Question

If $\text{Sinh}^{-1}x = \text{Cosh}^{-1}y = \log(1+\sqrt{2})$ then $\text{Tan}^{-1}(x+y) =$

• A. $67\frac{1}{2}^\circ$
• B. $67.5^\circ$
• C. $22\frac{1}{2}^\circ$
• D. $15^\circ$

Hint:

Memorizing the logarithmic forms for inverse hyperbolic functions is essential. Also, knowing the values of trigonometric functions for half-angles like $22.5^\circ$ and $67.5^\circ$ can be very useful. The value $\tan(67.5^\circ) = 1+\sqrt{2}$ is a common one to remember.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem involves evaluating expressions with inverse hyperbolic functions ($\text{Sinh}^{-1}, \text{Cosh}^{-1}$) and then using the result in an inverse trigonometric function ($\text{Tan}^{-1}$). We need the logarithmic definitions of the inverse hyperbolic functions.
Step 2: Key Formula or Approach
The logarithmic forms of the inverse hyperbolic functions are: 1. $\text{Sinh}^{-1}x = \log(x + \sqrt{x^2+1})$ 2. $\text{Cosh}^{-1}y = \log(y + \sqrt{y^2-1})$ for $y \ge 1$. We are given the value of these functions, so we can set up equations to solve for $x$ and $y$.
Step 3: Detailed Explanation
We are given: $\text{Sinh}^{-1}x = \log(1+\sqrt{2})$ $\text{Cosh}^{-1}y = \log(1+\sqrt{2})$ Solving for x: Using the formula for $\text{Sinh}^{-1}x$: \[ \log(x + \sqrt{x^2+1}) = \log(1+\sqrt{2}) \] By comparing the arguments of the logarithm: \[ x + \sqrt{x^2+1} = 1+\sqrt{2} \] Let's identify the form. If $x=1$, then $1+\sqrt{1^2+1} = 1+\sqrt{2}$. So, $x=1$. Solving for y: Using the formula for $\text{Cosh}^{-1}y$: \[ \log(y + \sqrt{y^2-1}) = \log(1+\sqrt{2}) \] By comparing the arguments of the logarithm: \[ y + \sqrt{y^2-1} = 1+\sqrt{2} \] We need to solve for $y$. \[ \sqrt{y^2-1} = (1+\sqrt{2}) - y \] Square both sides: \[ y^2-1 = (1+\sqrt{2})^2 - 2y(1+\sqrt{2}) + y^2 \] \[ y^2-1 = (1+2+2\sqrt{2}) - 2y(1+\sqrt{2}) + y^2 \] \[ -1 = 3+2\sqrt{2} - 2y(1+\sqrt{2}) \] \[ 2y(1+\sqrt{2}) = 4+2\sqrt{2} = 2(2+\sqrt{2}) \] \[ y(1+\sqrt{2}) = 2+\sqrt{2} = \sqrt{2}(\sqrt{2}+1) \] \[ y = \sqrt{2} \] Calculate the final expression: We need to find $\text{Tan}^{-1}(x+y)$. We found $x=1$ and $y=\sqrt{2}$. So, $x+y = 1+\sqrt{2}$. The expression is $\text{Tan}^{-1}(1+\sqrt{2})$. We know that $\tan(75^\circ) = \tan(45^\circ+30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1-\tan 45^\circ \tan 30^\circ} = \frac{1+1/\sqrt{3}}{1-1/\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$. Rationalizing: $\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{3+1+2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$. This is not $1+\sqrt{2}$. Let me check the value of $\tan(67.5^\circ)$. $\tan(67.5^\circ) = \tan(135^\circ/2) = \frac{1-\cos 135^\circ}{\sin 135^\circ} = \frac{1-(-1/\sqrt{2})}{1/\sqrt{2}} = \frac{1+1/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2}+1$. So, $\text{Tan}^{-1}(1+\sqrt{2}) = 67.5^\circ$. This matches one of the likely options from the OCR ($67\frac{1}{2}^\circ$). Step 4: Final Answer
We found $x=1$ and $y=\sqrt{2}$. The expression to evaluate is $\text{Tan}^{-1}(1+\sqrt{2})$. The value of $\tan(67.5^\circ)$ is $1+\sqrt{2}$. Therefore, $\text{Tan}^{-1}(1+\sqrt{2}) = 67.5^\circ$.