કિંમત શોધો: tan ^{ - 1}left(frac{{{c_1}x - y} | vedclass

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(C) આપેલ પદાવલિ $S = 	an ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}ight) + 	an ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}ight) +

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$	an ^{ - 1}\left(\frac{x}{y}ight)$

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(C) આપેલ પદાવલિ $S = 	an ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}ight) + 	an ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}ight) + 	an ^{ - 1}\left(\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}}ight) + ... + 	an ^{ - 1}\left(\frac{1}{{{c_n}}}ight)$ છે.પ્રથમ,પ્રથમ પદને ફરીથી લખતા: $	an ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}ight) = 	an ^{ - 1}\left(\frac{\frac{x}{y} - \frac{1}{c_1}}{1 + \frac{x}{y} \cdot \frac{1}{c_1}}ight) = 	an ^{ - 1}\left(\frac{x}{y}ight) - 	an ^{ - 1}\left(\frac{1}{c_1}ight)$.હવે,નિત્યસમ $	an ^{ - 1} a - 	an ^{ - 1} b = 	an ^{ - 1}\left(\frac{a - b}{1 + ab}ight)$ નો ઉપયોગ કરીને બાકીના પદોને દર્શાવો:$	an ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}ight) = 	an ^{ - 1}\left(\frac{1}{c_1}ight) - 	an ^{ - 1}\left(\frac{1}{c_2}ight)$.$	an ^{ - 1}\left(\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}}ight) = 	an ^{ - 1}\left(\frac{1}{c_2}ight) - 	an ^{ - 1}\left(\frac{1}{c_3}ight)$.આ પેટર્ન ચાલુ રાખતા,સામાન્ય પદ $	an ^{ - 1}\left(\frac{1}{c_{k-1}}ight) - 	an ^{ - 1}\left(\frac{1}{c_k}ight)$ મળે છે.આ બધાનો સરવાળો કરતા:$S = \left[	an ^{ - 1}\left(\frac{x}{y}ight) - 	an ^{ - 1}\left(\frac{1}{c_1}ight)ight] + \left[	an ^{ - 1}\left(\frac{1}{c_1}ight) - 	an ^{ - 1}\left(\frac{1}{c_2}ight)ight] + ... + \left[	an ^{ - 1}\left(\frac{1}{c_{n-1}}ight) - 	an ^{ - 1}\left(\frac{1}{c_n}ight)ight] + 	an ^{ - 1}\left(\frac{1}{c_n}ight)$.આ એક ટેલિસ્કોપિંગ શ્રેણી છે જ્યાં બધા મધ્યવર્તી પદો ઉડી જાય છે:$S = 	an ^{ - 1}\left(\frac{x}{y}ight) - 	an ^{ - 1}\left(\frac{1}{c_n}ight) + 	an ^{ - 1}\left(\frac{1}{c_n}ight) = 	an ^{ - 1}\left(\frac{x}{y}ight)$.

કિંમત શોધો: $\tan ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}\right) + \tan ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}\right) + \tan ^{ - 1}\left(\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}}\right) + ... + \tan ^{ - 1}\left(\frac{1}{{{c_n}}}\right)$

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