Question

3sinA=4cosA so find tansquareA_sinsquareA​

Answer 1

Step-by-step explanation:

Do you mean?

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

If so:

$3 \sin(x) = 4 \cos(x) \\ \frac{3 \sin(x) }{ \cos(x) } = 4 \\ \frac{ \sin(x) }{ \cos(x) } = \frac{4}{3} \\ \tan(x) = \frac{4}{3}$

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

(as tangent = opp/adj)

hence

Hypotenuse:

$\sqrt{{3}^{2} + {4}^{2} } = 5$

$\sin(x) = \frac{4}{5}$

Finally,

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