If A - 0- B - 0- A - B -3 and y -tan AtanB thenA- 1-31-3D- 1 - y - 2

The correct option is B y = tan A.tan(π/3 A) nbsp; ⇒ tan^2 A + √(3)(y 1) tan nbsp;A + y = 0 As tan A is real , D ≥ 0 ⇒ 3(y 1)^2 4y ≥ 0 ∴ nbsp;3y^2 10y ...

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Standard XII

Mathematics

Basic Trigonometric Identities

If A ≥ 0, B ≥...

Question

If A $\geq 0, B \geq 0, A + B = \frac{π}{3} a n d y = t a n A t a n B$ then

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$1\leq y\leq 2$

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Solution

The correct option is B

$y = t a n A . t a n (\frac{π}{3} - A)$

$\Rightarrow$ $t a n^{2} A + \sqrt{3} (y - 1) t a n A + y = 0$

As tan A is real , D $\geq 0 \Rightarrow 3 (y - 1)^{2} - 4 y \geq 0$

$∴$ $3 y^{2} - 10 y + 3 \geq 0 \Rightarrow y \leq \frac{1}{3} o r y \geq 3$

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If A ≥0,B≥0,A+B=π3 and y=tanAtanB then

The correct option is B y=tanA.tan(π3−A) ⇒ tan2A+√3(y−1)tanA+y=0 As tan A is real , D ≥0⇒3(y−1)2−4y≥0 ∴ 3y2−10y+3≥0⇒y≤13 or y≥3

If A $\geq 0, B \geq 0, A + B = \frac{π}{3} a n d y = t a n A t a n B$ then

The correct option is B

$y = t a n A . t a n (\frac{π}{3} - A)$

$\Rightarrow$ $t a n^{2} A + \sqrt{3} (y - 1) t a n A + y = 0$

As tan A is real , D $\geq 0 \Rightarrow 3 (y - 1)^{2} - 4 y \geq 0$

$∴$ $3 y^{2} - 10 y + 3 \geq 0 \Rightarrow y \leq \frac{1}{3} o r y \geq 3$