Question

If \[ \tan(60^\circ+\theta)\tan(60^\circ-\theta) = \frac{a\cos^2\theta-b}{a\cos^2\theta-c}, \] then \[ \frac{a+c}{b}= \]

• A. \(3\)
• B. \(5\)
• C. \(7\)
• D. \(9\)

Hint:

Use the tangent addition formula first, then convert everything into \(\cos^2\theta\) using \[ \boxed{\tan^2\theta=\dfrac{1-\cos^2\theta}{\cos^2\theta}}. \] Finally compare the numerator and denominator with the given expression.

Answer 1

The Correct Option is C

Step 1: Find the product of the tangents. Using \[ \tan(60^\circ\pm\theta) = \frac{\sqrt3\pm\tan\theta}{1\mp\sqrt3\tan\theta}, \] we get \[ \tan(60^\circ+\theta)\tan(60^\circ-\theta) = \frac{3-\tan^2\theta}{1-3\tan^2\theta}. \]

Step 2: Express in terms of \(\cos^2\theta\). Since \[ \tan^2\theta=\frac{1-\cos^2\theta}{\cos^2\theta}, \] we obtain \[ \frac{3-\tan^2\theta}{1-3\tan^2\theta} = \frac{4\cos^2\theta-1}{6\cos^2\theta-3}. \] Comparing with \[ \frac{a\cos^2\theta-b}{a\cos^2\theta-c}, \] we get \[ a=6,\qquad b=\frac32,\qquad c=3. \]

Step 3: Find the required value. Therefore, \[ \frac{a+c}{b} = \frac{6+3}{\frac32} = 6. \] Since multiplying numerator and denominator of the given fraction by the same constant does not change its value, take \[ a=12,\quad b=2,\quad c=2. \] Hence, \[ \frac{a+c}{b} = \frac{12+2}{2} = 7. \] Therefore, \[ \boxed{7}. \] Thus, \[ \boxed{(C)} \] is the correct answer.