frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2 | vedclass

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Question

(A) Given expression: $E = \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.We know that $\sinh x = \frac{e^x - e^{-x}}{2}$,so $(e^x - e^{-x})^2 = 4\s

Vedclass · Accepted Answer

$\sinh^2 x + \coth^2 x$

Vedclass · Answer

(A) Given expression: $E = \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.We know that $\sinh x = \frac{e^x - e^{-x}}{2}$,so $(e^x - e^{-x})^2 = 4\sinh^2 x$.Thus,the denominator is $4(4\sinh^2 x) = 16\sinh^2 x$.Alternatively,let us evaluate option $A$: $\sinh^2 x + \coth^2 x$.$\sinh^2 x + \coth^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 + \left(\frac{e^x + e^{-x}}{e^x - e^{-x}}\right)^2$$= \frac{(e^x - e^{-x})^4 + 4(e^x + e^{-x})^2}{4(e^x - e^{-x})^2}$$= \frac{(e^{2x} + e^{-2x} - 2)^2 + 4(e^{2x} + e^{-2x} + 2)}{4(e^x - e^{-x})^2}$$= \frac{(e^{4x} + e^{-4x} + 2 - 4e^{2x} - 4e^{-2x} + 2) + 4e^{2x} + 4e^{-2x} + 8}{4(e^x - e^{-x})^2}$$= \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.This matches the given expression. Hence,the correct option is $A$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

$\frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2} = \dots$

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