\(\dfrac{ 1-tan^245°}{1+tan^245°} \) =

(A) tan 90°            

(B) 1                    

(C) sin 45°            

(D) 0

Asked by Pragya Singh | 1 year ago |  100

1 Answer

Solution :-

 Right answer is (D) 0 .

Substitute the of tan 45° in the given equation

tan 45° = 1

\( \dfrac{ 1-tan^245°}{1+tan^245°}=\dfrac{(1-1^2)}{(1+1^2)}\)

\(\dfrac{0}{2}\) = 0

The solution of the above equation is 0.

Answered by Abhisek | 1 year ago

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