যদি sinθ = (a-b)/(a+b) হয়, তবে tan (π/2 - θ/2) এর মান কত?

দেওয়া আছে, \( \sin\theta = \frac{a-b}{a+b} \).

আমরা জানি, \( \cos\theta = \sqrt{1 - \sin^2\theta} \).

সুতরাং, \( \cos\theta = \sqrt{1 - \left(\frac{a-b}{a+b}\right)^2} \)

\(= \sqrt{1 - \frac{(a-b)^2}{(a+b)^2}} \)

\(= \sqrt{\frac{(a+b)^2 - (a-b)^2}{(a+b)^2}} \)

\(= \sqrt{\frac{a^2 + 2ab + b^2 - (a^2 - 2ab + b^2)}{(a+b)^2}} \)

\(= \sqrt{\frac{4ab}{(a+b)^2}} \)

\( \therefore \cos\theta = \frac{2\sqrt{ab}}{a+b} \)

এখন, আমরা জানি, \( \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} \)।

আবার, \( \tan(\pi/2 - x) = \cot x \).

সুতরাং, \( \tan\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \cot\left(\frac{\theta}{2}\right) \)

আমরা জানি, \( \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \)

\( = \sqrt{\frac{1 - \frac{2\sqrt{ab}}{a+b}}{1 + \frac{2\sqrt{ab}}{a+b}}} \)

\( = \sqrt{\frac{\frac{a+b - 2\sqrt{ab}}{a+b}}{\frac{a+b + 2\sqrt{ab}}{a+b}}} \)

\( = \sqrt{\frac{a+b - 2\sqrt{ab}}{a+b + 2\sqrt{ab}}} \)

\( = \sqrt{\frac{(\sqrt{a} - \sqrt{b})^2}{(\sqrt{a} + \sqrt{b})^2}} \)

\( = \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} \)

তাহলে, \( \cot\left(\frac{\theta}{2}\right) = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} \)

এখন, \( \tan\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \cot\left(\frac{\theta}{2}\right) = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} \cdot \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} \)

\( = \frac{(\sqrt{a} + \sqrt{b})^2}{a - b} = \frac{a + 2\sqrt{ab} + b}{a - b} \)

অন্যভাবে,

\( \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)

\( = \sqrt{\frac{1+\frac{a-b}{a+b}}{1-\frac{a-b}{a+b}}} = \sqrt{\frac{\frac{a+b+a-b}{a+b}}{\frac{a+b-a+b}{a+b}}} \)

\( = \sqrt{\frac{2a}{2b}} = \sqrt{\frac{a}{b}} \)

আবার, \( \tan(\frac{\pi}{2}-\frac{\theta}{2}) = \cot(\frac{\theta}{2}) = \pm \sqrt{\frac{1+\cos \theta}{\1-\cos \theta}} \)

\( \tan(\frac{\pi}{2}-\frac{\theta}{2}) = \sqrt{\frac{a}{b}} \)। 🤔

আমরা জানি, \( \tan(\pi/2 - \theta/2) = \cot(\theta/2) \). 😃

এবং \( \cot(\theta/2) = \pm \sqrt{\frac{1 + \cos\theta}{1 - \cos\theta}} \) 🥰

\( = \pm \sqrt{\frac{1 + \frac{2\sqrt{ab}}{a+b}}{1 - \frac{2\sqrt{ab}}{a+b}}} \)

\( = \pm \sqrt{\frac{\frac{a+b+2\sqrt{ab}}{a+b}}{\frac{a+b-2\sqrt{ab}}{a+b}}} \)

\( = \pm \sqrt{\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}} \)

\( = \pm \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \pm \frac{(\sqrt{a}+\sqrt{b})^2}{a-b} = \pm \frac{a+b+2\sqrt{ab}}{a-b} \)

যদি \( \sin\theta = \frac{a-b}{a+b} \) হয়, তবে \( \tan\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \pm\sqrt{\frac{a}{b}} \) হবে।