Question

If $\sec^{2}\theta+\tan^{2}\theta=7, 0<\theta<\frac{\pi}{2}$, then $\tan 2\theta$ is equal to

• A. 1
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $-\sqrt{3}$
• E. 0

Hint:

Logic Tip: Alternatively, since $\tan\theta = \sqrt{3}$ in the first quadrant, $\theta$ must be exactly $60^{\circ}$ ($\frac{\pi}{3}$). Therefore, $\tan 2\theta = \tan(120^{\circ}) = -\sqrt{3}$. Finding the angle directly is often faster than using the formula.

Answer 1

The Correct Option is D

Concept:
We can solve for a single trigonometric function by utilizing the fundamental Pythagorean identity: $\sec^2\theta - \tan^2\theta = 1$ (or $\sec^2\theta = 1 + \tan^2\theta$). Then, we can use the double angle formula for tangent: $$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$
Step 1: Convert the equation to a single trigonometric variable.
Given: $\sec^2\theta + \tan^2\theta = 7$ Substitute $\sec^2\theta$ with $1 + \tan^2\theta$: $$(1 + \tan^2\theta) + \tan^2\theta = 7$$ $$1 + 2\tan^2\theta = 7$$
Step 2: Solve for $\tan\theta$.
$$2\tan^2\theta = 6$$ $$\tan^2\theta = 3$$ Taking the square root gives $\tan\theta = \pm\sqrt{3}$. Since $0<\theta<\frac{\pi}{2}$ (the first quadrant), all trigonometric ratios are positive. Thus: $$\tan\theta = \sqrt{3}$$
Step 3: Apply the double angle formula for tangent.
We need to find $\tan 2\theta$: $$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$ Substitute $\tan\theta = \sqrt{3}$ and $\tan^2\theta = 3$: $$\tan 2\theta = \frac{2(\sqrt{3})}{1 - 3}$$ $$\tan 2\theta = \frac{2\sqrt{3}}{-2}$$ $$\tan 2\theta = -\sqrt{3}$$