Question : If $A=30^{\circ}$, then find the value of $\frac{(2 \tan A)}{\left(1-\tan^2 A\right)}$.
Option 1: $4 \sqrt{3}$
Option 2: $\frac{3}{\sqrt{3}}$
Option 3: $3$
Option 4: $2 \sqrt{3}$
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Correct Answer: $\frac{3}{\sqrt{3}}$
Solution : Given: $A=30^{\circ}$ And we know, $\tan 30° = \frac{1}{\sqrt3}$ So, $\frac{(2 \tan A)}{\left(1-\tan ^2 A\right)}$ $= \frac{2\tan 30°}{1-\tan^2 30°}$ $= \frac{2\times \frac{1}{\sqrt3}}{1-\frac{1}{3}} = \frac{2\times 3}{2\times \sqrt3} = \frac{3}{\sqrt 3}$ Hence, the correct answer is $\frac{3}{\sqrt3}$.
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Question : If $A=60^{\circ}$ and $B=30^{\circ}$, find the value of $\frac{(\tan A-\tan B)}{(1+\tan A \tan B)}$.
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Question : Find the value of $\cos 0^{\circ}+\cos 30^{\circ}-\tan 45^{\circ}+\operatorname{cosec} 60^{\circ}+\cot 90^{\circ}$.
Question : If $\frac{x-x\tan^{2}30^{\circ}}{1+\tan^{2}30^{\circ}}=\sin^{2}30^{\circ}+4\cot^{2}45^{\circ}-\sec^{2}60^{\circ}$, then value of $x$ is:
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