Question

In a right triangle ABC, right angled at B, the ratio of AB to AC is 1:√2. Find the
values of
(i)2 tan A/1+tan²A
(i)2tanA/1-tan²A​

Answer 1

ANSWER:

1: AB : AC = 1 : 2⇒ACAB=21 

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we have

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2⇒(2x)2=x2+BC2

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2⇒(2x)2=x2+BC2⇒BC2=2x2−x2=x2

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2⇒(2x)2=x2+BC2⇒BC2=2x2−x2=x2⇒BC=x

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2⇒(2x)2=x2+BC2⇒BC2=2x2−x2=x2⇒BC=x∴tanA=ABBC=xx=1

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2⇒(2x)2=x2+BC2⇒BC2=2x2−x2=x2⇒BC=x∴tanA=ABBC=xx=1we have,

AB : AC = 1 : 2⇒ACAB=21 ∴ AB = x and AC = 2x, for some x.By Pythagoras theorem, we haveAC2=AB2+BC2⇒(2x)2=x2+BC2⇒BC2=2x2−x2=x2⇒BC=x∴tanA=ABBC=xx=1we have,1+tan2A2tanA=1+12×1=22, = 1

2:

AB:AC=1:2

AB:AC=1:2AB=kAC=2k

AB:AC=1:2AB=kAC=2kwhere k= constant of proportionality

AB:AC=1:2AB=kAC=2kwhere k= constant of proportionality∴ BC2=AC2−AB2=3k2−k2=k2

AB:AC=1:2AB=kAC=2kwhere k= constant of proportionality∴ BC2=AC2−AB2=3k2−k2=k2∴ BC=k

AB:AC=1:2AB=kAC=2kwhere k= constant of proportionality∴ BC2=AC2−AB2=3k2−k2=k2∴ BC=ktanA=ABBC=kk=1

AB:AC=1:2AB=kAC=2kwhere k= constant of proportionality∴ BC2=AC2−AB2=3k2−k2=k2∴ BC=ktanA=ABBC=kk=11−tan2A2tanA=1−12×1=02= under fined