Question

\(\operatorname{Tanh}^{-1}(\sin\theta) =\)

• A. \(\operatorname{Sinh}^{-1}(\operatorname{cosec}\theta) \)
• B. \(\operatorname{Sinh}^{-1}(\sec\theta) \)
• C. \(\operatorname{Cosh}^{-1}(\operatorname{cosec}\theta) \)
• D. \(\operatorname{Cosh}^{-1}(\sec\theta) \)

Hint:

To solve problems involving inverse hyperbolic functions, it's crucial to remember their logarithmic forms. \(\operatorname{Tanh}^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\) \(\operatorname{Sinh}^{-1}(x) = \ln(x + \sqrt{x^2+1})\) \(\operatorname{Cosh}^{-1}(x) = \ln(x + \sqrt{x^2-1})\) Often, algebraic manipulation using trigonometric identities is required to transform one form into another.

Answer 1

The Correct Option is D

Step 1: Recall the logarithmic definition of \( \operatorname{Tanh}^{-1}(x) \).
The definition of the inverse hyperbolic tangent function is: \[ \operatorname{Tanh}^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right), \quad \text{for } |x|<1 \] In this problem, \( x = \sin\theta \). So we must have \( |\sin\theta| < 1 \), which implies \( \theta \neq \pm \frac{\pi}{2} + k\pi \) for any integer \( k \). Substitute \( x = \sin\theta \) into the definition: \[ \operatorname{Tanh}^{-1}(\sin\theta) = \frac{1}{2} \ln\left(\frac{1+\sin\theta}{1-\sin\theta}\right) \] Step 2: Manipulate the expression inside the logarithm.
To simplify the fraction inside the logarithm, multiply the numerator and the denominator by \( (1+\sin\theta) \): \[ \operatorname{Tanh}^{-1}(\sin\theta) = \frac{1}{2} \ln\left(\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\right) \] \[ = \frac{1}{2} \ln\left(\frac{(1+\sin\theta)^2}{1^2 - \sin^2\theta}\right) \] Recall the trigonometric identity \( 1 - \sin^2\theta = \cos^2\theta \): \[ = \frac{1}{2} \ln\left(\frac{(1+\sin\theta)^2}{\cos^2\theta}\right) \] This can be written as: \[ = \frac{1}{2} \ln\left(\left(\frac{1+\sin\theta}{\cos\theta}\right)^2\right) \] Using the logarithm property \( n \ln A = \ln A^n \): \[ = \ln\left(\frac{1+\sin\theta}{\cos\theta}\right) \] Separate the terms in the fraction: \[ = \ln\left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right) \] Recall that \( \frac{1}{\cos\theta} = \sec\theta \) and \( \frac{\sin\theta}{\cos\theta} = \tan\theta \): \[ \operatorname{Tanh}^{-1}(\sin\theta) = \ln(\sec\theta + \tan\theta) \] Step 3: Compare the result with the logarithmic definitions of other inverse hyperbolic functions.
Let's consider the relevant inverse hyperbolic functions from the options:
\( \operatorname{Sinh}^{-1}(x) = \ln(x + \sqrt{x^2+1}) \) \( \operatorname{Cosh}^{-1}(x) = \ln(x + \sqrt{x^2-1}) \), for \( x \ge 1 \). 
Let's test option (4), which is \( \operatorname{Cosh}^{-1}(\sec\theta) \).
Using the definition of \( \operatorname{Cosh}^{-1}(x) \) with \( x = \sec\theta \): \[ \operatorname{Cosh}^{-1}(\sec\theta) = \ln(\sec\theta + \sqrt{\sec^2\theta-1}) \] Recall the trigonometric identity \( \sec^2\theta - 1 = \tan^2\theta \): \[ \operatorname{Cosh}^{-1}(\sec\theta) = \ln(\sec\theta + \sqrt{\tan^2\theta}) \] \[ = \ln(\sec\theta + |\tan\theta|) \] For the equality to hold (i.e., \( \ln(\sec\theta + \tan\theta) \)), we must assume that \( \tan\theta \ge 0 \). 
This assumption is typically made in such problems where a single-valued answer is expected, and it aligns with the domain requirements for \( \operatorname{Cosh}^{-1}(\sec\theta) \) (where \( \sec\theta \ge 1 \), implying \( \cos\theta > 0 \), which in turn implies \( \tan\theta \ge 0 \) in the principal value ranges). 
Thus, \( \operatorname{Tanh}^{-1}(\sin\theta) = \ln(\sec\theta + \tan\theta) \) matches \( \operatorname{Cosh}^{-1}(\sec\theta) \). 
The final answer is \( \boxed{\operatorname{Cosh}^{-1}(\sec\theta)} \).