Let's break down the steps shown in the image and simplify the expression.

The original question (labeled as F3) states: $\text{Prove that } \sinh^{-1}(\sin\theta) = \log\left(\cot\frac{\theta}{2}\right).$

Work done in the image:

1. It starts with: $\sinh^{-1}(\sin\theta) = \log\left[\sin\theta + \sqrt{\sin^2\theta + 1}\right]$

2. Then simplifies the square root expression: $\sqrt{\sin^2\theta + 1} = \cos\theta$ Since $\sin^2\theta + \cos^2\theta = 1$, this identity is used to reduce the square root.

3. This gives: $\log(\sin\theta + \cos\theta)$

4. The next steps rewrite $\sin\theta + \cos\theta$ in terms of $\csc\theta$ and $\sec\theta$: $\log\left(\csc\theta + \sec\theta\right)$

5. Finally, it simplifies into the form: $\log\left(\frac{\sec\theta + \csc\theta}{\sec\theta \cdot \csc\theta}\right)$

This final expression still needs further verification or transformation to match $\log\left(\cot\frac{\theta}{2}\right)$, though the general structure is almost there. One would need to relate this to the half-angle identity for cotangent.

Would you like to dive deeper into this derivation or have more detailed steps? Let me know!

Here are 5 related questions to expand your understanding:

1. What is the identity for $\sinh^{-1}(x)$ in terms of logarithms?
2. How does the identity $\cos^2\theta + \sin^2\theta = 1$ assist in simplifying expressions?
3. What is the half-angle identity for $\cot(\theta/2)$?
4. How can $\csc\theta$ and $\sec\theta$ be expressed in terms of sine and cosine?
5. What are the key steps to transform $\sinh^{-1}(\sin\theta)$ into a logarithmic expression?

Tip: Always look for trigonometric identities or hyperbolic identities when simplifying complex trigonometric and inverse expressions!