Question

Prove the following :
(a) ${\cosh }^{2}x-{\sinh }^{2}x=1$
(b) $\sinh 2x=2\sinh x\cosh x$
(c) $\cosh 2x={\cosh }^{2}x+{\sinh }^{2}x$
(d) ${\tanh }^{2}x=1-{\mathrm{sech}}^{2}x$

Answer 1

If x is any real or complex number, the hyperbolic $\sin$ of x & hyperbolic cosine of x is defines by-

$\sin hx=\frac{1}{2}(e^x-e^{-x})$ …………..(A)

$\cos hx=\frac{1}{2}(e^x+e^{-x})$ ………….(B)

(i) $\cos h^{2}x-\sin h^{2}x=1$

LHS

$\Rightarrow \cos h^{2}x-\sin h^{2}x={\left[\frac{e^x+e^{-x}}{2}\right]}^2-{\left[\frac{e^x-e^{-x}}{2}\right]}^2$

$e^x\times e^{-x}=e^{x-x}=e^o=1$ $=\frac{(e^{2x}+e^{-2x}+2e^x.e^{-x})(e^{2x}+e^{-1x}-2e^x.e^{-x})}{4}$

$=\frac{e^{2x}+e^{-2x}+2e^o-e^{2x}-e^{-2x}=2e^o}{4}$

$=\frac{4e^o}{4}=e^o=1$ RHS.

(ii) $\sin h2x=2\sin hx\cos hx$

RHS$\Rightarrow 2\sin hx\cos hx$

$\Rightarrow 2\left(\frac{e^x-e^{-x}}{2}\right)\times \left(\frac{e^x+e^{-x}}{2}\right)$ Using $(a-b)(a+b)=a^2-b^2$

$\Rightarrow \frac{1}{2}(e^{2x}-e^{-2x})$ $\to$ This is similar except $\left(2x\right)$ in place of x in equation $2)$

$\Rightarrow \sin h\left(2x\right)$ LHS.

iii) $\cos h^{2}x=\cos h^{2}x+\sin h^{2}x$

RHS$\Rightarrow \cos h^{2}x+\sin h^{2}x$

$={\left[\frac{e^x+e^{-x}}{2}\right]}^2+{\left[\frac{e^x+e^{-x}}{2}\right]}^2$

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

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

$=\left(\frac{e^{2x}+e^{-2x}}{2}\right)$ $\to$ This is similar to equation B except $\left(2x\right)$ in place of x.

$=\cos h\left(2x\right)$ LHS

iv) $\tan h^{2}x=1-\sec h^{2}x$

LHS$\Rightarrow \tan h^{2}x$

$=\frac{\sin h^{2}x}{\cos h^{2}x}=\frac{{\left(\frac{e^x-e^{-x}}{2}\right)}^2}{{\left(\frac{e^x+e^{-x}}{2}\right)}^2}={\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)}^2$

RHS$\Rightarrow 1-\sec h^{2}x$

$\Rightarrow 1-\frac{1}{\cos h^{2}x}$

$\Rightarrow 1-\frac{1}{{\left(\frac{e^x+e^{-x}}{2}\right)}^2}$

$\Rightarrow 1-\frac{4}{(e^x+e^{-x}{)}^2}$

$\Rightarrow \frac{e^{2x}+e^{-2x}+2e^o-4}{(e^x+e^{-x}{)}^2}=\frac{(e^x{)}^2+(e^{-x}{)}^2-2e^x.e^{-x}}{(e^x+e^{-x}{)}^2}=\frac{(e^x-e^{-x}{)}^2}{(e^x+e^{-x}{)}^2}$

LHS$=$RHS.