(1-cos2θ+sin2θ)/(1+cos2θ+sin2θ) = কত?

🤔 প্রশ্ন: \(\frac{1-\cos2\theta+\sin2\theta}{1+\cos2\theta+\sin2\theta}\) = কত?

💡 সলিউশন:

আমরা জানি, \( \cos2\theta = 1 - 2\sin^2\theta \) এবং \( \sin2\theta = 2\sin\theta\cos\theta \)।

তাহলে,

\(\frac{1-\cos2\theta+\sin2\theta}{1+\cos2\theta+\sin2\theta} = \frac{1-(1-2\sin^2\theta)+2\sin\theta\cos\theta}{1+(1-2\sin^2\theta)+2\sin\theta\cos\theta}\)

\(= \frac{1-1+2\sin^2\theta+2\sin\theta\cos\theta}{1+1-2\sin^2\theta+2\sin\theta\cos\theta}\)

\(= \frac{2\sin^2\theta+2\sin\theta\cos\theta}{2-2\sin^2\theta+2\sin\theta\cos\theta}\)

\(= \frac{2\sin\theta(\sin\theta+\cos\theta)}{2(1-\sin^2\theta)+2\sin\theta\cos\theta}\)

আমরা জানি, \( \cos^2\theta + \sin^2\theta = 1 \), সুতরাং \( 1 - \sin^2\theta = \cos^2\theta \)।

\(= \frac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos^2\theta+2\sin\theta\cos\theta}\)

\(= \frac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos\theta(\cos\theta+\sin\theta)}\)

\(= \frac{\sin\theta}{\cos\theta}\)

\(= \tan\theta\)

সুতরাং, \(\frac{1-\cos2\theta+\sin2\theta}{1+\cos2\theta+\sin2\theta} = \tan\theta\). 🎉