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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.\(\frac{π}4\) |
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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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