Question

1+sin^2A =3sinAcosA, prove tanA =1 or 1/2 ?

Answer 1

1+Sin²A= 3SinA Cos A.

Cos²A+Sin²A+Sin²A = 3SinA CosA [ 1 = Sin²+Cos A]

Cos²A+2Sin²A = 3SinA CosA....(1)

DIVIDE (1) BY COS²A we get,

1+2Tan²A = 3 TanA

1+2Tan²A - 3 TanA = 0

(2TanA-1) ( TanA-1) = 0

2TanA -1 = 0. TanA -1= 0

2TanA = 1. TanA =1

TanA = 1/2. TanA = 1

Answer 2

Answer:

→ tanA = 1/2 or 1 .

Step-by-step explanation:

1 + sin²A = 3sinAcosA .

[ Dividing both side by cos²A ] .

==> (1 + sin²A)/(cos²A) = 3sinAcosA/cos²A .

==> sec²A + tan²A = 3tanA .

==> 1 + tan²A + tan²A = 3tanA .

==> 1 + tan²A + tan²A - 3tanA = 0 .

==> 2tan²A - 3tanA + 1 = 0 .

==> 2tan²A - 2tanA - tanA + 1 = 0 .

==> 2tanA( tanA - 1 ) - 1( tanA - 1 ) = 0 .

==> ( 2tanA - 1 ) ( tan A - 1 ) = 0 .

==> 2tanA - 1 = 0 or tan A - 1 = 0.

•°• tanA = 1/2 or 1 .

✔✔ Hence, it is proved ✅✅.

THANKS