n in N માટે,જો cot ^{-1} 3+cot ^{-1} 4+cot ^{ | vedclass

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(D) આપેલ સમીકરણ: $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$.$x > 0$ માટે $\cot ^{-1} x = 	an ^{-1} (1/x)$ નો ઉપયોગ કરીને $

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(D) આપેલ સમીકરણ: $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$.$x > 0$ માટે $\cot ^{-1} x = 	an ^{-1} (1/x)$ નો ઉપયોગ કરીને $	an ^{-1}$ માં ફેરવતા:$	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{4}+	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{n}=\frac{\pi}{4}$.પ્રથમ,$	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{4} = 	an ^{-1} \left(\frac{1/3+1/4}{1-(1/3)(1/4)}ight) = 	an ^{-1} \left(\frac{7/12}{11/12}ight) = 	an ^{-1} \frac{7}{11}$ મેળવીએ.હવે,$	an ^{-1} \frac{1}{5}$ ઉમેરતા: $	an ^{-1} \frac{7}{11}+	an ^{-1} \frac{1}{5} = 	an ^{-1} \left(\frac{7/11+1/5}{1-(7/11)(1/5)}ight) = 	an ^{-1} \left(\frac{46/55}{48/55}ight) = 	an ^{-1} \frac{46}{48} = 	an ^{-1} \frac{23}{24}$.તેથી,$	an ^{-1} \frac{23}{24} + 	an ^{-1} \frac{1}{n} = \frac{\pi}{4}$.$	an ^{-1} \frac{1}{n} = 	an ^{-1} 1 - 	an ^{-1} \frac{23}{24} = 	an ^{-1} \left(\frac{1-23/24}{1+(1)(23/24)}ight) = 	an ^{-1} \left(\frac{1/24}{47/24}ight) = 	an ^{-1} \frac{1}{47}$.આમ,$n = 47$.

$n \in N$ માટે,જો $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$ હોય,તો $n$ ની કિંમત ......... છે.

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