Question

If $\sinh^{-1}(x) = \log 3$ and $\cosh^{-1}(y) = \log\left(\frac{3}{2}\right)$, then $\tanh^{-1}(x - y) =$

• A. $\log\left(\frac{5}{\sqrt{3}}\right)$
• B. $\log\left(\frac{2}{\sqrt{3}}\right)$
• C. $\log\left(\frac{4}{3}\right)$
• D. $\frac{1}{2}\log\left(\frac{5}{3}\right)$

Hint:

Use $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$, $\cosh^{-1} y = \ln(y + \sqrt{y^2 - 1})$, and $\tanh^{-1} z = \frac{1}{2} \ln\left(\frac{1 + z}{1 - z}\right)$ to solve hyperbolic equations. Verify numerically if needed.

Answer 1

The Correct Option is D

We are given the following equations:

• $\sinh^{-1}(x) = \log 3$
• $\cosh^{-1}(y) = \log\left(\frac{3}{2}\right)$

We are tasked with finding $\tanh^{-1}(x - y)$.

Step 1: Express $x$ and $y$ in terms of hyperbolic functions.

We start by recalling the definitions of inverse hyperbolic functions:

• $\sinh^{-1}(x) = z$ means $x = \sinh(z)$.
• $\cosh^{-1}(y) = z$ means $y = \cosh(z)$.

From the given equation $\sinh^{-1}(x) = \log 3$, we can deduce:

$x = \sinh(\log 3)$

Using the identity for $\sinh(z)$, we have:

$\sinh(\log 3) = \frac{e^{\log 3} - e^{-\log 3}}{2} = \frac{3 - \frac{1}{3}}{2} = \frac{\frac{9}{3} - \frac{1}{3}}{2} = \frac{8}{6} = \frac{4}{3}$

Thus, $x = \frac{4}{3}$.

Next, from the equation $\cosh^{-1}(y) = \log\left(\frac{3}{2}\right)$, we can deduce:

$y = \cosh\left(\log\left(\frac{3}{2}\right)\right)$

Using the identity for $\cosh(z)$, we have:

$\cosh(\log a) = \frac{a + \frac{1}{a}}{2}$

Thus:

$y = \cosh\left(\log\left(\frac{3}{2}\right)\right) = \frac{\frac{3}{2} + \frac{2}{3}}{2} = \frac{\frac{9}{6} + \frac{4}{6}}{2} = \frac{\frac{13}{6}}{2} = \frac{13}{12}$

Therefore, $y = \frac{13}{12}$.

Step 2: Calculate $x - y$.

Now that we have $x = \frac{4}{3}$ and $y = \frac{13}{12}$, we compute:

$x - y = \frac{4}{3} - \frac{13}{12}$

To subtract these fractions, we first find a common denominator:

$x - y = \frac{16}{12} - \frac{13}{12} = \frac{3}{12} = \frac{1}{4}$

Thus, $x - y = \frac{1}{4}$.

Step 3: Compute $ \tanh^{-1}(x - y) $.

We are asked to find $\tanh^{-1}(x - y)$, which is $\tanh^{-1}\left(\frac{1}{4}\right)$. Recall that:

$\tanh^{-1}(z) = \frac{1}{2} \log\left(\frac{1 + z}{1 - z}\right)$

Applying this formula:

$\tanh^{-1}\left(\frac{1}{4}\right) = \frac{1}{2} \log\left(\frac{1 + \frac{1}{4}}{1 - \frac{1}{4}}\right) = \frac{1}{2} \log\left(\frac{\frac{5}{4}}{\frac{3}{4}}\right) = \frac{1}{2} \log\left(\frac{5}{3}\right)$

Final Answer:

The final answer is $\frac{1}{2} \log\left(\frac{5}{3}\right)$.