Let \(\mathrm{S}=\left[-\pi, \frac{\pi}{2}\right)-\left\{-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}\). Then the number of elements in the set \(A=\{\theta \in S: \tan \theta(1+\sqrt{5} \tan (2 \theta))=\sqrt{5}-\tan (2 \theta)\}\) is

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Trigonometric Equations

Let

S = [- π, \frac{π}{2}) - {- \frac{π}{2}, - \frac{π}{4}, - \frac{3 π}{4}, \frac{π}{4}}

. Then the number of elements in the set

A = {θ \in S : tan θ (1 + 5 tan (2 θ)) = 5 - tan (2 θ)}

Answer: 5

Rearranging the given equation:

tan θ + 5 tan (2 θ) tan θ = 5 - tan (2 θ)

. This simplifies to

tan θ + tan (2 θ) = 5 (1 - tan θ tan (2 θ))

. Using the tangent addition formula, this becomes

tan (3 θ) = 5

. The general solution is

3 θ = nπ + α

, where

tan α = 5

, giving

θ = \frac{nπ}{3} + \frac{α}{3}

. For

θ \in [- π, \frac{π}{2})

excluding the specified points, we find exactly 5 valid values of

θ

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer