4 tan^{-1} frac{1}{5} - tan^{-1} frac{1}{70} | vedclass

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(C) આપણે સૂત્ર $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$4 	an^{-1} \frac{1}{5} = 2 \left( 2 	an^{-1} \frac

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) આપણે સૂત્ર $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$4 	an^{-1} \frac{1}{5} = 2 \left( 2 	an^{-1} \frac{1}{5} ight) = 2 	an^{-1} \left( \frac{2/5}{1 - 1/25} ight) = 2 	an^{-1} \left( \frac{2/5}{24/25} ight) = 2 	an^{-1} \left( \frac{5}{12} ight)$.હવે,$2 	an^{-1} \frac{5}{12} = 	an^{-1} \left( \frac{2(5/12)}{1 - (5/12)^2} ight) = 	an^{-1} \left( \frac{5/6}{1 - 25/144} ight) = 	an^{-1} \left( \frac{5/6}{119/144} ight) = 	an^{-1} \left( \frac{5}{6} 	imes \frac{144}{119} ight) = 	an^{-1} \left( \frac{120}{119} ight)$.આગળ,આપણે $	an^{-1} \frac{1}{99} - 	an^{-1} \frac{1}{70} = 	an^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} ight) = 	an^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} ight) = 	an^{-1} \left( \frac{-29}{6931} ight) = -	an^{-1} \left( \frac{1}{239} ight)$ ની ગણતરી કરીએ.છેલ્લે,$	an^{-1} \frac{120}{119} - 	an^{-1} \frac{1}{239} = 	an^{-1} \left( \frac{120/119 - 1/239}{1 + (120/119)(1/239)} ight) = 	an^{-1} \left( \frac{(28680 - 119) / 28441}{(28441 + 120) / 28441} ight) = 	an^{-1} \left( \frac{28561}{28561} ight) = 	an^{-1}(1) = \frac{\pi}{4}$.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} = $

Similar Questions

જો $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$ હોય,તો $x$ ની કિંમત શોધો.

ધારો કે $S = \{x \in R : 0 < x < 1 \text{ અને } 2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$. જો $n(S)$ એ $S$ માં રહેલા ઘટકોની સંખ્યા દર્શાવતું હોય,તો:

કિંમત શોધો: $\operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$

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