જો tan ^{-1} x = frac{pi}{4} - tan ^{-1} left | vedclass

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(B) આપેલ સમીકરણ: $	an ^{-1} x = \frac{\pi}{4} - 	an ^{-1} \left( \frac{1}{3} ight)$ આપણે જાણીએ છીએ કે $\frac{\pi}{4} = 	an ^{-1}(1)$,તેથી: $	an

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$\frac{1}{2}$

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(B) આપેલ સમીકરણ: $	an ^{-1} x = \frac{\pi}{4} - 	an ^{-1} \left( \frac{1}{3} ight)$ આપણે જાણીએ છીએ કે $\frac{\pi}{4} = 	an ^{-1}(1)$,તેથી: $	an ^{-1} x = 	an ^{-1}(1) - 	an ^{-1} \left( \frac{1}{3} ight)$ સૂત્ર $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A - B}{1 + AB} ight)$ નો ઉપયોગ કરતા: $	an ^{-1} x = 	an ^{-1} \left( \frac{1 - \frac{1}{3}}{1 + 1 	imes \frac{1}{3}} ight)$ $	an ^{-1} x = 	an ^{-1} \left( \frac{\frac{2}{3}}{\frac{4}{3}} ight)$ $	an ^{-1} x = 	an ^{-1} \left( \frac{2}{4} ight)$ $	an ^{-1} x = 	an ^{-1} \left( \frac{1}{2} ight)$ તેથી,$x = \frac{1}{2}$.

જો $\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left( \frac{1}{3} \right)$ હોય,તો $x$ ની કિંમત શોધો.

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