in the right triangle ABC , right angle at B , the ratio of AB to AC is 1 : √2 find the values of tanA1+tan2A,and
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Solution
AB : AC = 1 : √2⇒ABAC=1√2 ∴ AB = x and AC = √2x, for some x. By Pythagoras theorem, we have AC2=AB2+BC2 ⇒(√2x)2=x2+BC2 ⇒BC2=2x2−x2=x2 ⇒BC=x ∴tanA=BCAB=xx=1 we have, tanA1+tan2A=11+12=12