1.

If tan A = \(\frac{a}{a+1}\) and tan B = \(\frac{1}{2a+1}\),then the value of A + B isA. 0 B. \(\frac{π}2\)C.\(\frac{π}3\)D.\(\frac{π}4\)

Answer»

put a = 0 

tan(A) = 0 

tan(B) =1

tan A + B = \(\frac{tanA+tanB}{1-tanAtanB}\)

tan A + B =  \(\frac{0+1}{1-0}\)

tan A + B = 1

tan A + B = tan-11

tan A + B = \(\frac{π}4\)



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