यदि operatorname{Tanh}^{-1} x = operatorname{ | vedclass

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(A) दिया गया है $\operatorname{Tanh}^{-1} x = \log \sqrt{5}$.परिभाषा $\operatorname{Tanh}^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right)$ का

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$\frac{\pi}{4}$

Vedclass · Answer

(A) दिया गया है $\operatorname{Tanh}^{-1} x = \log \sqrt{5}$.परिभाषा $\operatorname{Tanh}^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right)$ का उपयोग करते हुए,हमारे पास है $\frac{1}{2} \log \left( \frac{1+x}{1-x} \right) = \log \sqrt{5} = \frac{1}{2} \log 5$.अतः,$\frac{1+x}{1-x} = 5 \implies 1+x = 5-5x \implies 6x = 4 \implies x = \frac{2}{3}$.दिया गया है $\operatorname{Coth}^{-1} y = \log \sqrt{5}$.परिभाषा $\operatorname{Coth}^{-1} y = \frac{1}{2} \log \left( \frac{y+1}{y-1} \right)$ का उपयोग करते हुए,हमारे पास है $\frac{1}{2} \log \left( \frac{y+1}{y-1} \right) = \frac{1}{2} \log 5$.अतः,$\frac{y+1}{y-1} = 5 \implies y+1 = 5y-5 \implies 4y = 6 \implies y = \frac{3}{2}$.अब,$xy = \left( \frac{2}{3} \right) \left( \frac{3}{2} \right) = 1$.इसलिए,$\operatorname{Tan}^{-1}(xy) = \operatorname{Tan}^{-1}(1) = \frac{\pi}{4}$.

यदि $\operatorname{Tanh}^{-1} x = \operatorname{Coth}^{-1} y = \log \sqrt{5}$ है,तो $\operatorname{Tan}^{-1}(xy) = $

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