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

(A) cos 60°         

(B) sin 60°             

(C) tan 60°          

(D) sin 30°

Asked by Pragya Singh | 1 year ago |  92

1 Answer

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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