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) ${{\tanh }^{2}}x=1-{{\operatorname{sech}}^{2}}x$

Answer 1

Hint: To prove these identities, we will use the definition of $\sinh x$, $\cosh x$and $\tanh x$. We know that $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ , $\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ and $\tanh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$. We will substitute these values in one side of the equation and simplify it to obtain the other side of that equation.

Complete step-by-step solution:
(a) ${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$
We will consider the LHS of this equation,
LHS $={{\cosh }^{2}}x-{{\sinh }^{2}}x$
\[\begin{align}
  & ={{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}-\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right) \\
 & =\dfrac{{{e}^{2x}}+2{{e}^{x}}\cdot {{e}^{-x}}+{{e}^{-2x}}}{4}-\dfrac{{{e}^{2x}}-2{{e}^{x}}\cdot {{e}^{-x}}+{{e}^{-2x}}}{4} \\
 & =\dfrac{{{e}^{2x}}+2+{{e}^{-2x}}-{{e}^{2x}}+2-{{e}^{-2x}}}{4} \\
 & =\dfrac{4}{4} \\
 & =1 \\
\end{align}\]
$=$ RHS
Hence, proved.
(b) $\sinh 2x=2\sinh x\cosh x$
We will consider the RHS of this equation,
RHS $=2\sinh x\cosh x$
\[\begin{align}
  & =2\cdot \left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)\cdot \left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right) \\
 & =\dfrac{({{e}^{x}}-{{e}^{-x}})\cdot ({{e}^{x}}+{{e}^{-x}})}{2} \\
 & =\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{2} \\
 & =\sinh 2x \\
\end{align}\]
$=$ LHS
Hence, proved.
(c) $\cosh 2x={{\cosh }^{2}}x+{{\sinh }^{2}}x$
We will consider the RHS of this equation,
RHS $={{\cosh }^{2}}x+{{\sinh }^{2}}x$
Substituting and expanding the values, we get RHS as,
\[\begin{align}
  & ={{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}+{{\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)}^{2}} \\
 & =\dfrac{{{e}^{2x}}+2{{e}^{x}}\cdot {{e}^{-x}}+{{e}^{-2x}}}{4}+\dfrac{{{e}^{2x}}-2{{e}^{x}}\cdot {{e}^{-x}}+{{e}^{-2x}}}{4} \\
 & =\dfrac{{{e}^{2x}}+2+{{e}^{-2x}}+{{e}^{2x}}-2+{{e}^{-2x}}}{4} \\
 & =\dfrac{2{{e}^{2x}}+2{{e}^{-2x}}}{4} \\
 & =\dfrac{{{e}^{2x}}+{{e}^{-2x}}}{2} \\
 & =\cosh 2x \\
\end{align}\]
$=$ LHS
Hence, proved.
(d) ) ${{\tanh }^{2}}x=1-{{\operatorname{sech}}^{2}}x$
We know that $\operatorname{sech}x=\dfrac{1}{\cosh x}$. Now, we will consider the RHS of the given equation,
RHS $=1-{{\operatorname{sech}}^{2}}x$ \[=1-{{\left( \dfrac{1}{\cosh x} \right)}^{2}}\]
Substituting and expanding the values, we get
\[\begin{align}
  & =1-{{\left( \dfrac{1}{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)} \right)}^{2}} \\
 & =1-{{\left( \dfrac{2}{{{e}^{x}}+{{e}^{-x}}} \right)}^{2}} \\
 & =1-\dfrac{4}{{{e}^{2x}}+2+{{e}^{-2x}}} \\
 & =\dfrac{{{e}^{2x}}+2+{{e}^{-2x}}-4}{{{e}^{2x}}+2+{{e}^{-2x}}} \\
 & =\dfrac{{{e}^{2x}}-2+{{e}^{-2x}}}{{{e}^{2x}}+2+{{e}^{-2x}}} \\
 & =\dfrac{{{({{e}^{x}}-{{e}^{-x}})}^{2}}}{{{({{e}^{x}}+{{e}^{-x}})}^{2}}} \\
 & ={{\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right)}^{2}} \\
 & ={{\tanh }^{2}}x \\
\end{align}\]
$=$ LHS
Hence, proved.

Note: The calculations in this question involve exponentials. We should keep in mind the identities of indices while expanding the squares of exponential functions. It is very important to keep the signs of all terms in check, otherwise, we will not be able to prove the identities. It is crucial to know when to switch between definitions while simplifying equations involving trigonometric functions.