If for \theta \in[-\frac\pi3, 0] , the points (x, y)=(3 \tan (\theta+\frac\pi3), 2 \tan (\theta+\frac\pi6)) lie on x y+\alpha x+\beta y+\gamma=0 , then \alpha^2+\beta^2+\gamma^2 is equal to :

Trigonometric Ratio and Identites PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q35 with Solution

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Correct answer: A

Parametric point given

We are given $x=3\tan\left(\theta+\frac{\pi}{3}\right),\qquad y=2\tan\left(\theta+\frac{\pi}{6}\right)$ and these points satisfy $xy+\alpha x+\beta y+\gamma=0.$

We must find $\alpha^2+\beta^2+\gamma^2$ .

Introduce a simpler parameter

Let $t=\tan\theta.$ Since $\theta\in\left[-\frac\pi3,0\right]$ , the expressions are well-defined except at the endpoint where tangent blows up, but the identity relation we derive remains valid.

Now use tangent addition formulas:

$\tan\left(\theta+\frac{\pi}{3}\right)=\frac{\tan\theta+\sqrt3}{1-\sqrt3\tan\theta}=\frac{t+\sqrt3}{1-\sqrt3 t},$

$\tan\left(\theta+\frac{\pi}{6}\right)=\frac{\tan\theta+\frac1{\sqrt3}}{1-\frac{t}{\sqrt3}}=\frac{\sqrt3 t+1}{\sqrt3-t}.$

Hence $x=3\cdot \frac{t+\sqrt3}{1-\sqrt3 t},$ $y=2\cdot \frac{\sqrt3 t+1}{\sqrt3-t}.$

Express $t$ in terms of $x$

From $x=3\cdot \frac{t+\sqrt3}{1-\sqrt3 t},$ we get $x(1-\sqrt3 t)=3(t+\sqrt3).$ So $x-x\sqrt3 t=3t+3\sqrt3,$ $-(x\sqrt3+3)t=3\sqrt3-x,$ $t=\frac{x-3\sqrt3}{x\sqrt3+3}.$

Substitute into the expression for $y$

We have $y=2\cdot \frac{\sqrt3 t+1}{\sqrt3-t}.$

First compute:

=\frac{\sqrt3 x-9}{x\sqrt3+3}+1 =\frac{\sqrt3 x-9+x\sqrt3+3}{x\sqrt3+3} =\frac{2\sqrt3 x-6}{x\sqrt3+3} =\frac{2(\sqrt3 x-3)}{x\sqrt3+3}.$$ Also, $$\sqrt3-t=\sqrt3-\frac{x-3\sqrt3}{x\sqrt3+3} =\frac{\sqrt3(x\sqrt3+3)-(x-3\sqrt3)}{x\sqrt3+3} =\frac{3x+3\sqrt3-x+3\sqrt3}{x\sqrt3+3} =\frac{2x+6\sqrt3}{x\sqrt3+3} =\frac{2(x+3\sqrt3)}{x\sqrt3+3}.$$ Therefore $$y=2\cdot \frac{\frac{2(\sqrt3 x-3)}{x\sqrt3+3}}{\frac{2(x+3\sqrt3)}{x\sqrt3+3}} =2\cdot \frac{\sqrt3 x-3}{x+3\sqrt3}.$$ So $$y(x+3\sqrt3)=2(\sqrt3 x-3).$$ Expanding, $$xy+3\sqrt3 y=2\sqrt3 x-6,$$ $$xy-2\sqrt3 x+3\sqrt3 y+6=0.$$ Comparing with $$xy+\alpha x+\beta y+\gamma=0,$$ we get $$\alpha=-2\sqrt3,\qquad \beta=3\sqrt3,\qquad \gamma=6.$$ --- 5. **Compute the required value** $$\alpha^2+\beta^2+\gamma^2=(-2\sqrt3)^2+(3\sqrt3)^2+6^2$$ $$=12+27+36=75.$$ --- 6. **Check with options** The value is $$\boxed{75}$$ which corresponds to **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So they agree.

