cotalpha + cotbeta=a, tanalpha+tanbeta=b  এবং  alpha+beta=theta  হলে  cottheta=?  কত? 

Explanation:

Another Explanation (5): bài toán: cotα + cotβ = a, tanα + tanβ = b এবং α + β = θ হলে cotθ = ? সমাধান: আমরা জানি, cotθ = \(\frac{1}{tanθ}\) আবার, tanθ = tan(α + β) = \(\frac{tanα + tanβ}{1 - tanα tanβ}\) = \(\frac{b}{1 - tanα tanβ}\) সুতরাং, cotθ = \(\frac{1 - tanα tanβ}{b}\) ...(1) এখন, cotα + cotβ = a => \(\frac{1}{tanα}\) + \(\frac{1}{tanβ}\) = a => \(\frac{tanα + tanβ}{tanα tanβ}\) = a => \(\frac{b}{tanα tanβ}\) = a => tanα tanβ = \(\frac{b}{a}\) (1) নং সমীকরণে tanα tanβ এর মান বসিয়ে পাই, cotθ = \(\frac{1 - \frac{b}{a}}{b}\) = \(\frac{\frac{a-b}{a}}{b}\) = \(\frac{a-b}{ab}\) = \(\frac{a}{ab} - \frac{b}{ab}\) = \(\frac{1}{b} - \frac{1}{a}\) অতএব, cotθ = \(\frac{1}{b} - \frac{1}{a}\) 🎉🎉