[3px]\)বা, \(\left(\sec A + \tan A \right) \left( \sec A - \tan A \right) = 1 [ a^2 - b^2 = (a+b)(a-b) ] \\[5px]\)বা, \(\sec A - \tan A = \frac {1} {\sec A + \tan A} \\[7px]\)বা, \(\sec A - \tan A = \frac {1} { \frac {5} {2} } [ \sec A + \tan A = \frac {5} {2}\) এর মান বসিয়ে ]বা, \(\sec A - \tan A = 1 \times \frac {2} {5} \\[5px]\)সুতরাং, \(\sec A - \tan A = \frac {2} {5} { \left[ Ans.\right] }\)