Question

For $n\in\mathbb Z$, a set of values of $\theta$ satisfying \[ \sec\theta-1=(\sqrt2-1)\tan\theta \] is

• A. \[ 2n\pi-\frac{\pi}{4} \]
• B. \[ 2n\pi+\frac{\pi}{2} \]
• C. \[ (2n+1)\pi+\frac{\pi}{4} \]
• D. \[ 2n\pi+\frac{\pi}{4} \]

Hint:

Whenever $1-\cos\theta$ appears, convert it into a half-angle expression.

Answer 1

The Correct Option is D

Step 1: Change to sine and cosine.
Write $\sec\theta=\dfrac1{\cos\theta}$ and $\tan\theta=\dfrac{\sin\theta}{\cos\theta}.$ The equation $\sec\theta-1=(\sqrt2-1)\tan\theta$ becomes \[ \frac{1-\cos\theta}{\cos\theta}=(\sqrt2-1)\frac{\sin\theta}{\cos\theta}. \]
Step 2: Cancel the common bottom.
Multiplying both sides by $\cos\theta$ (and noting $\cos\theta\ne0$): \[ 1-\cos\theta=(\sqrt2-1)\sin\theta. \]
Step 3: Use half-angle forms.
Replace $1-\cos\theta=2\sin^2\dfrac{\theta}{2}$ and $\sin\theta=2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}.$
Step 4: Cancel $2\sin\dfrac{\theta}{2}$.
Dividing both sides by $2\sin\dfrac{\theta}{2}$ (for the non-trivial family): \[ \sin\frac{\theta}{2}=(\sqrt2-1)\cos\frac{\theta}{2}. \]
Step 5: Make a tangent.
Divide by $\cos\dfrac{\theta}{2}$: $\tan\dfrac{\theta}{2}=\sqrt2-1.$ Since $\tan\dfrac{\pi}{8}=\sqrt2-1$, we get $\dfrac{\theta}{2}=n\pi+\dfrac{\pi}{8}.$
Step 6: Solve for $\theta$.
Multiply by $2$: $\theta=2n\pi+\dfrac{\pi}{4}.$ \[ \boxed{\theta=2n\pi+\dfrac{\pi}{4}} \]