यदि sinh x = frac{5}{12} है,तो cosh frac{x}{2 | vedclass

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(D) हम जानते हैं कि $\cosh x = \sqrt{1 + \sinh^2 x}$।दिया गया है $\sinh x = \frac{5}{12}$,इसलिए $\cosh x = \sqrt{1 + (\frac{5}{12})^2} = \sqrt{1 + \fr

Vedclass · Accepted Answer

$\frac{5}{2 \sqrt{6}}$

Vedclass · Answer

(D) हम जानते हैं कि $\cosh x = \sqrt{1 + \sinh^2 x}$।दिया गया है $\sinh x = \frac{5}{12}$,इसलिए $\cosh x = \sqrt{1 + (\frac{5}{12})^2} = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12}$।सर्वसमिका $\cosh x = 2 \cosh^2 \frac{x}{2} - 1$ का उपयोग करने पर:$2 \cosh^2 \frac{x}{2} - 1 = \frac{13}{12}$$2 \cosh^2 \frac{x}{2} = \frac{13}{12} + 1 = \frac{25}{12}$$\cosh^2 \frac{x}{2} = \frac{25}{24}$$\cosh \frac{x}{2} = \sqrt{\frac{25}{24}} = \frac{5}{2 \sqrt{6}}$।

यदि $\sinh x = \frac{5}{12}$ है,तो $\cosh \frac{x}{2} =$

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