જો tan ^{-1}left[frac{1}{1+1 cdot 2}right]+ta | vedclass

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(D) આપણે જાણીએ છીએ કે $	an ^{-1}\left[\frac{1}{1+k(k+1)}ight] = 	an ^{-1}\left[\frac{(k+1)-k}{1+k(k+1)}ight] = 	an ^{-1}(k+1) - 	an ^{-1}(k)$.

Vedclass · Accepted Answer

$\frac{n}{n+2}$

Vedclass · Answer

(D) આપણે જાણીએ છીએ કે $	an ^{-1}\left[\frac{1}{1+k(k+1)}ight] = 	an ^{-1}\left[\frac{(k+1)-k}{1+k(k+1)}ight] = 	an ^{-1}(k+1) - 	an ^{-1}(k)$.સરવાળાના દરેક પદ માટે આ સૂત્ર લાગુ પાડતા:$	an ^{-1}\left[\frac{1}{1+1 \cdot 2}ight] = 	an ^{-1}(2) - 	an ^{-1}(1)$$	an ^{-1}\left[\frac{1}{1+2 \cdot 3}ight] = 	an ^{-1}(3) - 	an ^{-1}(2)$...$	an ^{-1}\left[\frac{1}{1+n(n+1)}ight] = 	an ^{-1}(n+1) - 	an ^{-1}(n)$આ પદોનો સરવાળો કરતા,આપણને ટેલિસ્કોપિંગ શ્રેણી મળે છે:$S = (	an ^{-1}(2) - 	an ^{-1}(1)) + (	an ^{-1}(3) - 	an ^{-1}(2)) + \cdots + (	an ^{-1}(n+1) - 	an ^{-1}(n))$$S = 	an ^{-1}(n+1) - 	an ^{-1}(1)$સૂત્ર $	an ^{-1}(A) - 	an ^{-1}(B) = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$ નો ઉપયોગ કરતા:$S = 	an ^{-1}\left(\frac{(n+1)-1}{1+(n+1)(1)}ight) = 	an ^{-1}\left(\frac{n}{1+n+1}ight) = 	an ^{-1}\left(\frac{n}{n+2}ight)$આપેલ છે કે $S = 	an ^{-1}(x)$,તેથી $x = \frac{n}{n+2}$.

જો $\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]+\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]+\cdots+\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right]=\tan ^{-1}[x]$ હોય,તો $x=$

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