JEE Main 2026 — Mathematics Trigonometry

Consider the general term $\dfrac{\sin x}{\cos 3x}$cos3xsinx​.

We can rewrite this by multiplying and dividing by $2\cos x$2cosx:

$$\dfrac{2\sin x \cos x}{2\cos 3x \cos x} = \dfrac{\sin 2x}{2\cos 3x \cos x}$$2cos3xcosx2sinxcosx​=2cos3xcosxsin2x​

Using $\sin 2x = \sin(3x - x) = \sin 3x \cos x - \cos 3x \sin x$sin2x=sin(3x−x)=sin3xcosx−cos3xsinx, we get:

$$\dfrac{\sin 3x \cos x - \cos 3x \sin x}{2\cos 3x \cos x} = \dfrac{1}{2} \left( \dfrac{\sin 3x}{\cos 3x} - \dfrac{\sin x}{\cos x} \right) = \dfrac{1}{2} (\tan 3x - \tan x)$$2cos3xcosxsin3xcosx−cos3xsinx​=21​(cos3xsin3x​−cosxsinx​)=21​(tan3x−tanx)

Applying this identity to each term of $A$A:

$$\dfrac{\sin 3^\circ}{\cos 9^\circ} = \dfrac{1}{2} (\tan 9^\circ - \tan 3^\circ)$$cos9∘sin3∘​=21​(tan9∘−tan3∘)

$$\dfrac{\sin 9^\circ}{\cos 27^\circ} = \dfrac{1}{2} (\tan 27^\circ - \tan 9^\circ)$$cos27∘sin9∘​=21​(tan27∘−tan9∘)

$$\dfrac{\sin 27^\circ}{\cos 81^\circ} = \dfrac{1}{2} (\tan 81^\circ - \tan 27^\circ)$$cos81∘sin27∘​=21​(tan81∘−tan27∘)

Adding these terms, the series telescopes:

$$A = \dfrac{1}{2} (\tan 9^\circ - \tan 3^\circ + \tan 27^\circ - \tan 9^\circ + \tan 81^\circ - \tan 27^\circ)$$A=21​(tan9∘−tan3∘+tan27∘−tan9∘+tan81∘−tan27∘)

$$A = \dfrac{1}{2} (\tan 81^\circ - \tan 3^\circ)$$A=21​(tan81∘−tan3∘)

We are given $B = \tan 81^\circ - \tan 3^\circ$B=tan81∘−tan3∘.

Substituting $B$B into the equation for $A$A:

$$A = \dfrac{1}{2} B$$A=21​B

$$\Rightarrow \dfrac{B}{A} = 2$$⇒AB​=2

Answer: $2$2