Question

3sinA=4cosA so find tansquareA_sinsquareA​

Answer 1

Explanation:

Do you mean?

`\frac{ { \tan(x) }^(2) }{ \ { \sin(x) }^(2) }`

If so:

`3 \sin(x) = 4 \cos(x) \\ (3 \sin(x) )/( \cos(x) ) = 4 \\ ( \sin(x) )/( \cos(x) ) = (4)/(3) \\ \tan(x) = (4)/(3)`

Did the question specify which quadrant? I'm just going to assume that it's the first

(as tangent = opp/adj)

hence

Hypotenuse:

`\sqrt{{3}^(2) + {4}^(2) } = 5`

`\sin(x) = (4)/(5)`

Finally,

`\frac{ { \tan(x) }^(2) }{ { \sin(x) }^(2) } = \frac{ { (4)/(3) }^(2) }{ { (4)/(5) }^(2) } \\ = 2 (7)/(9)`