Question

If \[ u=\log\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right), \] then \[ \tanh\frac{u}{2} = \]

• A. \[ \tan\frac{\theta}{2} \]
• B. \[ \cot\frac{\theta}{2} \]
• C. \[ \sec\frac{\theta}{2} \]
• D. \[ \sin\frac{\theta}{2} \]

Hint:

Remember: $\tanh\frac{u}{2}=\frac{e^u-1}{e^u+1}$.

Answer 1

The Correct Option is A

Step 1: Switch from log to exponential.
We are given $u=\log\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right).$ Taking the exponential of both sides, $e^u=\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right).$
Step 2: Recall the hyperbolic tangent.
By definition $\tanh\dfrac{u}{2}=\dfrac{e^{u/2}-e^{-u/2}}{e^{u/2}+e^{-u/2}}$, which can also be written as $\dfrac{e^u-1}{e^u+1}.$
Step 3: Put in $e^u$.
So $\tanh\dfrac{u}{2}=\dfrac{\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)-1}{\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)+1}.$
Step 4: Expand the tangent of a sum.
Let $t=\tan\dfrac{\theta}{2}.$ Then $\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)=\dfrac{1+t}{1-t}$ because $\tan\dfrac{\pi}{4}=1.$
Step 5: Substitute and simplify.
Top: $\dfrac{1+t}{1-t}-1=\dfrac{2t}{1-t}.$ Bottom: $\dfrac{1+t}{1-t}+1=\dfrac{2}{1-t}.$ Dividing, the $\dfrac{2}{1-t}$ parts cancel and leave $t.$
Step 6: State the result.
So $\tanh\dfrac{u}{2}=t=\tan\dfrac{\theta}{2}.$ \[ \boxed{\tan\dfrac{\theta}{2}} \]