Question

If 4sinx=3cosx then sec^2x/4[1-tan^2x] equal to

Answer 1

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

Now, solving the problem:

$\frac{ { \sec }^{2} x}{4(1 - { \tan}^{2} x)} \\ \\ = \frac{1 + { \tan}^{2} x}{4(1 - { \tan}^{2} x)} \\ \\ = \frac{1 + { (\frac{3}{4}) }^{2} }{4(1 - {( \frac{3}{4}) }^{2}) } \\ \\ = \frac{1 + \frac{9}{16} }{4(1 - \frac{9}{16} )} \\ \\ = \frac{ \frac{25}{16} }{4( \frac{7}{16}) } \\ \\ = \frac{25}{16} \times \frac{16}{4 \times 7} \\ \\ = \frac{25}{28}$