Question

If $1+\sin^2(2A)=3\sin A\cos A$, then what are the possible values of $\tan A$?

• A. $1/4,\,2$
• B. $1/6,\,3$
• C. $1/2,\,1$
• D. $1/8,\,4$

Hint:

When given a trigonometric identity with $\sin A$ and $\cos A$, divide by $\cos^2 A$ to convert into a quadratic in $\tan A$.

Answer 1

The Correct Option is C

We are given \[ 1+\sin^2A=3\sin A\cos A. \] Divide both sides by $\cos^2 A$ (valid for $\cos A\neq0$): \[ \frac{1}{\cos^2A}+\tan^2A=3\tan A. \] But $\frac{1}{\cos^2A}=1+\tan^2A$. So \[ 1+\tan^2A+\tan^2A=3\tan A \quad\Rightarrow\quad 1+2\tan^2A=3\tan A. \] Hence quadratic: \[ 2\tan^2A-3\tan A+1=0. \] Solve: \[ \tan A=\frac{3\pm \sqrt{9-8}}{4}=\frac{3\pm1}{4}. \] So $\tan A=\tfrac12$ or $1$. \[ \boxed{\tfrac12,\,1} \] 

Answer 2

Instead of dividing through by \(\cos^2A\) and substituting \(\sec^2A=1+\tan^2A\), we can rewrite the equation as a homogeneous expression in \(\sin A,\cos A\) and factor it directly, then check each option.

1. Option A (\(\tfrac14,2\)): Replacing \(1\) with \(\sin^2A+\cos^2A\), the equation \(1+\sin^2A=3\sin A\cos A\) becomes \[ \sin^2A+\cos^2A+\sin^2A=3\sin A\cos A\;\Rightarrow\;2\sin^2A-3\sin A\cos A+\cos^2A=0. \] This factors as \((2\sin A-\cos A)(\sin A-\cos A)=0\), giving \(\tan A=\tfrac12\) or \(\tan A=1\), not \(\tfrac14\) or \(2\), so this option is incorrect.
2. Option B (\(\tfrac16,3\)): Neither value satisfies the factored equation above, so this option is incorrect.
3. Option C (\(\tfrac12,1\)): From \((2\sin A-\cos A)=0\) we get \(\tan A=\tfrac12\), and from \((\sin A-\cos A)=0\) we get \(\tan A=1\); both satisfy the equation exactly, matching this option.
4. Option D (\(\tfrac18,4\)): Neither value satisfies the factored equation, so this option is incorrect.

Factoring the homogeneous quadratic in \(\sin A,\cos A\) confirms \(\tan A=\tfrac12\) or \(1\).

Hence, the correct answer is option C: \(\dfrac12,\,1\).