Question

# ^^20(5)(b)^^20\quad j\cos (\alpha+j\beta) \cos (\alpha+j \beta)=\cos \alpha \cos (j \beta)-\sin \alpha \sin (j \beta) j \cos (\alpha+j \beta)=j \cos \alpha \cdot \cos (j \beta)-\sin \alpha j \sin (j \beta) \text

{ Now } j \sin (j \beta)=j \cdot\left(\frac{e^{j(j \beta)}-e^{-j(j \beta)}}{2 j}\right) =\frac{1}{2}\left(e^{-\beta}-e^{\beta}\right) =-\frac{1}{2}\left(e^{\beta}-e^{-\beta}\right) =-\sinh \beta \& \quad j \cos (j \beta)=j \cdot\left(\frac{e^{j(j \beta)}+e^{-j(j \beta)}}{2}\right) =j \cdot\left(\frac{e^{-\beta}+e^{\beta}}{2}\right) =j \cosh \beta 10 ; \quad j \cos (\alpha+j \beta)=j \cos \alpha \cosh \beta-\sin \alpha(-\sinh \beta) i \cos (\alpha+j \beta)=\sinh \beta \sin \alpha+j \cosh \beta(01 \alpha (a+b)^{2} e^{-2 x}=(c \cosh x+c \sinh x)^{2} e^{-2 x} =c^{2}\left(\frac{e^{x}+e^{-x}}{2}+\frac{e^{x}-e^{-x}}{2}\right)^{2} e^{-2 x} =c^{2}\left(e^{x}\right)^{2} e^{-2 x}=c^{2} a^{2}-b^{2}=(c \cosh x)^{2}-(c \sinh x)^{2} =c^{2}\left(\cosh ^{2} x-\sinh ^{2} x\right) =c^{2}(\cos h x+\sinh x)(\cos h x-\sinh x) =c^{2}\left(\frac{e^{k}+e^{-x}}{2}+\frac{e^{x} e^{-x}}{2}\right)\left(\frac{e^{x}+e^{-x}}{2}-\frac{e^{x}-e^{-x}}{2}\right) =c^{2}\left(e^{x}\right)\left(e^{-x}\right) =c^{2} \cdot \operatorname{so}(a+b)^{2} e^{-2 x}=a^{2}-b^{2}

Fig: 1

Fig: 2

Fig: 3

Fig: 4

Fig: 5

Fig: 6

Fig: 7

Fig: 8

Fig: 9

Fig: 10

Fig: 11

Fig: 12

Fig: 13

Fig: 14

Fig: 15

Fig: 16

Fig: 17

Fig: 18

Fig: 19

Fig: 20

Fig: 21