જો tanh x = frac{1}{2} હોય,તો sinh 2x - text{ | vedclass

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(B) આપેલ છે,$	anh x = \frac{1}{2}$.આપણે જાણીએ છીએ કે $	anh^2 x = 1 - 	ext{sech}^2 x$,તેથી $	ext{sech}^2 x = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4}

Vedclass · Accepted Answer

$\frac{11}{15}$

Vedclass · Answer

(B) આપેલ છે,$	anh x = \frac{1}{2}$.આપણે જાણીએ છીએ કે $	anh^2 x = 1 - 	ext{sech}^2 x$,તેથી $	ext{sech}^2 x = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.આમ,$	ext{sech } x = \frac{\sqrt{3}}{2}$ અને $\cosh x = \frac{2}{\sqrt{3}}$.કારણ કે $	anh x = \frac{\sinh x}{\cosh x}$,તેથી $\sinh x = 	anh x \cdot \cosh x = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.હવે,$\sinh 2x = 2 \sinh x \cosh x = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = \frac{4}{3}$.અને $	ext{sech } 2x = \frac{1}{\cosh 2x} = \frac{1}{\cosh^2 x + \sinh^2 x} = \frac{1}{(\frac{2}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2} = \frac{1}{\frac{4}{3} + \frac{1}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}$.તેથી,$\sinh 2x - 	ext{sech } 2x = \frac{4}{3} - \frac{3}{5} = \frac{20 - 9}{15} = \frac{11}{15}$.

જો $\tanh x = \frac{1}{2}$ હોય,તો $\sinh 2x - \text{sech } 2x = $

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