Question

If 1 + Sin²Φ = 3SinΦCosΦ, then prove that tanΦ = 1 or 1/2​

Answer 1

Answer:

Step-by-step explanation:

1 + Sin²Ф = 3SinФCosФ

=>  1/Cos²Ф + Sin²Ф/ Cos²Ф = 3SinФCosФ/Cos²Ф

=>  Sec²Ф + Tan²Ф= 3 TanФ

=>  1 + 2Tan²Ф = 3TanФ

=>  2Tan²Ф - 3TanФ +1 = 0

=>   2Tan²Ф -2tanФ - tanФ + 1 =0

=>  2tanФ (tanФ- 1) -1 (TanФ - 1) = 0

=>  (2TanФ - 1)(TanФ - 1) = 0

So, either

2TanФ-1=0

=>  TanФ= 1/2

or

TanФ-1 =0

=>  TanФ = 1

Hence proved

Answer 2

We have,
1 + Sin²Φ = 3SinΦCosΦ

Let's divide both sides by Cos²Φ,

Which will give,

Sec²Φ + tan²Φ = 3tanΦ. -----(i)


Using identity,

Sec²Φ= tan²Φ +1 ---(ii)

Equating using (ii) in (i)

tan²Φ +1 + tan²Φ = 3tanΦ.

2tan²Φ - 3tanΦ +1 = 0


Solving this quadratic equation,

2 tan²Φ - 2tanΦ -tanΦ +1 =0

2tanΦ( tanΦ-1) - (tanΦ-1) =0

(2tanΦ-1)(tanΦ-1) =0

Getting the roots will give,

tanΦ = 1

and

2tanΦ-1=0
2tanΦ=1
tanΦ = 1/2




Hence proved.