<p>If \(\frac{1+\sin 2x}{1-\sin 2x} = \tan^2(a+x)\) for all \(x\) then the numerically smallest value of \(a\) is</p>
Step-by-Step Solution
Key Concept: Recognize that 1+sin2x = (sinx+cosx)² and 1-sin2x = (sinx-cosx)², then simplify the fraction to get tan²(a+x) by matching with the tangent addition formula structure.
<p><strong>Step 1:</strong> Simplify the left side using sum-of-squares identities.</p><p>1 + sin2x = sin²x + cos²x + 2sinx·cosx = (sinx + cosx)²</p><p>1 - sin2x = sin²x + cos²x - 2sinx·cosx = (sinx - cosx)²</p><p><strong>Step 2:</strong> Form the ratio.</p><p>$$\frac{1+\sin 2x}{1-\sin 2x} = \frac{(\sin x + \cos x)^2}{(\sin x - \cos x)^2} = \left(\frac{\sin x + \cos x}{\sin x - \cos x}\right)^2$$</p><p><strong>Step 3:</strong> Simplify the inner fraction by dividing numerator and denominator by cosx.</p><p>$$\frac{\sin x + \cos x}{\sin x - \cos x} = \frac{\tan x + 1}{\tan x - 1}$$</p><p><strong>Step 4:</strong> Apply tangent addition formula. Recognize that:</p><p>$$\frac{\tan x + 1}{\tan x - 1} = \frac{\tan x + \tan\frac{\pi}{4}}{1 - \tan x \tan\frac{\pi}{4}} = \tan\left(x + \frac{\pi}{4}\right)$$</p><p><strong>Step 5:</strong> Therefore:</p><p>$$\tan^2\left(x + \frac{\pi}{4}\right) = \tan^2(a+x)$$</p><p><strong>Step 6:</strong> This means: $a + x = x + \frac{\pi}{4} + n\pi$ for integer n, giving $a = \frac{\pi}{4} + n\pi$</p><p><strong>Step 7:</strong> For numerically smallest value, use n = 0 and n = -1:</p><p>a = π/4 ≈ 0.785 (n=0) or a = -3π/4 ≈ -2.356 (n=-1)</p><p>The smallest in magnitude is $a = \frac{\pi}{4}$</p><p>∴ Answer: A</p>
Correct Answer: A