$$\dfrac{ 2\;tan\;30°}{1-tan^230°}$$ =

(A) cos 60°

(B) sin 60°

(C) tan 60°

(D) sin 30°

Asked by Pragya Singh | 1 year ago |  92

##### Solution :-

Right answer is  (C)  tan 60° is correct.

Substitute the of tan 30° in the given equation

tan 30° = $$\dfrac{1}{\sqrt{3}}$$

$$\dfrac{ 2\;tan\;30°}{1-tan^230°}=2( \dfrac{1}{\sqrt{3}})$$

$$\dfrac{ 2\;tan\;30°}{1-tan^230°}=1-( \dfrac{1}{\sqrt{3}})^2$$

=  $$1-\dfrac{1}{3}$$

$$\dfrac{ 2\;tan\;30°}{1-tan^230°}=( \dfrac{2}{\sqrt{3}})$$

$$\dfrac{ 2\;tan\;30°}{1-tan^230°}= \dfrac{2}{3}$$

The value of the given equation is equivalent to tan 60°.

Answered by Abhisek | 1 year ago

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