Question

For $0 < c \leqslant \pi ,{\sinh ^{ - 1}}(\cot x) = $
A.$\log (\cot \frac{x}{2})$
B.$\log (\tan \frac{x}{2})$
C.$\log (1 + \cot x)$
D.$\log (1 + \tan x)$

Answer 1

Hint: We know that ${\sinh ^{ - 1}}x = \log \left( {x + \sqrt {1 + {x^2}} } \right)$ and now we need to replace x by cot x and simplify using the identities like $\cos e{c^2}x - {\cot ^2}x = 1$,$\cot x = \frac{{\cos x}}{{\sin x}}$ and $\cos ecx = \frac{1}{{\sin x}}$,$2{\cos ^2}x = 1 + \cos 2x$ and $\sin 2x = 2\sin x\cos x $ to obtain the required value.

Complete step-by-step answer:
We are given an inverse hyperbolic function.
We know that ${\sinh ^{ - 1}}x = \log \left( {x + \sqrt {1 + {x^2}} } \right)$
Here we are asked to find ${\sinh ^{ - 1}}(\cot x)$
We have cot x in the place of x
So by applying the formula above we get
\[
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\cot x + \sqrt {1 + {{(\cot x)}^2}} } \right) \\
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\cot x + \sqrt {1 + ({{\cot }^2}x)} } \right) \\
\]
By the identity $\cos e{c^2}x - {\cot ^2}x = 1$
We get, $\cos e{c^2}x = 1 + {\cot ^2}x$
Therefore ,
\[
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\cot x + \sqrt {\cos e{c^2}x} } \right) \\
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\cot x + \cos ecx} \right) \\
\]
And here we know that $\cot x = \frac{{\cos x}}{{\sin x}}$and$\cos ecx = \frac{1}{{\sin x}}$
Substituting in the above equation, we get
\[
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\frac{{\cos x}}{{\sin x}} + \frac{1}{{\sin x}}} \right) \\
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\frac{{\cos x + 1}}{{\sin x}}} \right) \\
\]
Now let's use the identity.$2{\cos ^2}x = 1 + \cos 2x$.and $\sin 2x = 2\sin x\cos x$in the above equation
\[
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\frac{{2{{\cos }^2}\frac{x}{2}}}{{2\sin \frac{x}{2}\cos \frac{x}{2}}}} \right) \\
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\frac{{\cos \frac{x}{2}}}{{\sin \frac{x}{2}}}} \right) \\
   \Rightarrow {\sinh ^{ - 1}}(\cot x) = \log \left( {\cot \frac{x}{2}} \right) \\
\]
Therefore the correct option is A.

Note: Hyperbolic functions also satisfy identities analogous to those of the ordinary trigonometric functions and have important physical applications. For example, the hyperbolic cosine function may be used to describe the shape of the curve formed by a high-voltage line suspended between two towers
In mathematics, the inverse hyperbolic functions are the inverse functions of the hyperbolic functions.
For a given value of a hyperbolic function, the corresponding inverse hyperbolic function provides the corresponding hyperbolic angle