If for θ ∈ [-π/3, 0], the points (x, y) = (3 tan(θ + π/3), 2 tan(θ + π/6)) lie on xy + αx + βy + γ = 0, then α^2 + β^2 + γ^2 is equal to

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q1 – JEE-MAIN 2025

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Home Maths Trigonometric FunctionsQuestion #494

If for

θ \in [- \frac{π}{3}, 0]

, the points

(x, y) = (3 \tan (θ + \frac{π}{3}), 2 \tan (θ + \frac{π}{6}))

lie on

x y + α x + β y + γ = 0

, then

α^{2} + β^{2} + γ^{2}

is equal to

Clue 1 of 3: Trigonometric Substitution

Identify the relationship between the arguments of the tangent functions to simplify the expressions for $x$ and $y$ .

🧠 Full Solution Path

Step 1: Express

x

and

y

in terms of tangent functions and find their relationship✦ Active

Let $A = θ + \frac{π}{3}$ and $B = θ + \frac{π}{6}$ . From the given information, we have $x = 3 \tan A$ and $y = 2 \tan B$ . Observe the difference between the angles $A$ and $B$ :

A - B = (θ + \frac{π}{3}) - (θ + \frac{π}{6}) = \frac{π}{3} - \frac{π}{6} = \frac{π}{6}

Now, apply the tangent function to this difference:

\tan (A - B) = \tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}

Step 2: Apply the tangent subtraction formula and substitute

x

and

y

○ Expand

Using the tangent subtraction formula $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ , and substituting $\tan A = \frac{x}{3}$ and $\tan B = \frac{y}{2}$ :

\frac{\frac{x}{3} - \frac{y}{2}}{1 + \frac{x}{3} \cdot \frac{y}{2}} = \frac{1}{\sqrt{3}}

Simplify the expression:

\frac{\frac{2 x - 3 y}{6}}{1 + \frac{x y}{6}} = \frac{1}{\sqrt{3}} ⟹ \frac{2 x - 3 y}{6 + x y} = \frac{1}{\sqrt{3}}

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid sign errors.

Step 3: Rearrange the equation to find

α, β, γ

and calculate

α^{2} + β^{2} + γ^{2}

○ Expand

Cross-multiply and rearrange the equation to match the form $x y + α x + β y + γ = 0$ :

\sqrt{3} (2 x - 3 y) = 6 + x y

⟹ 2 \sqrt{3} x - 3 \sqrt{3} y = 6 + x y

⟹ x y - 2 \sqrt{3} x + 3 \sqrt{3} y + 6 = 0

Comparing this with the given equation $x y + α x + β y + γ = 0$ , we identify the coefficients:

α = - 2 \sqrt{3}

β = 3 \sqrt{3}

γ = 6

Finally, calculate $α^{2} + β^{2} + γ^{2}$ :

α^{2} = (- 2 \sqrt{3})^{2} = 4 \times 3 = 12

β^{2} = (3 \sqrt{3})^{2} = 9 \times 3 = 27

γ^{2} = 6^{2} = 36

α^{2} + β^{2} + γ^{2} = 12 + 27 + 36 = 75

💡 Teacher's Secret Hint

Double-check the squaring of terms involving square roots.

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Maths Question 1 – JEE-MAIN 2025

Clue 1 of 3: Trigonometric Substitution

🧠 Full Solution Path