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View written solutionFree

Correct answer: A

Parametric point given

We are given $x=3\tan\left(\theta+\frac{\pi}{3}\right),\qquad y=2\tan\left(\theta+\frac{\pi}{6}\right)$ and these points satisfy $xy+\alpha x+\beta y+\gamma=0.$

We must find $\alpha^2+\beta^2+\gamma^2$ .

Introduce a simpler parameter

Let $t=\tan\theta.$ Since $\theta\in\left[-\frac\pi3,0\right]$ , the expressions are well-defined except at the endpoint where tangent blows up, but the identity relation we derive remains valid.

Now use tangent addition formulas:

$\tan\left(\theta+\frac{\pi}{3}\right)=\frac{\tan\theta+\sqrt3}{1-\sqrt3\tan\theta}=\frac{t+\sqrt3}{1-\sqrt3 t},$

$\tan\left(\theta+\frac{\pi}{6}\right)=\frac{\tan\theta+\frac1{\sqrt3}}{1-\frac{t}{\sqrt3}}=\frac{\sqrt3 t+1}{\sqrt3-t}.$

Hence $x=3\cdot \frac{t+\sqrt3}{1-\sqrt3 t},$ $y=2\cdot \frac{\sqrt3 t+1}{\sqrt3-t}.$

Express $t$ in terms of $x$

From $x=3\cdot \frac{t+\sqrt3}{1-\sqrt3 t},$ we get $x(1-\sqrt3 t)=3(t+\sqrt3).$ So $x-x\sqrt3 t=3t+3\sqrt3,$ $-(x\sqrt3+3)t=3\sqrt3-x,$ $t=\frac{x-3\sqrt3}{x\sqrt3+3}.$

Substitute into the expression for $y$

We have $y=2\cdot \frac{\sqrt3 t+1}{\sqrt3-t}.$

First compute:

=\frac{\sqrt3 x-9}{x\sqrt3+3}+1 =\frac{\sqrt3 x-9+x\sqrt3+3}{x\sqrt3+3} =\frac{2\sqrt3 x-6}{x\sqrt3+3} =\frac{2(\sqrt3 x-3)}{x\sqrt3+3}.$$ Also, $$\sqrt3-t=\sqrt3-\frac{x-3\sqrt3}{x\sqrt3+3} =\frac{\sqrt3(x\sqrt3+3)-(x-3\sqrt3)}{x\sqrt3+3} =\frac{3x+3\sqrt3-x+3\sqrt3}{x\sqrt3+3} =\frac{2x+6\sqrt3}{x\sqrt3+3} =\frac{2(x+3\sqrt3)}{x\sqrt3+3}.$$ Therefore $$y=2\cdot \frac{\frac{2(\sqrt3 x-3)}{x\sqrt3+3}}{\frac{2(x+3\sqrt3)}{x\sqrt3+3}} =2\cdot \frac{\sqrt3 x-3}{x+3\sqrt3}.$$ So $$y(x+3\sqrt3)=2(\sqrt3 x-3).$$ Expanding, $$xy+3\sqrt3 y=2\sqrt3 x-6,$$ $$xy-2\sqrt3 x+3\sqrt3 y+6=0.$$ Comparing with $$xy+\alpha x+\beta y+\gamma=0,$$ we get $$\alpha=-2\sqrt3,\qquad \beta=3\sqrt3,\qquad \gamma=6.$$ --- 5. **Compute the required value** $$\alpha^2+\beta^2+\gamma^2=(-2\sqrt3)^2+(3\sqrt3)^2+6^2$$ $$=12+27+36=75.$$ --- 6. **Check with options** The value is $$\boxed{75}$$ which corresponds to **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So they agree.

Trigonometric Ratio and Identites question

Trigonometric Ratio and Identites question

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Trigonometric Ratio and Identites question

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