tan 2 alpha cdot tan left(30^{circ}-alpharigh | vedclass

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(C) Let $x = 30^{\circ}-\alpha$ and $y = 60^{\circ}-\alpha$. Then $x+y = 90^{\circ}-2\alpha$. We know that $	an(x+y) = 	an(90^{\circ}-2\alpha) = \co

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(C) Let $x = 30^{\circ}-\alpha$ and $y = 60^{\circ}-\alpha$. Then $x+y = 90^{\circ}-2\alpha$. We know that $	an(x+y) = 	an(90^{\circ}-2\alpha) = \cot 2\alpha = \frac{1}{	an 2\alpha}$. Using the formula $	an(x+y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we have: $\frac{	an x + 	an y}{1 - 	an x 	an y} = \frac{1}{	an 2\alpha}$. Cross-multiplying gives: $	an 2\alpha (	an x + 	an y) = 1 - 	an x 	an y$. Rearranging the terms: $	an 2\alpha 	an x + 	an 2\alpha 	an y + 	an x 	an y = 1$. Substituting back $x = 30^{\circ}-\alpha$ and $y = 60^{\circ}-\alpha$,we get: $	an 2\alpha 	an(30^{\circ}-\alpha) + 	an 2\alpha 	an(60^{\circ}-\alpha) + 	an(60^{\circ}-\alpha) 	an(30^{\circ}-\alpha) = 1$.

$\tan 2 \alpha \cdot \tan \left(30^{\circ}-\alpha\right)+\tan 2 \alpha \cdot \tan \left(60^{\circ}-\alpha\right)+\tan \left(60^{\circ}-\alpha\right) \cdot \tan \left(30^{\circ}-\alpha\right)$ is equal to

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