In $△ A B C$ ,
the value of $t a n 2 A + t a n 2 B + t a n 2 C =$ ?

\frac{s i n 2 A s i n 2 B s i n 2 C}{c o t 2 A c o t 2 B c o t 2 C}

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\frac{c o s 2 A c o s 2 B c o s 2 C}{s i n 2 A s i n 2 B s i n 2 C}

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\frac{c o s 2 A c o s 2 B c o s 2 C}{c o t 2 A c o t 2 B c o t 2 C}

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\frac{s i n 2 A s i n 2 B s i n 2 C}{c o s 2 A c o s 2 B c o s 2 C}

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Solution

The correct option is D $\frac{s i n 2 A s i n 2 B s i n 2 C}{c o s 2 A c o s 2 B c o s 2 C}$
$2 A + 2 B + 2 C = 180^{\circ}$

$\Rightarrow 2 A + 2 B = 180^{\circ} - 2 C$

$t a n (2 A + 2 B) = t a n (180^{\circ} - 2 C)$

$t a n (2 A + 2 B) = - t a n 2 C$

We know that

$t a n (2 A + 2 B) = \frac{2 t a n A + 2 t a n B}{1 - t a n 2 A t a n 2 B}$

So,

$\frac{t a n 2 A + t a n 2 B}{1 - t a n 2 A t a n 2 B} = - t a n 2 C$

Cross multiplication with each other, we get

$t a n 2 A + t a n 2 B = - t a n 2 C + t a n 2 A t a n 2 B t a n 2 C$

$t a n A + t a n 2 B + t a n 2 C = t a n 2 A t a n 2 B t a n 2 C = \frac{s i n 2 A s i n 2 B s i n 2 C}{c o s 2 A c o s 2 B c o s 2 C}$

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