tan(45° + A) + tan(45° – A)) এর মান কত? 

Explanation:

Another Explanation (5): দেওয়া আছে, tan(45° + A) + tan(45° – A) এর মান নির্ণয় করতে হবে। আমরা জানি, \(tan(a + b) = \frac{tan(a) + tan(b)}{1 - tan(a)tan(b)}\) এবং \(tan(a - b) = \frac{tan(a) - tan(b)}{1 + tan(a)tan(b)}\) সুতরাং, \(tan(45° + A) = \frac{tan(45°) + tan(A)}{1 - tan(45°)tan(A)} = \frac{1 + tan(A)}{1 - tan(A)}\) এবং \(tan(45° - A) = \frac{tan(45°) - tan(A)}{1 + tan(45°)tan(A)} = \frac{1 - tan(A)}{1 + tan(A)}\) এখন, \(tan(45° + A) + tan(45° - A) = \frac{1 + tan(A)}{1 - tan(A)} + \frac{1 - tan(A)}{1 + tan(A)}\) =\(\frac{(1 + tan(A))^2 + (1 - tan(A))^2}{(1 - tan(A))(1 + tan(A))}\) =\(\frac{1 + 2tan(A) + tan^2(A) + 1 - 2tan(A) + tan^2(A)}{1 - tan^2(A)}\) =\(\frac{2 + 2tan^2(A)}{1 - tan^2(A)}\) =\(\frac{2(1 + tan^2(A))}{1 - tan^2(A)}\) আমরা জানি, \(sec^2(A) = 1 + tan^2(A)\) এবং \(cos(2A) = \frac{1 - tan^2(A)}{1 + tan^2(A)}\) সুতরাং, \(tan(45° + A) + tan(45° - A) = 2 \cdot \frac{1 + tan^2(A)}{1 - tan^2(A)}\) = \(2 \cdot \frac{1}{cos(2A)}\) = \(2sec(2A)\) অতএব, \(tan(45° + A) + tan(45° – A) = 2sec(2A)\) 🥳