જો x = operatorname{sech}^{-1} frac{1}{2} + t | vedclass

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(C) આપેલ છે કે,$x = \operatorname{sech}^{-1} \frac{1}{2} + 	anh^{-1} \frac{1}{2}$.લોગરીધમિક સ્વરૂપોનો ઉપયોગ કરતા: $\operatorname{sech}^{-1} z = \ln \

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$\frac{4 \sqrt{3} + 3}{3}$

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(C) આપેલ છે કે,$x = \operatorname{sech}^{-1} \frac{1}{2} + 	anh^{-1} \frac{1}{2}$.લોગરીધમિક સ્વરૂપોનો ઉપયોગ કરતા: $\operatorname{sech}^{-1} z = \ln \left( \frac{1 + \sqrt{1 - z^2}}{z} ight)$ અને $	anh^{-1} z = \frac{1}{2} \ln \left( \frac{1 + z}{1 - z} ight)$.$z = \frac{1}{2}$ માટે,$\operatorname{sech}^{-1} \frac{1}{2} = \ln (2 + \sqrt{3})$ અને $	anh^{-1} \frac{1}{2} = \ln \sqrt{3}$.તેથી,$x = \ln (2 + \sqrt{3}) + \ln \sqrt{3} = \ln (2\sqrt{3} + 3)$.હવે,$\cosh x = \frac{e^x + e^{-x}}{2}$.$e^x = 2\sqrt{3} + 3$ હોવાથી,$e^{-x} = \frac{2\sqrt{3} - 3}{3}$.તેથી,$\cosh x = \frac{(2\sqrt{3} + 3) + \frac{2\sqrt{3} - 3}{3}}{2} = \frac{8\sqrt{3} + 6}{6} = \frac{4\sqrt{3} + 3}{3}$.

જો $x = \operatorname{sech}^{-1} \frac{1}{2} + \tanh^{-1} \frac{1}{2}$ હોય,તો $\cosh x =$

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